Scalar and Vector

What is the main difference between scalars and vectors ? Solution

1. Time
2. Distance
3. Magnitude
4. Direction
5. No difference

Scalar measurements have only have a magnitude (no direction). Vectors have both magnitude and direction.

For example 45 meters West is a vector quantity.

The following list consists of scalar and vector quantities. Determine which are the four scalar quantities. Solution

1. Distance
2. Velocity
3. Time
4. Displacement
5. Mass
6. Speed
7. Acceleration

Scalars: Distance, Time, Mass, Speed...
Scalar quantities do not have direction, and consist only of a magnitude.

The following list consists of scalar and vector quantities. Determine which are the four vector quantities. Solution

1. Distance
2. Velocity
3. Time
4. Displacement
5. Position
6. Speed
7. Acceleration

Vector quantities have both direction and magnitude.

The three vectors are velocity, displacement, position, and acceleration. (Force is another vector, not shown here).

Calculating Average Speed

An Ironman® triathlon is a race consisting of three events: swim, bike, and run. If Sarah McSpeedy completed each event with the distances and times in the following table, calculate her average speed in the race, in m/s. Solution

Event	Distance (km)	Time (min)
Swimming	3.86	76
Cycling	180	385
Running	42.2	294

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m/s

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Convert kilometers to meters, and minutes to seconds: $\begin{align} Σ\ distance & = 3.86\ km + 180\ km + 42.2\ km \\ \\ & = 226.06\ km × 1000\ m/km \\ \\ & = 226,060\ m \\ \\ Σ\ time & = 76\ min + 385\ min + 294\ min \\ \\ & = 755\ min × 60\ s/min \\ \\ & = 45,300\ s \\ \\ \end{align}$ Total distance divided by total time: $\begin{align} s_{avg} & = \dfrac{Σ\ distance}{Σ\ time} \\ \\ & = \dfrac{226,060\ m}{45,300\ s} \\ \\ & = 4.99\ m/s \\ \\ \end{align}$

Calculating Average Speed

A car travels around a circular track with an average speed of 216 km/hour. If the time to travel around the track 10 times is 837.76 s, determine the radius of the track. (Circumference = 2$\pi$r) Solution

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m

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Convert speed units into m/s: 216 km/hr ÷ 3.6 = 60 m/s

The circumference is the distance: multiply by 10 to get the total distance. $\begin{align} \text{Distance} & = \text{Average Speed} × \text{Time} \\ \\ \text{Circumference} × \text{Number of Times Around} & = \text{Average Speed} × \text{Time} \\ \\ 2\,\pi\, r × 10 & = \text{Average Speed} × \text{Time} \\ \\ r & = \cfrac{\text{Average Speed} × \text{Time}}{(2\pi)(\text{Laps})} \\ \\ & = \cfrac{60m/s × 837.76s}{(2\pi)(10)} \\ \\ & = 800 \text{m} \\ \\ \end{align}$

Average and Instantaneous Speed

It is possible for the instantaneous speed to be much higher or lower than the average speed. Solution

1. True
2. False

Instantaneous speed is the speed at a precise moment in time. Since the average speed is an average that can result from many different instantaneous speeds, then the instantaneous speed can be a vary wide range of values lower or higher than the average speed calculated.

For example if you were to walk most of the distance to some place and sprint the last 10 meters, then the average speed would likely be much lower than the instant you were sprinting the fastest speed.

Calculating Average Velocity

A cat chases its tail in a circle with a radius of 0.25 m for a period of 10.0 s. If the cat ends up at the exact place as it started, the cat's average velocity equals zero. Solution

1. True
2. False

Average Velocity is a change in position divided by time, and depends on the total displacement, rather than the distance. If the change in position equals zero, then the average velocity also equals zero.

Kinematics and Mass

A bowling ball with a mass of ten times the mass of a golf ball, is dropped from rest at the same time as the golf ball, from the same height. Neglecting air resistance, the bowling ball will hit the ground first because it is heavier. Solution

1. True
2. False

Mass does not matter, the both the bowling and golf balls will hit the ground at the same time. None of the kinematics equations have mass in them. They will fall with the same time have the same final velocity on impact at the ground.

Gravity is the acceleration causing the motion of the two balls, and assuming this is on Earth, then it is the same for both balls.

Motion Graph Calculations

Determine the average velocity of the motion in the following position-time graph from 0s - 15s. Solution

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m/s

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Velocity is a vector quantity that involves displacement. $\begin{align} \overrightarrow{v} & = \dfrac{\text{Change in Position}}{\text{Change in Time}} \\ \\ & = \dfrac{\Delta \overrightarrow{d}}{\Delta t} \\ \\ & = \dfrac{\overrightarrow{d}_2 - \overrightarrow{d}_1}{t_2 - t_1} \quad\quad\quad\quad\quad\quad \text{Same as slope} \\ \\ & = \dfrac{30m - 10m}{15s - 0s} \\ \\ & = \dfrac{20m}{15s} \\ \\ & = 1.33\ m/s \\ \\ \end{align}$

Describing Motion Graphs

At what time(s) is velocity equal to zero? Solution

1. 1s - 4s
2. 6s and 10s
3. 16s only
4. 1s - 4s, and 16s

Velocity is zero where the y-axis indicates 0 m/s, which is located at 16 seconds.

Describing Motion Graphs

In which portion(s) of the graph is the magnitude of acceleration, uniform and non-zero? Solution

1. I only
2. II only
3. II and III
4. IV only
5. None of the above

• The slope on a velocity-time graph represents the acceleration.
• The flat line in phase I indicates a slope of zero, so the acceleration is zero.
• Phase II has a negative slope, which is negative acceleration.
• Phase III has a positive slope, which is positive acceleration.
• Phase IV is a curve, which is a non-uniform acceleration as it constantly changes and would have different instantaneous accelerations along the curve.

Motion Graph Calculations

Determine the average acceleration of the motion in the following velocity-time graph from 0s - 12s. Solution Video

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m/s²

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Average acceleration is just from the first point at 0s to the last point at 12s. $\begin{align} \overrightarrow{a} & = \dfrac{\text{Change in Velocity}}{\text{Change in Time}} \\ \\ & = \dfrac{\Delta \overrightarrow{v}}{\Delta t} \\ \\ & = \dfrac{\overrightarrow{v}_2 - \overrightarrow{v}_1}{t_2 - t_1} \quad\quad\quad\quad\quad\quad \text{Same as slope} \\ \\ & = \dfrac{-50\ m/s - 50\ m/s}{12\ s - 0\ s} \\ \\ & = \dfrac{-100\ m/s}{12\ s} \\ \\ & = -8.33\ m/s^2 \\ \\ \end{align}$

Motion Graph Calculations

Determine the displacement of the motion in the following velocity-time graph from 0s - 16s. Solution Video

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m

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$\begin{align} Area & = \text{Area Rectangle} \ + \ \text{Area Triangle} \\ \\ & = (\Delta Length) × (\Delta Height) \ + \ \dfrac{1}{2}(\Delta Base)×(\Delta Height) \\ \\ & = (16 - 0) × (20 - 0) + \dfrac{1}{2}(16 - 0) × (50 - 20) \\ \\ & = 16s × 20\ m/s \ + \ \dfrac{1}{2}(16\ s) × (30\ m/s) \\ \\ & = 320\ m + 240\ m \\ \\ & = 560\ m \\ \\ \end{align}$

Motion Graph Calculations

Determine the change in velocity of the motion in the following acceleration-time graph from 1s - 9s. Solution

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m/s

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Change in velocity = area under acceleration time function. $\begin{align} Area & = (\Delta Length) × (\Delta Height) \\ \\ & = (\Delta x) × (\Delta y) \\ \\ \Delta \overrightarrow{v} & = (\Delta t) × (\overrightarrow{a}) \\ \\ & = (t_2 - t_1) × (\overrightarrow{a}) \\ \\ & = (9s - 1s) × (-30\ m/s^2) \\ \\ & = (8s) × (-30\ m/s^2) \\ \\ & = -240\ m/s \\ \\ \end{align}$

Getting Used to North, East, South, West

People walking with a compass heading of [E 20˚ S], would be equivalent to: Solution

1. [W 20˚ N]
2. [S 20˚ E]
3. [S 70˚ E]
4. [S 70˚ W]

Drawing a sketch of the compass heading on a N-E-S-W system shows that the two headings overlap and are the same: [S 70˚ E] and [E 20˚ S].

Determine the resultant velocity of the velocities: 25 km/hr [N 50˚ E], 30 km/hr [S 40˚ E], and 60 km/hr [W].

Break into components... $\begin{align} & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ & = \\ \\ \end{align}$

Average Velocity

In one hour, a boat travels 6 km North, then 4 km East, and then 3 km South. The average velocity is: Solution

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km/hr [N 53˚ E]

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Relative Velocity

A person is walking 1 m/s East, on a subway car that is travelling 20 m/s West. Determine the relative velocity of the person with respect to the ground. Solution Video

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m/s [West]

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Here I choose to define West as positive, and East as negative. $\begin{align} \overrightarrow{v}_{relative} & = \overrightarrow{v}_{subway} + \overrightarrow{v}_{person} \\ \\ & = +20\ m/s\ \text{West}\ +\ (+1\ m/s\ \text{East}) \\ \\ & = +20\ m/s\ \text{West}\ +\ (-1\ m/s\ \text{West}) \\ \\ & = +19\ m/s\ \text{West} \\ \\ \end{align}$

Kinematics Equations (Non-Uniform Motion, Uniform Acceleration)

Solve the following questions without using acceleration.

If an object accelerates from 2 m/s to 22 m/s over 5.0 s, determine the distance travelled. Solution

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m

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Givens:
v₁ = 2 m/s
v₂ = 22 m/s
t = 5.0 s
Find d $\begin{align} d & = \cfrac{v_1 + v_2}{2} × t \\ \\ & = \cfrac{2\ m/s + 22\ m/s}{2} × 5.0s \\ \\ & = \cfrac{24\ m/s}{2} × 5.0\ s \\ \\ & = 12\ m/s × 5.0\ s \\ \\ & = 60\ m \\ \\ \end{align}$

A bike accelerates over 0.5 km for 1 minute 40 seconds. Determine the initial speed if the final speed is 3.6 km/hr. Solution

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m/s

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Givens:
d = 0.5 km = 500m
t = 1:40 min = 100s
v₂ = 3.6 km/hr ÷ 3.6 = 1.0 m/s
Find v₁ $\begin{align} d & = \dfrac{v_1 + v_2}{2} × t \\ \\ \dfrac{d}{t} & = \dfrac{v_1 + v_2}{2} \\ \\ \dfrac{v_1 + v_2}{2} & = \dfrac{d}{t} \\ \\ v_1 + v_2 & = \dfrac{d}{t} × 2 \\ \\ v_1 & = \left(\dfrac{d}{t} \ × \ 2\right) - v_2 \\ \\ & = \left(\dfrac{500m}{100s} \ × \ 2\right) - 1.0\ m/s \\ \\ & = \left(5\ m/s \ × \ 2\right) - 1.0\ m/s \\ \\ & = \left(10\ m/s\right) - 1.0\ m/s \\ \\ & = 9\ m/s \\ \\ \end{align}$

Kinematics Equations (Non-Uniform Motion, Uniform Acceleration)

Solve the following questions without using distance.

