Practice Questions: Energy

SECTION I: MULTIPLE CHOICE

Directions

Select the best answer for each of the following questions. Each question is followed by four possible answers, labeled A through D. Choose the single best response unless otherwise directed.

Question 1

A 2.0 kg block slides across a horizontal surface with an initial speed of 8.0 m/s and comes to rest after traveling 16 m. What is the magnitude of the average frictional force acting on the block?

1. 2.0 N
2. 4.0 N
3. 8.0 N
4. 16 N

Question 2

A student performs an experiment in which a cart of mass m is released from rest at the top of a ramp of height h. The cart rolls down the ramp and reaches the bottom with speed v. If the student repeats the experiment with a cart of mass 2m released from the same height, how will the speed at the bottom compare to v?

1. The speed will be \(\frac{v}{2}\)
2. The speed will be \(\frac{v}{\sqrt{2}}\)
3. The speed will be \(v\)
4. The speed will be \(2v\)

Question 3

What is the work done by the force as the object moves from x = 0 m to x = 5 m?

1. 16 J
2. 24 J
3. 32 J
4. 40 J

Question 4

A ball is thrown vertically upward with an initial speed \(v_0\). At its maximum height, which of the following statements about the ball's energy is correct? (Assume air resistance is negligible.)

1. The kinetic energy is at its maximum value and the gravitational potential energy is zero.
2. The kinetic energy is zero and the gravitational potential energy is at its maximum value.
3. Both the kinetic energy and gravitational potential energy are at their maximum values.
4. Both the kinetic energy and gravitational potential energy are zero.

Question 5

Which of the following expressions correctly represents the speed of the block immediately after it loses contact with the spring?

1. \(v = \sqrt{\frac{k}{m}}x\)
2. \(v = \sqrt{\frac{k}{2m}}x\)
3. \(v = \frac{k}{m}x\)
4. \(v = \sqrt{\frac{2k}{m}}x\)

Question 6

A pendulum consists of a mass attached to a string of length \(L\). The mass is pulled to one side so that the string makes an angle θ with the vertical, then released from rest. At the lowest point of its swing, what is the relationship between the pendulum's kinetic energy and the initial gravitational potential energy (measured relative to the lowest point)?

1. The kinetic energy equals the initial potential energy only if θ is small.
2. The kinetic energy is always greater than the initial potential energy.
3. The kinetic energy equals the initial potential energy regardless of the angle θ.
4. The kinetic energy is always less than the initial potential energy due to energy loss.

Question 7

Data Table: Cart on Incline

A student releases carts of different masses from the same height on an inclined plane and measures their speeds at the bottom.

Which conclusion is best supported by the data?

1. The speed at the bottom is directly proportional to the mass of the cart.
2. The speed at the bottom is inversely proportional to the mass of the cart.
3. The speed at the bottom is approximately independent of the mass of the cart.
4. The kinetic energy at the bottom is the same for all carts.

Question 8

A box is pushed across a rough horizontal floor at constant velocity by a horizontal force \(F\). Which of the following statements about the energy transformations is correct?

1. The work done by force \(F\) increases the kinetic energy of the box.
2. The work done by force \(F\) is converted entirely into gravitational potential energy.
3. The work done by force \(F\) is converted into thermal energy due to friction.
4. No net work is done on the box because its velocity is constant.

Question 9

Two identical balls are released from the same height. Ball A is dropped straight down, while Ball B is thrown horizontally. Assuming air resistance is negligible, how do the speeds of the two balls compare just before they hit the ground?

1. Ball A has a greater speed because it travels a shorter distance.
2. Ball B has a greater speed because it has horizontal velocity.
3. Ball A has a greater speed because all its energy goes into vertical motion.
4. The balls have the same speed because they fall through the same vertical height.

Question 10

What is the total work done on the object during the time interval from t = 0 s to t = 5 s?

1. 12 J
2. 24 J
3. 36 J
4. 48 J

Question 11

A car traveling at speed \(v\) on a level road applies its brakes and skids to a stop over a distance \(d\). If the car had been traveling at speed \(2v\) under identical conditions, over what distance would it skid before stopping?

1. \(d\)
2. \(2d\)
3. \(4d\)
4. \(8d\)

Question 12

What is the speed of Cart 2 immediately after the collision?

1. \(\frac{v}{2}\)
2. \(v\)
3. \(\frac{3v}{2}\)
4. \(2v\)

Question 13

A roller coaster car of mass \(m\) starts from rest at height \(h\) above the ground. If the car has speed \(v\) at ground level, which of the following