Determine the acceleration if something speeds up from 12 m/s to 32 m/s over 10 s. Solution

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m/s²

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Givens:
v₁ = 12 m/s
v₂ = 32 m/s
t = 10 s
Find $\overrightarrow{a}$ $\begin{align} \overrightarrow{a} & = \dfrac{\overrightarrow{v}_2 - \overrightarrow{v}_1}{t} \\ \\ & = \dfrac{32\ m/s - 12\ m/s}{10\ s} \\ \\ & = \dfrac{20\ m/s}{10\ s} \\ \\ & = 2\ m/s^2 \\ \\ \end{align}$

A racecar driver accelerates uniformly at 5 m/s² over 15s to a final speed of 100 m/s. Determine the initial speed. Solution

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m/s

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Givens:
a = 5 m/s²
t = 15 s
v₂ = 100 m/s
Find v₁ $\begin{align} \overrightarrow{a} & = \dfrac{\overrightarrow{v}_2 - \overrightarrow{v}_1}{t} \\ \\ \overrightarrow{v}_2 & = \overrightarrow{v}_1 + \overrightarrow{a}\Delta t \\ \\ \overrightarrow{v}_1 & = \overrightarrow{v}_2 - \overrightarrow{a}\Delta t \\ \\ & = 100\ m/s - (5\ m/s^2)(15s) \\ \\ & = 100\ m/s - 75\ m/s \\ \\ & = 25\ m/s \\ \\ \end{align}$

Kinematics Equations (Non-Uniform Motion, Uniform Acceleration)

Solve the following questions without using initial velocity.

An object accelerates over a distance of 839.5 m at 3 m/s² [E] in 23 s. Find the final velocity. Solution

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m/s [E]

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Givens:
$\overrightarrow{a}$ = 3 m/s² [E]
t = 23s
d = 839.5m
Find $\overrightarrow{v}_2$ $\begin{align} \Delta d & = \overrightarrow{v}_2 t + \dfrac{1}{2}\overrightarrow{a}t^2 \\ \\ \Delta d - \dfrac{1}{2}\overrightarrow{a}t^2 & = \overrightarrow{v}_2 t \\ \\ \overrightarrow{v}_2 t & = \Delta d - \dfrac{1}{2}\overrightarrow{a}t^2 \\ \\ \overrightarrow{v}_2 & = \dfrac{\Delta d}{t} - \dfrac{1}{2} \dfrac{\overrightarrow{a}t^2}{t} \\ \\ & = \dfrac{\Delta d}{t} - \dfrac{1}{2}\overrightarrow{a}t \\ \\ & = \dfrac{839.5m}{23s} - \dfrac{1}{2}(3\ m/s^2)(23s) \\ \\ & = 36.5\ m/s - 34.5\ m/s \\ \\ & = 2.0\ m/s [E] \\ \\ \end{align}$

An object travelling at an unknown initial velocity comes to rest over a displacement of 73.5 m [W]. If the acceleration is 3 m/s², determine the time this takes. Solution

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s

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Givens:
d = 73.5m [W]
a = 3 m/s²
$\overrightarrow{v}_2$ = 0
Find t $\begin{align} \Delta d & = \overrightarrow{v}_2 t + \dfrac{1}{2}\overrightarrow{a}t^2 \\ \\ \Delta d & = (0) t + \dfrac{1}{2}\overrightarrow{a}t^2 \\ \\ \Delta d & = \dfrac{1}{2}\overrightarrow{a}t^2 \\ \\ 2(\Delta d) & = \overrightarrow{a}t^2 \\ \\ t^2 & = \dfrac{2(\Delta d)}{\overrightarrow{a}} \\ \\ \sqrt{t^2} & = \sqrt{\dfrac{2(\Delta d)}{\overrightarrow{a}}} \\ \\ t & = \sqrt{\dfrac{2(73.5m)}{3\ m/s^2}} \\ \\ & = \sqrt{49s^2} \\ \\ & = 7\ s \\ \\ \end{align}$

An object accelerates at 6 m/s² over 85.0 m to a final velocity of 2 m/s. Determine the time this takes. Solution

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s

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Givens:
d = 85.0m
a = 6 m/s²
$\overrightarrow{v}_2$ = 2 m/s
Find t $\begin{align} \Delta d & = \overrightarrow{v}_2 t + \dfrac{1}{2}\overrightarrow{a}t^2 \\ \\ 85 & = (2\ m/s)\,t + \dfrac{1}{2}(6\ m/s^2)\,t^2 \\ \\ 85 & = 2t + 3t^2 \\ \\ 0 & = 3t^2 + 2t - 85 \quad\quad\quad\quad\quad\quad \text{Use Quadratic Equation} \\ \\ \\ \\ \\ \\ \\ \\ t & = {-b\pm\sqrt{b^2-4ac} \over 2a} \quad\quad\quad\quad\quad a=3, \ b=2, \ c=-85\\ \\ & = {-(2)\pm\sqrt{(2)^2-4(3)(-85)} \over 2(3)} \\ \\ & = {-(2)\pm\sqrt{4 + 1020} \over 6} \\ \\ & = {-(2)\pm\sqrt{1024} \over 6} \\ \\ & = {-(2)\pm 32 \over 6} \\ \\ & = {-(2) + 32 \over 6} \\ \\ & = 5 s \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \text{Time cannot be negative} \\ \\ \end{align}$

Kinematics Equations (Non-Uniform Motion, Uniform Acceleration)

Solve the following questions without using final velocity.

Find the initial velocity of an object that accelerates 5 m/s² over 52m in 4s. Solution

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m/s

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Givens:
$\overrightarrow{a}$ = 5 m/s²
d = 52m
t = 4s
Find $\overrightarrow{v}_1$ $\begin{align} \Delta d & = \overrightarrow{v}_1 t + \dfrac{1}{2}\overrightarrow{a} t^2 \\ \\ \overrightarrow{v}_1 t & = \Delta d - \dfrac{1}{2}\overrightarrow{a} t^2 \\ \\ \overrightarrow{v}_1 & = {\Delta d - \dfrac{1}{2}\overrightarrow{a} t^2 \over t} \\ \\ & = {52\ m - \dfrac{1}{2}(5\ m/s^2) (4\ s)^2 \over 4\ s} \\ \\ & = {52\ m - 40\ m \over 4\ s} \\ \\ & = {12\ m \over 4\ s} \\ \\ & = 3\ m/s \\ \\ \end{align}$

Determine the time for an object to accelerate 3 m/s² from rest over 37.5 m. Solution

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s

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Givens:
$\overrightarrow{a}$ = 3 m/s²
d = 37.5m
$\overrightarrow{v}_1$ = 0 m/s
Find t $\begin{align} \Delta d & = \overrightarrow{v}_1 t + \dfrac{1}{2}\overrightarrow{a} t^2 \\ \\ \Delta d & = (0) t + \dfrac{1}{2}\overrightarrow{a} t^2 \\ \\ \Delta d & = \dfrac{1}{2}\overrightarrow{a} t^2 \\ \\ 2(\Delta d) & = \overrightarrow{a} t^2 \\ \\ t^2 & = \dfrac{2(\Delta d)}{\overrightarrow{a}} \\ \\ t & = \sqrt{\dfrac{2(\Delta d)}{\overrightarrow{a}}} \\ \\ & = \sqrt{\dfrac{2(37.5\ m)}{3\ m/s^2}} \\ \\ & = \sqrt{25s^2} \\ \\ & = 5\ s \\ \\ \end{align}$

Kinematics Equations (Non-Uniform Motion, Uniform Acceleration)

Solve the following questions without using time.

Determine the displacement of an object that accelerates at 4 m/s² [S] from an initial velocity of 15 m/s [S] to a final velocity of 23 m/s [S]. Solution

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m [S]

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Givens:
$\overrightarrow{a}$ = 4 m/s² [S]
$\overrightarrow{v}_1$ = 15 m/s [S]
$\overrightarrow{v}_2$ = 23 m/s [S]
Find $\overrightarrow{d}$ $\begin{align} \overrightarrow{v}_2^2 & = \overrightarrow{v}_1^2 + 2 \overrightarrow{a} \overrightarrow{d} \\ \\ 2 \overrightarrow{a} \overrightarrow{d} & = \overrightarrow{v}_2^2 - \overrightarrow{v}_1^2 \\ \\ \overrightarrow{d} & = \dfrac{\overrightarrow{v}_2^2 - \overrightarrow{v}_1^2}{2\overrightarrow{a}} \\ \\ & = \dfrac{(23\ m/s)^2 - (15\ m/s)^2}{2(4\ m/s^2)} \\ \\ & = \dfrac{304 m^2/s^2}{8\ m/s^2} \\ \\ & = 32\ m\ [S]\\ \\ \end{align}$

An object accelerates at 3.2 m/s² over 85m. Determine the initial velocity if the final velocity is 34.2 m/s. Solution

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m/s

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Givens:
$\overrightarrow{a}$ = 3.2\ m/s²
d = 85m
$\overrightarrow{v}_2$ = 34.2\ m/s
Find $\overrightarrow{v}_1$ $\begin{align} \overrightarrow{v}_2^2 & = \overrightarrow{v}_1^2 + 2 \overrightarrow{a} \overrightarrow{d} \\ \\ \overrightarrow{v}_1^2 & = \overrightarrow{v}_2^2 - 2 \overrightarrow{a} \overrightarrow{d} \\ \\ \sqrt{\overrightarrow{v}_1^2} & = \sqrt{\overrightarrow{v}_2^2 - 2 \overrightarrow{a} \overrightarrow{d}} \\ \\ \overrightarrow{v}_1 & = \sqrt{\overrightarrow{v}_2^2 - 2 \overrightarrow{a} \overrightarrow{d}} \\ \\ & = \sqrt{(34.2)^2 - 2 (3.2)(85)} \\ \\ & = 25\ m/s \\ \\ \end{align}$

Theory: Gravity and Galileo

Two objects of same size but different mass are dropped from a tower at the same initial speed. Ignore air resistance. Solution

1. The heavier object will hit the ground first
2. The lighter object will hit the ground first
3. The objects will hit the ground at the same time because of gravity
4. The objects will hit the ground at the same time because their size and materials are the same

Ignoring air resistance, the acceleration of falling objects (any masses) is the same because the acceleration (gravity) is the same, 9.81 m/s² [Down] and the initial velocities are the same.

As you can see in the equation below, the final velocities of the different masses would be the same (because initial velocity, acceleration, and displacement are the same) so the masses would hit the ground at the same time. $v_2^2 = v_1^2 + 2ad$

Gravity and g's

An acceleration of $5(\overrightarrow{g})$ [↓] equals Solution

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m/s² [↓]

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In this case, g does not represent grams, rather it represents the unit of gravity, -9.81 m/s².

So $5(\overrightarrow{g})$ would be $5 × -9.81 m/s^2$

Projectile Motion: Combining Uniform and Non-Uniform Motion due to Gravity

A baseball is thrown horizontally off a cliff and takes 4.0 s to reach the ground. It travels a distance of 25 m from the base of the cliff.

How high is the cliff where the ball was thrown? Solution

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m

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The vertical component of the motion of the baseball has the same time as the whole projectile motion. So we will solve for height based on the purely vertical component of the motion.
Givens:
v_{1 y} = 0 m/s
t = 4 s
a = -9.81 m/s²
Find height, h $\begin{align} h & = v_1 t + \frac{1}{2}at^2 \\ \\ & = \cancel{(0)(4)}^0 + \frac{1}{2}(-9.81)(4)^2 \\ \\ & = 78.48\ m \\ \\ \end{align}$

Determine final velocity of the baseball in the horizontal x-direction. Solution

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m/s

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In the horizontal (x) component direction--motion is uniform, or constant velocity:
Givens:
d_x = 25 m
t = 4 s
Find v_x $\begin{align} v & = \frac{d}{t} \\ \\ v & = \frac{25}{4} \\ \\ & = 6.25\ m/s \\ \\ \end{align}$

Determine final velocity of the baseball in the vertical y-direction. Solution

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m/s

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In the vertical (y) component direction:
(Note that the time is the same in the x and y components) Givens:
v_{1 y} = 0 m/s
t = 4 s
a = -9.81 m/s²
Find v_{2 y} $\begin{align} v_2 & = v_1 + at \\ \\ & = 0 + (-9.81)(4) \\ \\ & = 39.24\ m/s \\ \\ \end{align}$

Determine the total velocity of the baseball at the instant before it hits the ground. Solution

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m/s

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This question uses values calculated previously (final velocity components in the 'x' and 'y' direction). You must calculate these (not shown here) if you haven't already, hence the significance of this question. Use vector addition of the x and y velocities found previously. Use pythagorean theorem: $\begin{align} a^2 & = b^2 + c^2 \\ \\ v_T^2 & = v_x^2 + v_y^2 \\ \\ v_T^2 & = 6.25^2 + 39.24^2 \\ \\ v_T^2 & = 1,578.84 \\ \\ v_T & = \sqrt{1,578.84} \\ \\ v_T & = 39.7\ m/s \\ \\ \end{align}$

Mass and Weight

The S.I. unit of mass to use in physics calculations is the: Solution

1. Pound
2. Ounce
3. Gram
4. Kilogram

Kilogram

In physics, the quantities mass and weight Solution

1. Are the same thing, and are equal magnitudes
2. Differ by a factor of gravity
3. Have the same units
4. Are both scalar quantities

Weight = Mass × Gravity

Newtons's Laws of Motion: Force, Mass, and Acceleration

Choose the correct proportionality between force, mass, and acceleration. Solution

1. Acceleration is directly proportional to force, and mass is directly proportional to acceleration.
2. Acceleration is inversely proportional to force, and mass is directly proportional to acceleration.
3. Acceleration is inversely proportional to force, and mass is inversely proportional to acceleration.
4. Acceleration is directly proportional to force, and mass is inversely proportional to acceleration.

F = m • a
Increasing acceleration (with constant mass) increases the force.
Increasing mass (with constant force) decreases the acceleration.

An object with mass 8m is accelerated with a net force 4F. If the same force is applied to a smaller mass, 2m, then the new acceleration would be: Solution

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times higher

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$\begin{align} 4F & = 8ma \\ \\ a & = \dfrac{1}{2}\dfrac{F}{m} \\ \\ \end{align}$ $\begin{align} 4F & = 2ma \\ \\ a & = 2\dfrac{F}{m} \\ \\ \end{align}$ $\begin{align} & = \dfrac{\left( 2\dfrac{F}{m} \right)}{\left( \frac{1}{2}\dfrac{F}{m} \right)} \\ \\ & = 4 \\ \\ \end{align}$

Simple Force Intro

The force required to accelerate a frictionless, 1000 kg rollercoaster from rest to 80 m/s in 10 seconds is: Solution

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N

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F = ma = ... = 8000 N

Coefficients of Friction µ_k

The coefficients of kinetic (µ_k) and static (µ_s) friction do not have units. Solution

1. True
2. False

The equation F_f = F_µ × F_N shows that the units of force cancel (Newton), leaving behind the unit-less coefficient of friction.

Forces

Determine the force required to accelerate a 0.10 kg hockey puck from 1.0 m/s to 60 m/s over 0.25 s. Solution

F =

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N

Hint Unavailable

Givens:
m = 0.10 kg
v₁ = 1.0 m/s
v₂ = 60 m/s
t = 0.25 s
Find F Calculate acceleration first: $\begin{align} \vec{v_2} & = \vec{v_1} + \vec{a}t \\ \\ \vec{a} & = \dfrac{\vec{v_2} - \vec{v_1}}{t} \\ \\ \vec{a} & = \dfrac{60\ m/s - 1\ m/s}{0.25\ s} \\ \\ \vec{a} & = 236\ m/s^2 \\ \\ \end{align}$ Then calculate net force: $\begin{align} F & = (m)(\vec{a}) \\ \\ F & = (0.10\ kg)(236\ m/s^2) \\ \\ F & = 23.6\ N \\ \\ \end{align}$

Forces

Your physics teacher stands on a scale in an elevator to demonstrate the effect of acceleration on weight. At rest, the scale reads 80 kg, going downwards the scale reads 60kg. Determine the acceleration of the elevator. Solution

a =

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m/s² down

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Givens:
m_(actual) = 80 kg
m_(observed) = 60 kg
Find a The observed mass on the scale is an indication of the reactive Normal Force (F_N). At rest the normal force equals the force due to gravity. When the elevator accelerates downward the normal force equals: $\begin{align} F_N & = F_g - F_a \\ \\ (m_{observed})(g) & = (m_{actual})(g) - (m_{actual})(\vec{a}) \\ \\ (60\ kg)(g) & = (80\ kg)(g) - (80\ kg)(\vec{a}) \\ \\ (80\ kg)(\vec{a}) & = (80\ kg)(g) - (60\ kg)(g) \\ \\ (80\ kg)(\vec{a}) & = (20\ kg)(9.81\ m/s^2) \\ \\ \vec{a} & = \dfrac{(20\ kg)(9.81\ m/s^2)}{(80\ kg)} \\ \\ \vec{a} & = 2.45\ m/s^2 \\ \\ \end{align}$ (Side note: an elevator accelerating upward has a normal force = F_g + F_a)

Forces

A rescue helicopter airlifts a 200 kg moose that was trapped in the middle of a thin sheet of ice on a lake. The only cable available in the helicopter can support a maximum force of 3000 N. Determine the maximum acceleration the helicopter can rise upwards. Solution

a =

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m/s² up

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Givens:
m = 200 kg
F_max = 3000 N
Find a_max The tension in the cable is based on the force due to gravity of the moose and the applied force of the acceleration of the helicopter. $\begin{align} F_{tension\ max} & = +F_{A\ up} - (F_{gravity\ down}) \\ \\ F_{tension\ max} & = F_{A\ up} - F_{gravity\ down} \\ \\ F_{tension\ max} & = (m)(\vec{a}) - (m)(g) \\ \\ \vec{a} & = \dfrac{F_{tension\ max} + (m)(g)}{(m)} \\ \\ \vec{a} & = \dfrac{3000\ N + (200\ kg)(-9.81\ m/s^2)}{(200\ kg)} \\ \\ \vec{a} & = 5.19\ m/s^2 \\ \\ \end{align}$

Forces With Friction

A 1.0 kg block of sandpaper is pushed at a constant speed over wood with a force of 11.0 N. Determine the coefficient of kinetic friction between the wood and the sandpaper in this situation. Solution

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$\begin{align} F_{NET} & = 0 \\ \\ F_{applied}\ -\ F_{kinetic \ friction} & = 0 \\ \\ F_{applied} & = F_{kinetic \ friction} \\ \\ 11.0\ N & = \mu_k \cdot (F_N) \\ \\ 11.0\ N & = \mu_k \cdot (m \, g) \\ \\ \mu_k & = \dfrac{11.0\ N}{(1.0\ kg)(9.81\ m/s^2)} \\ \\ & = 1.12 \\ \\ \end{align}$

Forces With Friction

A 100 kg cube pallet of bricks in the back of a 2000 kg pickup truck. Accelerating from rest, the driver notices that the load of bricks starts to slide dangerously when the truck reaches an acceleration of ±0.37 m/s².

Determine the coefficient of static friction between the cube pallet and the bed of the pickup truck. Solution

µ_s =

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Givens:
m_bricks = 100 kg
m_truck = 2000 kg
v₁ = 0 m/s
a = ±0.37 m/s²
Find F_f At the maximum point of acceleration, the force of static friction of the cube pallet of bricks is equal to the net force of the pickup truck: $\begin{align} F_{friction\ bricks} & = F_{net} \\ \\ (\mu_{s})[F_N] & = (m_{total})(\vec{a}) \\ \\ (\mu_{s})[(m_{bricks})(g)] & = (m_{total})(\vec{a}) \\ \\ \mu_{s} & = \dfrac{(m_{total})(\vec{a})}{(m_{bricks})(g)} \\ \\ \mu_{s} & = \dfrac{(2000\ kg + 100\ kg)(±0.37\ m/s^2)}{(100\ kg)(9.81\ m/s^2)} \\ \\ \mu_{s} & = 0.79 \\ \\ \end{align}$

Describe what the person could do to the brick load to be able to accelerate faster in the pickup truck. Solution

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See that increasing the mass of the bricks would increase the acceleration... $\begin{align} F_{friction\ bricks} & = F_{net} \\ \\ (\mu_{s})[F_N] & = (m_{total})(\vec{a}) \\ \\ (\mu_{s})[(m_{bricks})(g)] & = (m_{total})(\vec{a}) \\ \\ \vec{a} & = \dfrac{(\mu_{s})(m_{bricks})(g)}{(m_{total})} \\ \\ \vec{a} & = \dfrac{(\mu_{s})(m_{bricks})(g)}{m_{bricks} + m_{truck}} \\ \\ \end{align}$

Forces With Friction

You accelerate your Porsche at 14.72 m/s² on a perfectly level surface without spinning out the wheels. Assuming drag and kinetic friction is negligible, what is the coefficient of static friction between the tire treads and the road surface? Solution

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$\begin{align} F_{applied} & = F_{static \ friction} \\ \\ m \cdot a & = \mu_s \cdot (F_{normal}) \\ \\ m \cdot a & = \mu_s \cdot (m\,g) \quad\quad\quad\quad\quad\quad \text{m cancels} \\ \\ a & = \mu_s \cdot g \\ \\ \mu_s & = \dfrac{a}{g} \\ \\ \mu_s & = \dfrac{14.72\ m/s^2}{9.81\ m/s^2} \\ \\ \mu_s & = 1.50 \\ \\ \end{align}$

Static Friction of Rolling Car Tire

The force of static friction of a car tire is in the same direction as the movement of the car. Solution

1. True
2. False

True, this is what allows the car to drive forward... force of static friction and direction of car acceleration are in the same direction.

Static Friction

A conveyor belt is moving a box along a flat surface. The box slips as it gets moving and then becomes stationary on the moving conveyor belt. During this time the forces of static and kinetic friction are always in the same direction as the movement of the conveyor belt. Solution

1. True
2. False

The forces of static and kinetic friction carry the box forward... same direction.

Forces

An Amazon™ fulfillment center sorts packages by routing them onto different conveyor belt tracks. The conveyor belts move at a constant 2.0 m/s and the coefficient of kinetic friction of 0.66 exists between the conveyor belt surface and a package. When packages are lowered onto the conveyor belt from rest, how long does it take for a package to stop sliding? Solution

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s

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Givens:
v₁ = 0 m/s
v₂ = 2 m/s
µ_k = 0.66
Find time Force due to friction is in the same direction as the net force on the package - they are equal. First get acceleration with net force equation: $\begin{align} F_f & = F_{net} \\ \\ (\mu_k)(F_N) & = (m)(a) \\ \\ (\mu_k)(m)(g) & = (m)(a) \\ \\ a & = \dfrac{(\mu_k)\cancel{(m)}(g)}{\cancel{(m)}} \\ \\ a & = (0.66)(9.81\ m/s^2) \\ \\ \end{align}$ Now that you have acceleration, calculate the time with kinematics: $\begin{align} v_2 & = v_1 + at \\ \\ t & = \dfrac{v_2 - v_1}{a} \\ \\ t & = \dfrac{2\ m/s - 0\ m/s}{(0.66)(9.81\ m/s^2)} \\ \\ t & = 0.309\ s \\ \\ \end{align}$

Inclined Planes

Increasing the angle of inclination on an inclined plane will decrease the weight. Solution

1. True
2. False

Weight (mg) is constant.

For a stationary object, increasing the angle of inclination on an inclined plane will: Solution

1. Increase weight
2. Increase the coefficient of static friction
3. Decrease the normal force
4. Decrease the force of friction

Weight and the coefficient of static friction are constant. The force of friction increases.

Inclined Planes with Friction

A block rests on the side of a hill that is angled at 10˚ to the horizontal. Calculate the coefficient of static friction, µ_s. Solution Video

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$\begin{align} \text{Determine the y-component of}\ F_g: \quad\quad\quad\quad\quad F_{g(y)} & = F_g\cdot cos\theta \\ \\ \text{Set equal to the normal force}\ F_N: \quad\quad\quad\quad\quad\quad F_N & = F_g\cdot cos\theta \\ \\ \\ \\ \\ \\ \text{Determine the x-component of}\ F_g: \quad\quad\quad\quad\quad F_{g(x)}& = F_g\cdot sin\theta \\ \\ \\ \\ \\ \\ \\ \\ F_{static\ friction} & = F_{g(x)} \\ \\ \mu_s \cdot (F_N) & = F_g\cdot sin\theta \\ \\ \mu_s \cdot (F_g\cdot cos\theta) & = F_g\cdot sin\theta \quad\quad\quad\quad\quad F_g\ \text{cancels} \\ \\ \mu_s \cdot (cos\theta) & = sin\theta \\ \\ \mu_s & = \dfrac{sin\theta}{cos\theta} \\ \\ \mu_s & = \dfrac{sin10}{cos10} \\ \\ \mu_s & = 0.18 \\ \\ \end{align}$

Inclined Planes Without Friction

A world-class skier can ski down a 60˚ hill with negligible friction. Determine the skier's acceleration down the hill. Solution

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m/s²

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$\begin{align} F_{NET} & = F_{g(x)} \\ \\ & = (F_g) \cdot sin \theta \\ \\ m\,a & = (m\,g) \cdot sin \theta \quad\quad\quad\quad\quad\quad m \ cancels \\ \\ a & = g \cdot sin \theta \\ \\ & = (9.81\ m/s^2) \, sin(60) \\ \\ a & = 8.50 \, m/s^2 \\ \\ \end{align}$

A 20 kg sled is being pulled up a hill inclined at 10˚ to the horizontal. The sled is pulled by a string with a tension force of 10.0 N that makes an angle of 20˚ with the surface of the inclined plane. Determine the normal force of the sled. Solution

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N

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$\begin{align} F_{normal} & = F_{y-component\ of\ gravity} \ - \ F_{y-component\ of\ tension} \\ \\ F_N & = F_{g(y)} - F_{T(y)} \\ \\ & = F_g sin\theta_1 - F_T sin\theta_2\\ \\ & = (m\,g) sin\theta_1 - F_T sin\theta_2\\ \\ & = (20\ kg)(9.81\ m/s^2)\, sin(10) - (1.0 N)\, sin(20) \\ \\ & = 189.8\ N \\ \\ \end{align}$

Gravitation Equations

Which of the following is the formula used for determining gravitational field strength, g? Solution

1. $\dfrac{GM}{r}$
2. $\dfrac{GM}{r^2}$
3. $\dfrac{GM_1m_2}{r}$
4. $\dfrac{GM_1m_2}{r^2}$
5. $m \times g$

The units of gravitational field strength (g) are ms^-1 or Nkg^-1.

Gravitation

Using the constants below, determine the force of gravity on a 20,000kg space station orbiting 200 km above the surface of the Earth. Solution mass_Earth = 5.98 × 10²⁴ kg
radius_Earth = 6.4 × 10⁶ m
Gravitational constant, G = 6.67 × 10^-11 Nm²/kg²

F_g =

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N

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$\begin{align} F_g & = \dfrac{GM_1m_2}{r^2} \\ \\ F_g & = \dfrac{GM_{Earth}\,m_{object}}{r^2} \\ \\ F_g & = \dfrac{GM_1m_2}{(r_{Earth}\ +\ altitude)^2} \\ \\ & = \dfrac{(6.67 × 10^{-11})(5.98 × 10^{24})(20000)}{(6.4 × 10^6\ +\ 200 × 10^3)^2} \\ \\ & = 1.83 × 10^{5}\ N \\ \\ \end{align}$

Force of Gravity Theory

An interplanetary space shuttle is launched from Earth to a planet with three times the mass of the Earth, and half the radius. Compared to the force of gravity on Earth, the force of gravity on the other planet is: Solution

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times higher

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$\begin{align} F_g & = \dfrac{GM_1m_2}{r^2} \\ \\ & = \dfrac{G(M_{Planet})\,m_{object}}{(r)^2} \\ \\ & = \dfrac{G(3M_{Planet})\,m_{object}}{\left(\tfrac{1}{2}r\right)^2} \\ \\ & = \dfrac{3}{\left(\tfrac{1}{2}\right)^2} \\ \\ & = \dfrac{3}{\tfrac{1}{4}} \\ \\ & = 3 × \dfrac{4}{1} \\ \\ & = 12 \\ \\ \end{align}$

Work SI Base Units

Use the equation for work due to gravity to derive the equivalent in base units (kg, m, s). Solution W = F_g · d

1. $\dfrac{kg · s^2}{m}$
2. $\dfrac{kg · m}{s^2}$
3. $\dfrac{kg · s}{m^2}$
4. $\dfrac{kg · m^2}{s^2}$

$\begin{align} W & = F_g · d \\ \\ & = m · g · d \\ \\ \\ \\ & = kg · \tfrac{m}{s^2} · m \\ \\ & = \dfrac{kg · m^2}{s^2} \\ \\ \end{align}$

Work

Determine the mass of an object that can be lifted a height of 350 cm, with an energy of 5000 J. Solution

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kg

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350cm = 3.5m $\begin{align} W & = F \cdot d \\ \\ W & = mg \cdot d \\ \\ m & = \dfrac{W}{gd} \\ \\ m & = \dfrac{5000J}{(9.8)(3.5)} \\ \\ m & = \dfrac{5000J}{(9.8)(3.5)} \\ \\ m & = 145.8\ kg \\ \\ \end{align}$

Work on an Incline

A cart is being pushed with a net 10 N force up a frictionless incline (parallel to the face of the incline) that has a vertical distance of 40 m, and horizontal distance of 30 m, and a ramp length of 50 m. Determine the work necessary to push the cart to the top of the incline. Solution

W =

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J

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Use the displacement in the same direction as the force. Work is conservative, meaning the magnitude does not depend on the path taken. $\begin{align} W & = F · d \\ \\ & = 10 · 50 \\ \\ & = 500\ J \\ \\ \end{align}$

Work

Calculate the height a 10 kg object can be lifted on the moon (moon's gravity = 1.6 m/s²) if the object can be lifted 1,000 cm on earth with the same energy. Solution

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m

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1,000cm = 10m

moon's gravity = 1.6 m/s²

earth's gravity = 9.8 m/s²

Work = (Force) × Distance = (Mass × Gravity) × Distance. $\begin{align} Work\ on\ Moon & = Work\ on\ Earth \\ \\ m ·g_m· d_m & = m ·g_e ·d_e \\ \\ (10kg)·(1.6)·(d_m) & = (10kg)·(9.8)·(10m) \\ \\ (10kg)·(1.6)·(d_m) & = 980J \\ \\ (1.6)·(d_m) & = 980J \\ \\ d_m & = \dfrac{980J}{1.6} \\ \\ d_m & = 61.25\ m \\ \\ \end{align}$

Work

A strong person can lift and hold a 50 kg stone at waist-height, 1m above the ground. Determine the work done by the strong person to hold the stone there for 10 minutes. Solution

W =

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J

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While there is a vertical force (F_g = mg), there is no displacement parallel to this force.

The stone does not move, displacement = 0 $\begin{align} W & = F · d \\ \\ W & = \cancel{F · (0)} \\ \\ W & = 0\ J \\ \\ \end{align}$ Work is the energy applied by an applied force over a change in position. Work is zero when the displacement parallel to the force is zero.

Work Theory

Work equals zero in which of the following situations? Solution

1. Work due to gravity on a hockey puck sliding with constant velocity across a frictionless surface.
2. Work done on an object that experiences no forces.
3. Work due to the normal force of a computer resting on the flat surface of a table.

1. II only
2. III only
3. I and III
4. I, II, and III

W = F · d

1. Work equals zero when the distance direction is not parallel to the force direction. Force of gravity is vertical while the distance travelled is horizontal, or the distance in the same direction is zero, so work is zero.
2. Force equals zero, so work equals zero.
3. Distance equals zero, so work equals zero.

Mechanical Energy

An apple is dropped from a height of 10 m. The maximum speed of the apple occurs when the height equals zero.

1. True
2. False

Which of the following is correct?

1. gravitational potential energy - kinetic energy = total mechanical energy
2. kinetic energy - gravitational potential energy = total mechanical energy
3. total mechanical energy = gravitational potential energy + kinetic energy
4. total mechanical energy = gravitational potential energy × kinetic energy

A ball is thrown into the air. As the ball rises the magnitude of mechanical energy does not change. Solution

1. True
2. False

Mechanical energy is constant the whole way because it is always a sum of gravitational potential and kinetic energies.

The apple is held 10.0 m above the ground with a total mechanical energy of 10.0 J. Determine the gravitational potential energy and kinetic energy at the points in the following table. Solution

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Height	Gravitational Potential Energy	Kinetic Energy	Total Mechanical Energy
10.0 m	J	J	10 J
5.0 m	J	J	10 J
0 m	J	J	10 J

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total mechanical energy = gravitational potential energy + kinetic energy

Height	Gravitational Potential Energy	Kinetic Energy	Total Mechanical Energy
10.0 m	10 J	0 J	10 J
5.0 m	5 J	5 J	10 J
0 m	0 J	10 J	10 J

Mechanical Energy

A 200 g ball rolls 2.0 m/s off a table, which is 1.2 m above the ground.

Determine the kinetic energy at the moment the ball leaves the tabletop. Solution

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J

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200 g = 0.2 kg $\begin{align} kinetic\ energy & = \dfrac{1}{2}mv^2 \\ \\ & = \dfrac{1}{2}(0.2\ kg)(2\ m/s)^2 \\ \\ & = 0.4\ J \\ \\ \end{align}$

Determine the gravitational potential energy at the moment the ball leaves the tabletop. [Use g = 9.8 m/s²] Solution

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J

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200 g = 0.2 kg $\begin{align} gravitational\ potential\ energy & = mgh \\ \\ & = (0.2\ kg)(9.8\ m/s^2)(1.2\ m) \\ \\ & = 2.35\ J \end{align}$

Total mechanical energy = 2.75 J Solution

1. True
2. False

total mechanical energy = gravitational potential energy + kinetic energy

= 2.35 J + 0.4 J

= 2.75 J

The maximum speed of the ball is: Solution

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m/s

The maximum speed occurs at the ground, when total mechanical energy is 0% gravitational potential energy and 100% kinetic energy.

The maximum speed occurs at the ground, when total mechanical energy is 0% gravitational potential energy and 100% kinetic energy. Set total mechanical energy equal to kinetic energy: solve for v: $\begin{align} total\ energy & = kinetic\ energy \\ \\ 2.75\ J & = \dfrac{1}{2}(0.2\ kg)v^2 \\ \\ 2(2.75) & = \dfrac{1}{\cancel{2}}(0.2)v^2 \\ \\ 5.5 & = (0.2)v^2 \\ \\ v^2 & = \dfrac{5.5}{0.2} \\ \\ v & = \sqrt{27.5} \\ \\ v & = 5.2\ m/s \\ \\ \end{align}$

Work Due to Friction on an Incline Plane

A 10 kg box slides down an incline at a constant velocity from a height of 12 m. The coefficient of kinetic friction of the surface is 0.2 and the plane is inclined at 50˚ to the horizontal.

The work due to friction on an incline plane is calculated with the following equation, where h is the vertical height in meters. Solution Work_friction = Force_friction × height

1. True
2. False

Works is nonconservative, but it must use the distance, d of the face of the plane that the box slides on.

Work_friction = Force_friction × distance

Calculate the work due to friction. Solution

W =

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J

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$\begin{align} F_f & = \mu_k · F_n \\ \\ F_f & = \mu_k · mg cos50˚ \\ \\ F_f & = (0.2) · (10)(10) cos50˚ \\ \\ F_f & = 12.9\ N \\ \\ sin50˚ & = \dfrac{height}{distance} \\ \\ distance & = \dfrac{12}{sin50˚} \\ \\ d & = 15.7\ m \\ \\ \end{align}$ $\begin{align} W_f & = F_f · d \\ \\ & = (12.9\ N)·(15.7\ m) \\ \\ & = 201.8\ J \\ \\ \end{align}$

Energy Conservation

A 500 kg roller-coaster car is perched 20.0 m above ground, at the start of a steep drop.

The following three quantities are equal in magnitude.

Potential Energy at the Top

Kinetic Energy at the Bottom

Total Mechanical Energy

1. True
2. False

Assuming the track is frictionless, find the speed at the bottom of the drop. Solution

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m/s

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$\begin{align} ENERGY_{top} & = ENERGY_{bottom} \\ \\ PE & = KE \\ \\ mgh & = \dfrac{1}{2}mv^2 \\ \\ 2\cancel{m}gh & = \cancel{m}v^2 \\ \\ \sqrt{2gh} & = \sqrt{v^2} \\ \\ \sqrt{2gh} & = v \\ \\ v & = \sqrt{2gh} \\ \\ v & = \sqrt{2(9.81)(20)} \\ \\ v & = 19.8\ m/s \\ \\ \end{align}$

Total mechanical energy equals 9.8 × 10⁴ J. Solution

1. True
2. False

Speed at the bottom (v) calculated previously using conservation of energy. $\begin{align} total\ mechanical\ energy & = PE_{top} = KE_{bottom} \\ \\ & = \dfrac{1}{2}mv^2 \\ \\ & = \dfrac{1}{2}500(19.8)^2 \\ \\ & = 98,010\ J \\ \\ \end{align}$

---

Can be calculated with PE_top $\begin{align} total\ mechanical\ energy & = PE_{top} = KE_{bottom} \\ \\ & = mgh \\ \\ & = (500)(9.81)(20) \\ \\ & = 98,100\ J \\ \\ \end{align}$

Calculate how high the roller-coaster car would roll up a flat incline on the other end of the track if the track is angled at 60˚ to the horizontal and the coefficient of kinetic friction on the incline is 0.20 Solution

h =

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m

Work due to friction uses the incline distance = height/(sin60˚)

$\begin{align} F_N & = mg cos 60˚ \\ \\ F_N & = (500)(10)cos 60˚ \\ \\ F_N & = 5000cos60˚ \\ \\ \\ \\ sin60˚ & = \dfrac{height}{distance} \\ \\ sin60˚ & = \dfrac{h}{d} \\ \\ d & = \dfrac{h}{sin60˚} \\ \\ \end{align}$ $\begin{align} total\ energy\ lost & = total\ mechanical\ energy \\ \\ PE_{top} + Work_{friction} & = 98,010 \\ \\ mgh + \mu_k·F_N·d & = 98,010 \\ \\ (500)(10)h + (0.2)(5000cos60˚)\left(\dfrac{h}{sin60˚}\right) & = 98,010 \\ \\ 5000h + 577.4h & = 98,010 \\ \\ 5577.4h & = 98,010 \\ \\ h & = 17.6\ m \\ \\ \end{align}$

Energy Conservation

A roller-coaster goes around a loop in the track with a maximum height 50m above the ground, at a speed of 20 m/s. It then goes around a second loop at 30 m/s.

Is the maximum height of the second loop greater, or less than the maximum height of the first loop? Solution

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The maximum height of the second loop is less than the first because the roller-coaster is moving faster due to the conservation of total mechanical energy.

Calculate the maximum height of the second loop. Solution

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m

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The total mechanical energy (TME) of loop 2 equals the total mechanical energy (TME) of loop 1.

TME = KE + PE $\begin{align} TME_2 & = TME_1 \\ \\ KE_2 + PE_2 & = KE_1 + PE_1 \\ \\ \dfrac{1}{2}mv_2^2 + mgh_2 & = \dfrac{1}{2}mv_1^2 + mgh_1 \\ \\ \dfrac{1}{2}\cancel{m}v_2^2 + \cancel{m}gh_2 & = \dfrac{1}{2}\cancel{m}v_1^2 + \cancel{m}gh_1 \\ \\ \dfrac{1}{2}(30)^2 + (9.8)h_2 & = \dfrac{1}{2}(20)^2 + (9.8)(50) \\ \\ (9.8)h_2 & = \dfrac{1}{2}(20)^2 + (9.8)(50) - \dfrac{1}{2}(30)^2 \\ \\ 9.8h_2 & = 240 \\ \\ h_2 & = 24.5\ m \\ \\ \end{align}$

Conservation of Energy and Efficiency

Determine the law of conservation of energy.

1. Heat is the transfer of energy
2. Energy cannot be created or destroyed, only converted from one form into another
3. Heat moves across a temperature gradient from high to low
4. None of the above

A student pedals a bicycle on the ground with a total of 400 J of energy. If the student and the bicycle combined weigh 80 kg, and the speed is 3 m/s, determine the percent efficiency of this situation. Solution

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%

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$\begin{align} KE & = \dfrac{1}{2}mv^2 \\ \\ & = \dfrac{1}{2}(80)(3)^2 \\ \\ & = 360J \\ \\ \\ \\ efficiency & = \dfrac{energy\ used}{total\ initial\ energy} × 100\% \\ \\ & = \dfrac{360J}{400J} × 100\% \\ \\ & = 0.9 × 100\% \\ \\ & = 90\% \\ \\ \end{align}$

Power

A television uses 100 W of power over 60 minutes. Calculate the energy consumed. Solution

E =

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J

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E = (Power)(time) = (100 W)(3600 s) = 360,000 J

Power

An elevator weighs 8000 kg when fully loaded and must travel a height of 80 m from the ground to the top floor of a building. The company needs the elevator to make it to the top in 20 seconds.

Determine the minimum power of the motor that must be used, assuming 100% efficiency. Solution

P =

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W

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$\begin{align} gravitational\ potential\ energy & = mgh \\ \\ & = (8000)(9.8)(80) \\ \\ & = 6.27 × 10^6\ J \\ \\ \\ \\ power & = \dfrac{energy}{time} \\ \\ & = \dfrac{6.27 × 10^6\ J}{20\ s} \\ \\ & = 3.14 × 10^5\ W \\ \\ \end{align}$

If the efficiency of the motor was 75%, then the power of the motor used should be 25% higher than the theoretical power calculated in the previous section.

1. True
2. False

Determine the minimum power of the motor that must be used, assuming 75% efficiency. Solution

P =

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W

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$\begin{align} efficiency & = \dfrac{power\ used}{total\ power} × 100\% \\ \\ 75\% & = \dfrac{3.14 × 10^5\ W}{total\ power} × 100\% \\ \\ total\ power & = \dfrac{3.14 × 10^5\ W}{75\cancel{\%}} × 100\cancel{\%} \\ \\ & = 4.18 × 10^5\ W \\ \\ \end{align}$ (In reality the chosen motor should be considerably more powerful than this).

Energy & Power

Determine the longest run time of the electric-assisted bicycle setups below. Solution

1. Battery = 500 Wh, Motor = 300 W
2. Battery = 450 Wh, Motor = 250 W
3. Battery = 600 Wh, Motor = 320 W
4. Battery = 400 Wh, Motor = 260 W

Energy = Wh, (Watt·hours)
Power = W, (Watt) $\begin{align} E & = P \times t \\ \\ t & = \dfrac{E}{P} \\ \\ & = \dfrac{600\ \cancel{W}h}{320\ \cancel{W}} \\ \\ & = 1.875\ h \\ \\ \end{align}$

Two AA batteries in series, provides 3.0 Volts. If each battery cell is rated at 2.8 Ah (and the total is 2.8 Ah), determine the total energy in kJ. [1 Wh = 3.6 kJ] Solution

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kJ

Hint Unavailable

First determine the watt-hours 'Wh' $\begin{align} & = (3.0\ V)(2.8 Ah) \\ \\ & = 8.4\ Wh \end{align}$ Enegy, $\begin{align} & = 8.4\ \cancel{Wh} \times \dfrac{3.6\ kJ}{1\ \cancel{Wh}} \\ \\ & = 30.24\ kJ \end{align}$

Power and Velocity

The following derivation is correct. $\begin{align} power & = \dfrac{work}{time} \\ \\ & = \dfrac{force × distance}{time} \\ \\ & = force \ × \ \dfrac{distance}{time} \\ \\ & = force \ × \ velocity \\ \\ \end{align}$

1. True
2. False

A frictionless hockey puck travels at 10 m/s when a force of 50 N is applied to it. Determine the power applied. Solution

P =

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W

Hint Unavailable

$\begin{align} power & = \dfrac{work}{time} \\ \\ & = \dfrac{force × distance}{time} \\ \\ & = force × \dfrac{distance}{time} \\ \\ & = force × velocity \\ \\ & = 50\ N × 10\ m/s \\ \\ & = 500\ W \\ \\ \end{align}$

Heat Energy

Heat is the measure of energy in something. Solution

1. True
2. False

Heat is the transfer of energy from a higher to lower temperature.

Temperature is the measure of the average kinetic energy of molecules in a substance.

Heat Transfer

State the four methods of heat transfer from a flame on a stovetop to the inside of an egg boiling in a pot of water. Solution

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1) Convection and radiation from the flame through the air

2) Conduction through the metal

3) Convection through the water

4) Convection through the egg if it's liquid, convection if it's solid.

Heat Exchange

An increase in heat is indicated as: Solution

1. + q
2. - q
3. + T
4. + c

+ q, in relation to the energy input into the system.

Zeroth (0^th) Law of Thermodynamics

The zeroth law of thermodynamics states that if two objects are in thermal equilibrium with a third object, then the two objects are also in thermal equilibrium with eachother. Solution

1. True
2. False

Thermal equilibrium occurs when the temperatures are the same (average molecular kinetic energies are the same) and there is no net transfer of heat energy.

Thermal Energy

Substances with the same temperature have the same thermal energy. Solution

1. True
2. False

Thermal energy depends on the size of the substance. For example, it is easy for a "cold" swimming pool to contain more thermal energy than a hot cup of tea. Besides size, the other factor that matters is the type of substance.

Kelvin and Celsius

A 100 K increase in temperature is equal to Solution

1. 100˚C decrease in temperature
2. 100K decrease in temperature
3. 100˚C increase in temperature
4. 273K increase in temperature
5. 273˚C increase in temperature

While the change is the same, ∆K = ∆˚C The magnitudes are not the same, K ≠ ˚C

Specific Heat Theory

If the specific heat of aluminum is 2.01 J/(g·˚C), how much heat is required to raise the temperature of a 2 g block by 1˚C? Solution

Hint Clear Info

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J

Hint Unavailable

4.02 J

Specific Heat: q = m c ∆T

100 J of heat is added to a 25 g unknown block at 5.45˚C causing the block to heat to 10.00˚C. Determine the identity of the unknown sample.

1. Specific Heat Capacity of Gold = 0.129 J/(g·˚C)
2. Specific Heat Capacity of Sand = 0.290 J/(g·˚C)
3. Specific Heat Capacity of Glass = 0.840 J/(g·˚C)
4. Specific Heat Capacity of Concrete = 0.880 J/(g·˚C)
5. Specific Heat Capacity of Magnesium = 1.020 J/(g·˚C)

Specific Heat: q = m c ∆T

You want to make ice tea in the summertime by brewing a pot of 1.0L (1.0kg) of tea at 372K, and cooling it to the drinking temperature of 275K. You have an unlimited supply of 50g ice cubes at 263K. Determine the number of ice cubes to bring the ice tea to the perfect temperature. (c_water = 4.18 J·g^-1·˚C^-1, c_ice = 2.1 J·g^-1·˚C^-1, H_{f (water)} = 373 J/g) Solution

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cubes

Hint Unavailable

Since the tea and ice cubes reach the same temperature, the heat lost in the tea equals the heat gained in the ice cube.

Let the total mass of ice cubes be 50n (where 'n' is the number of ice cubes, and each cube is 50g) $\begin{align} q_{tea} & = -(q_{ice\ cube} + q_{cube\ phase\ change} + q_{cube\ water}) \\ \\ m·c·\Delta T & = -[m·c·\Delta T + m·\Delta H_f + m·c·\Delta T] \\ \\ 1000(4.18)(275-372) & = -[50n(2.1)(0 - -10) + 50n(373) + 50n(4.18)(2-0)] \\ \\ -405460 & = -50n[21 + 373 + 8.36] \\ \\ n & = \dfrac{-405460}{-50(402.36)} \\ \\ n & = 20.15 \\ \\ \end{align}$ Therefore (rounding up), a minimum of 21 ice cubes are needed.

Specific Heat, 2 Samples, Find mass: q = m c ∆T

The laboratory scale is broken and a chemist needs to figure out the mass of a known sample of pure silver. The chemist has a heat source, calorimeter, and thermometer to use so they decide to find the mass using calorimetry. Starting at room temperature, 25˚C, the sample place in the calorimeter, immersed in 0.45kg of water (4.18 J·g^-1·˚C^-1). It takes 142,875 J of heat to bring the sample to a boil at 100˚C. Given the specific heat capacity of silver is 0.240 J·g^-1·˚C^-1, determine the mass of the silver. Solution

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g

Hint Unavailable

Make sure to plug in 450 g for water, and not 0.45kg $\begin{align} q & = m_s c_s \Delta T + m_w c_w \Delta T \\ \\ q & = m_s c_s (T_f - T_i) + m_w c_w (T_f - T_i) \\ \\ 142875 & = (m_s)(0.240)(100 - 25) + (450)(4.18)(100 - 25) \\ \\ 142875 & = 18m_s + 141075 \\ \\ 142875 - 141075 & = 18m_s \\ \\ m_s & = \dfrac{(142875 - 141075)}{18} \\ \\ m_s & = 100\ g \\ \\ \end{align}$

Specific Heat, 2 Samples, Find T_f: q = m c ∆T

Equal masses of ethanol (2.46 J·g^-1·˚C^-1) and water (4.18 J·g^-1·˚C^-1) are mixed in a 700g solution at room temperature, 298K. Find the final temperature of the mixture when 124,031.88 J of heat is applied. Solution

T_f =

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K

Hint Unavailable

$\begin{align} q & = m_w c_w \Delta T + m_e c_e \Delta T \\ \\ q & = m_w c_w (T_f - T_i) + m_e c_e (T_f - T_i) \\ \\ 124,031.88 & = (350)(4.18)(T_f - 298) + (350)(2.46)(T_f - 298) \\ \\ \dfrac{124,031.88}{350} & = (4.18)(T_f - 298) + (2.46)(T_f - 298) \\ \\ \dfrac{124,031.88}{350} & = 4.18T_f - 1245.64 + 2.46T_f - 733.08 \\ \\ \dfrac{124,031.88}{350} + 1245.64 & + 733.08 = 6.64T_f \\ \\ 6.64T_f & = 2333.1 \\ \\ T_f & = 351.37\ K \\ \\ \end{align}$

Specific Heat, 2 Samples, Find T_f Mass Ratio: q = m c ∆T

A 1,050 ˚C glowing-hot block of gold (0.129 J·g^-1·˚C^-1) is placed in a calorimeter filled with nothing but air (1.02 J·g^-1·˚C^-1) at room temperature, 298K. If the mass of gold is 5 times the mass of air, find the final temperature in the calorimeter assuming no heat loss. Solution

T_f =

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˚C

Hint Unavailable

IMPORTANT: gain in heat = +q, loss of heat = -q $\begin{align} + q_{air} & = - q_{gold} \\ \\ + m_a c_a \Delta T & = -m_g c_g \Delta T \\ \\ m_a c_a (T_f - T_i) & = -m_g c_g (T_f - T_i) \\ \\ (1m)(1.02)(T_f - 25) & = -(5m)(0.129)(T_f - 1050) \\ \\ (1)(1.02)(T_f - 25) & = -(5)(0.129)(T_f - 1050) \\ \\ 1.02T_f - 25.5 & = -0.645T_f + 677.25 \\ \\ 1.02T_f + 0.645T_f & = + 677.25 + 25.5 \\ \\ 1.665T_f & = 702.75 \\ \\ T_f & = 422.07 ^{\circ}C \\ \\ \end{align}$

Types of Waves

Particles in longitudinal waves oscillate parallel to the direction of travel, while particles in transverse waves oscillate perpendicular to the direction of travel of the wave. Solution

1. True
2. False

For a longitudinal wave, imagine a sound wave traveling horizontally. The air particles also oscillate horizontally.

For a transverse wave, imagine a boat ('particle') stalled in the ocean with very large waves. The waves travel horizontally across the surface of the water, while the boat travels vertically.

Wave Terminology

Given the diagram below,

Which of the following is not one wavelength. Solution

1. A ↔ E
2. B ↔ F
3. D ↔ H
4. C ↔ G
5. None of the above

All the answer choices measure out one wavelength.

Name the points in the table. Solution

Hint Clear Info

Point	Name
B or F
D or H
vertical height of A to B, or C to D
A

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Hint Unavailable

Point	Name
B or F	Crest
D or H	Trough
vertical height of A to B, or C to D	Amplitude
A	Node

SI Units of Waves

What is the SI unit for frequency? Solution

1. m
2. s
3. Hz
4. A
5. Any of the above

Frequency is inversely proportional to period (s), while the SI unit hertz (Hz) is used in place of s^-1.

Period of a Pendulum

The equation for period of a pendulum (T), where l is length in meters, and g is gravity, in m/s².

The period of an oscillating pendulum depends on the mass attached to the end and the height the pendulum is pulled-back and released from. Solution

1. True
2. False

Notice that mass and height are both not included in the equation for period of a pendulum, therefore mass and height do not have an effect the period of a pendulum.

The only factors that affect the period of a pendulum are length and gravity.

If a simple pendulum has a period of 2.0 s, determine the length of the pendulum. Solution

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m

Hint Unavailable

g = 9.81 m/s²

Rearrange, isolate, and solve for length, l. $\begin{align} T & = 2\pi \sqrt{\dfrac{l}{g}} \\ \\ \dfrac{T}{2\pi } & = \sqrt{\dfrac{l}{g}} \\ \\ \left(\dfrac{T}{2\pi}\right)^2 & = \left(\sqrt{\dfrac{l}{g}}\right)^2 \\ \\ \left(\dfrac{T}{2\pi}\right)^2 & = \dfrac{l}{g} \\ \\ \left(\dfrac{T}{2\pi}\right)^2 × \ g & = l \\ \\ l & = g \cdot \left(\dfrac{T}{2\pi}\right)^2 \\ \\ & = (9.81) \left(\dfrac{2.0}{2\pi}\right)^2 \\ \\ & = 1\ m \\ \\ \end{align}$

Principle of Superposition

Which of the following would result in destructive interference between two waves?

1. When the troughs of two waves travelling in opposition directions meet.
2. When the crest of a wave overlaps the crest of another wave travelling in the same direction.
3. When the crest of a wave overlaps the trough of a wave traveling in the same, or opposite directions.
4. None of the above.

For total destructive interference to occur between two waves, what must be the same?

1. Amplitude
2. Frequency
3. Wavelength
4. ½λ Path Difference

1. II and IV
2. I, III, and IV
3. I and III
4. I, II, III, and IV

Standing Waves

A standing wave in a string oscillates with a speed of 1 m/s.

Standing waves are the result of an interference pattern.

1. True
2. False

Determine the distance between two adjacent nodes in a standing wave, in terms of wavelength, λ (lambda).

1. ⅛ λ
2. ¼ λ
3. ½ λ
4. λ

If two nodes are spaced 0.4 m apart, determine the frequency of the wave. Solution

f =

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Hz

Hint Unavailable

v = 1 m/s $\begin{align} λ & = node\ distance × 2 \\ \\ & = 0.4\ m × 2 \\ \\ & = 0.8\ m \\ \\ \end{align}$ $\begin{align} v & = ƒ \, λ \\ \\ ƒ & = \dfrac{v}{λ} \\ \\ & = \dfrac{1}{0.8} \\ \\ & = 1.25\ Hz \\ \\ \end{align}$

Beat Frequency

A 1,000 Hz tuning fork harmonizes with a second tuning fork, creating a 300 Hz beat frequency. Calculate the pitch (frequency) of the second tuning fork. Solution

$f_{beat} = \Big|f_1 - f_2\Big|$

1. 1,300 Hz only
2. 1,300 Hz or 1,600 Hz
3. 700 Hz only
4. 700 Hz or 1,300 Hz

$\begin{align} f_{beat} & = \Big|f_1 - f_2\Big| \\ \\ 300\ Hz & = \Big|1,000\ Hz - f_2\Big| \\ \\ 300\ Hz & = \pm \Big(1,000\ Hz - f_2\Big) \\ \\ \end{align}$ Solve the absolute value equation like this...
$\begin{align} 300\ Hz & = + \Big(1,000\ Hz - f_2\Big) \\ \\ 300 & = 1,000 - f_2 \\ \\ f_2 & = 700\ Hz \\ \\ \end{align}$ $\begin{align} 300\ Hz & = - \Big(1,000\ Hz - f_2\Big) \\ \\ 300 & = -1,000 + f_2 \\ \\ f_2 & = 1,300\ Hz \\ \\ \end{align}$ See that there are two possible answers...

Frequency and Length

The frequency in a standing wave on a string is 800 Hz.

Determine the new frequency if the length of the string is doubled. Solution

f' =

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Hz

Hint Unavailable

frequency, f is inversely proportional to length, L.

Doubling the length will decrease the frequency by a factor of ½... = 400 Hz

If this new frequency is the 4^th harmonic, determine the fundamental frequency, f₁. Solution

f₁ =

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Hz

Hint Unavailable

The n^th harmonic has frequency f_n = v/λ_n = (n)v/2L = (n)f₁
1^st, Fundamental frequency f₁ = v/λ1 = (1)v/2L = f₁
2^nd harmonic has frequency f₂ = v/λ₂ = (2)v/2L = 2f₁
3^rd harmonic has frequency f₃ = v/λ₃ = (3)v/2L = 3f₁
4^th harmonic has frequency f₄ = v/λ₄ = (4)v/2L = 4f₁
$\begin{align} f_4 & = 4 \cdot f_1 \\ \\ (400\ Hz) & = 4 \cdot f_1 \\ \\ f_1 & = \dfrac{400\ Hz}{4} \\ \\ f_1 & = 100\ Hz \\ \\ \end{align}$

Frequency

Standing waves in strings, and waves in open-end pipes share the same formula for frequency, where n is an integer {1, 2, 3, ...} Solution $f = \dfrac{n·v}{2L}$

1. True
2. False

Closed-end pipes use a different formula, where n is an odd integer: {1, 3, 5, ...}

$f = \dfrac{n·v}{4L}$

Open and Closed Pipes

Open pipes have the following resonant lengths (where lambda, λ is one wavelength): Solution ¼λ,   ¾λ,   1¼λ

1. True
2. False

Closed pipes have these resonant lengths.

Open pipes have: ½λ,   λ,   1½λ

Speed, Wavelength, and Frequency

A tweeter speaker plays a 2,000 Hz note with a 0.1725 m wavelength at the same time that the connected subwoofer plays a 90 Hz note. Calculate the wavelength of the subwoofer sound wave. Solution

λ =

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m

Hint Unavailable

You know that the speed of a sound wave depends on the temperature of the air. If the temperature is the same then the speed is the same.

First calculate the speed of sound $\begin{align} \vec{v}_{sound} & = f \lambda \\ \\ & = (0.1725)(2,000) \\ \\ & = 345\ m/s \end{align}$ Since the speaker is in the same room with the same air temperature, then the speed is the same for the subwoofer wave... $\begin{align} \vec{v}_{sound} & = f \lambda \\ \\ 345 & = (90)(\lambda) \\ \\ \lambda & = \dfrac{345}{90} \\ \\ & = 3.8\ m \end{align}$

Speed, Wavelength, and Frequency

The wavelength of a certain frequency of light is 5.0 × 10^-7 m in air, what is its frequency? (v_light = 3 x 10⁸m/s). Solution

f =

Hint Clear Info

× 10

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Hz

Hint Unavailable

$\begin{align} v & = f\lambda \\ \\ f & = \dfrac{v}{\lambda} \\ \\ & = \dfrac{3 × 10^8}{5 × 10^{-7}} \\ \\ & = \dfrac{3}{5} × \dfrac{10^8}{10^{-7}} \\ \\ & = 0.6 × 10^{8 - (-7)} \\ \\ & = 0.6 × 10^{8 + 7} \\ \\ & = 0.6 × 10^{15} \\ \\ & = 6.0 × 10^{14} \\ \\ \end{align}$

Calculations with Waves

A pendulum travels left and right for a total of 20 m, and completes 4 cycles, determine the amplitude. Solution

Hint Clear Info

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m

Hint Unavailable

Convert the distance travelled for 4 cycles, into the distance for 1 cycle: $\begin{align} & = \dfrac{20m}{4\ cycles} \\ \\ & = 5 \ \tfrac{m}{cycle} \\ \\ \end{align}$ Realize that the amplitude is ¼ of the total distance of a cycle. $\begin{align} & = ¼ × 5 \ \tfrac{m}{cycle} \\ \\ & = \dfrac{5}{4} \\ \\ \end{align}$ Therefore the amplitude is 1.25 m

Calculations with Waves

The 1st and 5th antinodes in a wave are 18 m apart. A total of 12 peaks pass through a point over the course of 4 seconds.

Determine the wavelength of the wave. Solution

λ =

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m

Hint Unavailable

Each adjacent antinode is ½λ.

$\begin{align} number\ of\ wavelengths & = antinode\ spaces × \dfrac{1}{2}λ \\ \\ & = (5-1) × \dfrac{1}{2}λ \\ \\ & = 4 × \dfrac{1}{2}λ \\ \\ & = 2λ \\ \\ \end{align}$ $\begin{align} 2λ & = 18\ m \\ \\ λ & = \dfrac{18\ m}{2} \\ \\ & = 9\ m \\ \\ \end{align}$

Determine the speed of the wave. Solution

v =

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m/s

Hint Unavailable

λ = 9 m $\begin{align} frequency & = \dfrac{number\ cycles}{time} \\ \\ & = \dfrac{12\ cycles}{4\ s} \\ \\ & = 3\ Hz \\ \\ \end{align}$ $\begin{align} v & = ƒ \, λ \\ \\ & = 3Hz × 9m \\ \\ & = 27 m/s \\ \\ \end{align}$

Intensity and Surface Area

You are located 50m from the speakers at a concert where the intensity is 0.0008 W/m². You decide to move closer to the stage and you are 5.0 m away from the speakers. Calculate the new intensity. Solution $\begin{align} I & = \dfrac{P}{A} \\ \\ & = \dfrac{P}{4\pi r^2} \end{align}$

Hint Clear Info

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w/m²

Hint Unavailable

You know that the sound dissipates in a spherical shape, with the distance, being radius, r...

Givens:
I₁ = 0.0008 W/m²
I₂ = ?
r₁ = 50 m
r₂ = 5.0 m
You also know that the power of the speaker system is constant. $\begin{align} P_1 & = P_2 \\ \\ (I_1)(4\pi r_1^2) & = (I_2)(4\pi r_2^2) \\ \\ (I_1)(\cancel{4\pi} r_1^2) & = (I_2)(\cancel{4\pi} r_2^2) \\ \\ (0.0008)(50^2) & = (I_2)(5^2) \\ \\ & = 0.08\ W/m^2 \\ \\ \end{align}$

Sound and Intensity

The intensity (W/m²) of the background noise in a restaurant, 60dB, is increased by a factor of 1,000. Determine the magnitude of the new intensity level in dB, by inspection. Solution

i =

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dB

Hint Unavailable

The intensity changes by a factor of ×10 when the decibels change by +10.

1000 = 10³ = 10 × 10 × 10... $60\ dB \xrightarrow{×\ 10} 70\ dB \xrightarrow{×\ 10} 80\ dB \xrightarrow{×\ 10} 90\ dB$ The change is + 30 dB.
Therefore the new intensity is 60 dB + 30 dB = 90 dB

Intensity (W/m2)	decibels (dB)
10-12	0
10-11	10
10-10	20
10-9	30
10-8	40
10-7	50
10-6	60
10-5	70
10-4	80
10-3	90
10-2	100
10-1	110
100 = 1	120
101	130
...	...

(90 dB is like the noise level of a food blender)

Calculate how many times louder a 100 dB sound is than a 50 dB sound. Solution $\text{times louder} = 10 ^\left(\tfrac{dB_H \,-\, dB_L}{10}\right)$

Hint Clear Info

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times louder

Hint Unavailable

$\begin{align} \text{times louder} & = 10 ^\left(\tfrac{dB_H \,-\, dB_L}{10}\right) \\ \\ & = 10 ^\left(\tfrac{100 \,-\, 50}{10}\right) \\ \\ & = 10 ^\left(\tfrac{50}{10}\right) \\ \\ & = 10 ^5 \\ \\ & = 100,000\ \text{times louder} \\ \\ \end{align}$

Calculate the dB intensity of a sound 100 times louder than 70 dB. Solution

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dB

Hint Unavailable

Solve for the higher intensity, dB_H $\begin{align} \text{times louder} & = 10 ^\left(\tfrac{dB_H \,-\, dB_L}{10}\right) \\ \\ 100 & = 10 ^\left(\tfrac{dB_H \,-\, 70}{10}\right) \\ \\ \end{align}$ When solving for a variable in the exponent, take the log of both sides of the equation. This will drip the exponent, multiplied down in-front... $\begin{align} \log(100) & = \log\left(10 ^\left(\tfrac{dB_H \,-\, 70}{10}\right) \right) \\ \\ \log(100) & = \left(\tfrac{dB_H \,-\, 70}{10}\right) \times \log(10) \\ \\ 2 & = \left(\tfrac{dB_H \,-\, 70}{10}\right) \times 1 \\ \\ \left(\tfrac{dB_H \,-\, 70}{10}\right) & = 2 \\ \\ dB_H \,-\, 70 & = 2(10) \\ \\ dB_H & = 20 + 70 \\ \\ dB_H & = 90\ dB \\ \\ \end{align}$

Chemistry in Physics

Which statement is not correct? Solution

1. Electrons are negatively charged particles surrounding an atom
2. Neutron particles have a neutral charged and are located in the nucleus of an atom
3. Protons are positively charged particles that are located in the nucleus of an atom
4. Uncharged atoms are called ions

Charged (positive or negative) atoms are called ions.

Charging an Object

An object is always positively charged when there are a high number of positively charged particles within the object. Solution

1. True
2. False

Not necessarily. An object with a neutral charge might have a high number of positively charged particles within it, but with an equal number of negatively charged particles.

An object is positively charged when: (number of positive charges) > (number of negative charges)

The Electrostatic Series

When copper is rubbed against fur, what will happen? Solution

Attraction of Electrons	Material
High attraction	Copper
↕	Ebonite
↕	Plastic
↕	Fur
↕	Glass
Low attraction	Acetate

1. Protons will move to the fur
2. Electrons will move to the fur
3. Protons will move to the copper
4. Electrons will move to the copper

Flow of electrons only.

Electric Field Lines

Electric field lines are always directed away from a positive charge and towards a negative charge.

1. True
2. False

Coulomb's Law

Coulomb's law is used to calculate the electric force between electrically charged particles.

Coulomb's law states that the electric force is proportional to the product of the charges and inversely proportional to the square of the separation distance. $F_e = \dfrac{kq_1 q_2 }{r^2}$

1. True
2. False

Doubling the distance between two charges, while doubling the magnitude of one of the charges would result in what change to the electric force (F_e)? Solution

1. 1 F_e
2. 2 F_e
3. ¼ F_e
4. ½ F_e

$\begin{align} F_e & = \dfrac{kq_1 q_2 }{r^2} \\ \\ & = \dfrac{k(2q_1) q_2 }{(2r)^2} \\ \\ & = \dfrac{k(2q_1) q_2 }{4r^2} \\ \\ & = \dfrac{2}{4} \cdot \dfrac{kq_1 q_2 }{r^2} \\ \\ & = \dfrac{1}{2} \cdot \dfrac{kq_1 q_2 }{r^2} \\ \\ \end{align}$ The result is F_e decreases by a factor of ½.

Elementary Charge (e)

The charge on an electron is -1.6 × 10^-19 C

The charge on a proton is +1.6 × 10^-19 C. Solution

1. True
2. False

The charge on a proton is equal in magnitude to the charge on an electron, with an opposite sign.

A charged object has a net charge of +2.72 × 10^-18 C. Which of the following statements is true? Solution

1. The object has 17 electrons in total
2. The object has 17 protons in total
3. The object has 17 more protons than electrons
4. The object has 17 more electrons than protons

e = +1.6 × 10^-19 C

q = +2.72 × 10^-18 C

Find N $\begin{align} total\ charge & = elementary\ charge × number\ of\ charges \\ \\ Q & = e × N \\ \\ N & = \dfrac{Q}{e} \\ \\ N & = \dfrac{+2.72 × 10^{-18}}{+1.6 × 10^{-19}} \\ \\ N & = +17 \\ \\ \end{align}$ We cannot determine the total number of protons, we can only say that there are 17 more protons than electrons in this charged object.

For example, if there were 18 protons and 1 electron, the net charge would still be +2.72 × 10^-18 C.

An unknown element has a nuclear charge of +2.24 × 10^-18 C. Determine the identity of the element. Solution

Element	Total Number of Protons
Carbon	6
Sodium	11
Silicon	14
Calcium	20

1. Carbon
2. Sodium
3. Silicon
4. Calcium

$\begin{align} N & = \dfrac{Q}{e} \\ \\ & = \dfrac{+2.24 × 10^{-18}}{+1.6 × 10^{-19}} \\ \\ & = 14 \\ \\ \end{align}$

AC and DC

Direct current is used in household wall outlets, and alternating current is used in batteries. Solution

1. True
2. False

Other way around:

Alternating current is used in household wall outlets, and direct current is used in batteries.

Naming Electrical Quantities

State the correct name when referring to the quantity: 10.0 V

1. Voltage
2. Electric Energy
3. Electric Potential Difference
4. Power Potential

Calculations with V

A 10 V battery powers a device with a 2 mA current. Calculate the energy applied over 1 minute. Solution

W =

Hint Clear Info

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J

Hint Unavailable

2 mA = 2 × 10^-3 A

1 minute = 60 s $\begin{align} Energy = W & = Charge × Electric\ Potential\ Difference \\ \\ W & = (Q) × V \\ \\ W & = (i \cdot t) × V \quad\quad\quad\quad\quad\quad\quad\quad [Q = i \cdot t] \\ \\ W & = (2 × 10^{-3} \cdot 60) × (10) \\ \\ W & = 1.2\ J \\ \\ \end{align}$

Conventional Current

Which direction does conventional current flow in the circuit: Solution

1. Left
2. Down
3. Clockwise
4. Counterclockwise

Conventional current is the flow of positive charges. The longer side of the cell is positive and the positive charges flow from the positive terminal, to the negative terminal.

Series and Parallel

Which two statements are true? (check both)

1. A is parallel to B
2. A is in series with B
3. C is in series with B
4. C is in series with A

Kirchhoff's Laws

Determine the missing potential differences and currents using Kirchhoff's laws. Solution

Hint Clear Info

i₁ =        A
V₂ =       V
V₃ =       V
i₃ =        A

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} i_1 & = i_0 = 8\ A \\ \\ V_2 & = V_0 - V_1 = 8\ V \\ \\ V_3 & = V_0 - V_1 = 8\ V \\ \\ i_3 & = i_0 - i_2 = 4\ A \\ \\ \end{align}$

Ohm's Law

Ohm's law can be applied as:

1. At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node
2. The sum of the electrical potential differences (voltage) around any closed network is zero
3. The current through a conductor between two points is directly proportional to the potential difference across the two points
4. An induced electromotive force (emf) always gives rise to a current whose magnetic field opposes the original change in magnetic flux.

Kirchhoff's and Ohm's Laws

Use Kirchhoff's law and Ohm's law (below) to solve the following circuits.

A simple combination of series and parallel loads. Solution

Hint Clear Info

V₁ =       V
i₃ =        A
i₁ =        A
V₃ =       V
R₁ =       Ω
R₂ =       Ω
R₃ =       Ω

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} V_1 & = V_0 = \underline{10\ V} \\ \\ \\ i_3 & = i_2 = \underline{3\ A} \\ \\ \\ i_1 & = i_T - i_3 = i_T - i_2 = 8 - 3 = \underline{5\ A} \\ \\ \\ V_3 & = V_0 - V_2 = 10 - 6 = \underline{4\ V} \\ \\ \\ \\ R_1 & = \dfrac{V_1}{i_1} = \dfrac{10\ V}{5\ A} = \underline{2\ Ω} \\ \\ R_2 & = \dfrac{V_2}{i_2} = \dfrac{6\ V}{3\ A} = \underline{2\ Ω} \\ \\ R_3 & = \dfrac{V_3}{i_3} = \dfrac{4\ V}{3\ A} = \underline{1.3\ Ω} \\ \\ \end{align}$

A more involved combination of series and parallel loads. Solution

Hint Clear Info

V₂ =       V
i₁ =        A
i₂ =        A
R₂ =       Ω
V₃ =       V
V₄ =       V
R₄ =       Ω
i₃ =        A
i_T =        A

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} V_2 & = V_T - V_1 = 10 - 6 = \underline{4\ V} \\ \\ i_1 & = \dfrac{V_1}{R_1} = \dfrac{6}{3} = \underline{2\ A} \\ \\ i_2 & = i_1 = \underline{2\ A} \\ \\ R_2 & = \dfrac{V_2}{i_2} = \dfrac{4}{2} = \underline{2\ Ω} \\ \\ V_3 & = V_T = \underline{10\ V} \\ \\ V_4 & = V_T = \underline{10\ V} \\ \\ R_4 & = \dfrac{V_4}{i_4} = \dfrac{10}{2} = \underline{5\ Ω} \\ \\ i_3 & = \dfrac{V_3}{R_3} = \dfrac{10}{1} = \underline{10\ A} \\ \\ i_T & = i_2 + i_3 + i_4 = 2 + 10 + 2 = \underline{14\ A} \\ \\ \end{align}$

Power

Power is the rate of energy use per unit of time.

True or false? $Power \ = \ \dfrac{\Delta E}{t} \ = \ iV \ = \ i^2R \ = \ \dfrac{V^2}{R}$

1. True
2. False

Which situation has the lowest power? Solution

1. A flow of 10 J of energy over a period of 4 seconds.
2. 2 A current across a 2 V potential difference.
3. 1 A current across a 3 Ω resistance.
4. 4 V potential difference across an 8 Ω resistance.

$\begin{align} \dfrac{\Delta E}{t} & = \dfrac{10\ J}{4\ s} = 2.5\ W \\ \\ \\ \\ iV & = (2\ A)(2\ V) = 4\ W \\ \\ \\ \\ i^2R & = (1\ A)^2(3\ Ω) = 3\ W \\ \\ \\ \\ \dfrac{V^2}{R} & = \dfrac{(4\ V)^2}{8\ Ω} = \dfrac{16}{8} = \underline{2\ W} \end{align}$

Magnetic Fields

Determine the incorrect statement about magnetic fields. Solution

1. Magnetic fields exert forces and are created around a magnet
2. Magnetic field lines are always parallel
3. Magnetic field lines never overlap other field lines
4. The density of field lines is directly proportional to strength and is strongest at the poles

Magnetic field lines are not straight, nor are they parallel. Field lines curve in 3D space.

Simple Magnetic Induction

The north pole (N) of a permanent magnetic is placed close to an iron rod. Determine the induced magnetic poles in the iron rod at the points (1) and (2). Solution

1. (1) = N and (2) = N
2. (1) = S and (2) = S
3. (1) = S and (2) = N
4. (1) = N and (2) = S

The north pole of the permanent magnet induces a south pole at the nearest end of the iron rod... (1) = S and (2) = N

Right Hand Rule of Electromagnetism

Determine the direction of the induced magnetic field at each point in some of the diagrams below.

The right hand grip rule is used for determining the direction of a magnetic field due to conventional current in a wire. Solution

1. True
2. False

True. Check it out...

Conventional current is defined as the flow of positive charges. Solution

1. True
2. False

While it is the delocalized (free) electrons in a metal that carries a current, by convention the flow of positive charges in the opposite direction is used.

A dot is used to indicate current coming out of the page, while an 'x' us used to indicate current going into the page. Solution

1. True
2. False

If the conventional current in a current-carrying wire travels to the right, at a location above the wire the magnetic field will point: Solution

1. To the right
2. To the left
3. Into the page
4. Out of the page
5. None of the above

If the conventional current in a current-carrying wire is directed into the page, at a location above the wire the magnetic field will point: Solution

1. To the right
2. To the left
3. Into the page
4. Out of the page
5. None of the above

Electromagnets (Solenoids)

A solenoid is a helical coil made of a conductive material, typically copper wire.

A solenoid forms a magnetic field when current flows through the coils in an electrical circuit.

1. True
2. False

The magnitude of magnetic field is highest in what part of a solenoid

1. The closest point to the outside of the coil
2. The center of the coils
3. At the start of the circuit
4. At the end of the circuit

Which of the following does not increase the strength of an electromagnet? Solution

1. Placing a (more) conductive metal core through the center of the solenoid
2. Increasing the current that flows through the solenoid
3. Increasing the number of coils in the solenoid
4. Increasing the diameter of the core of the coils

Increasing the diameter of the coil (making a bigger coil) spreads the magnetic field lines apart, decreasing the density of the field lines, lowering the magnetic field strength.

The right hand grip rule can be used for determining the direction of magnetic fields in a solenoid.

1. True
2. False

Determine the direction of the magnetic field in the core of the solenoid. (sorry for the bad diagram) Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Use the right hand grip rule around one of the coils. (See that the current comes from the side of the source with the longer bar)

To the left:

Determine the direction of the magnetic field in the core of the solenoid. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Use the right hand grip rule... (See that the current comes out at the top and goes in at the bottom)

To the right:

Force on a Current-Carrying Wire

A current carrying wire is placed in a magnetic field. Determine the direction of force. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Conventional current points into the page. Magnetic field is always directed from N to S, which in this case is to the left. Using the left hand rule for conventional current, the force is directed upwards.

AC and DC Power Generation

What is the main difference between an AC and DC power generation setup? Solution

1. DC uses more loops
2. AC requires a larger magnetic field
3. DC uses a split-ring commutator circuit connection rather than slip rings and brushes in AC
4. None of the above

The slip rings used in an AC generator maintain continual contact between the rotating loop and the rest of the circuit, causing an alternation in current direction.

A DC generator setup uses a split-ring commutator to connect the loop to the rest of the circuit. This maintains the same direction of current in the circuit.

Transformer

A physics student finds an old transformer that has 20 turns of wire on the primary (input) side and 10 turns on the secondary (induced) side. What kind of transformer is this? Solution

1. Step up
2. Step down
3. Mass spectrometer
4. No change in voltage

The ratio of emf's in the circuits is equal to the ratio of the number of turns.
p is primary, s is secondary. $\begin{align} \frac{V_s}{V_p} & = \frac{N_s}{N_p} = \frac{I_p}{I_s} \\ \\ \frac{V_s}{V_p} & = \frac{N_s}{N_p} \\ \\ \frac{V_s}{V_p} & = \frac{10}{20} \\ \\ \frac{V_s}{V_p} & = \frac{1}{2} \\ \\ V_s & = \frac{1}{2} \cdot V_p \\ \\ \end{align}$ The secondary voltage is half the primary voltage. Therefore this is a step down transformer.

The primary power equals the secondary power in all step up or step down transformers. Solution

1. True
2. False

$\begin{align} P_p & = P_s \\ \\ I_p \cdot V_p & = I_s \cdot V_s \\ \\ \frac{I_p}{I_s} & = \frac{V_s}{V_p} \\ \\ & ... \\ \\ \frac{V_s}{V_p} & = \frac{N_s}{N_p} = \frac{I_p}{I_s} \\ \\ \end{align}$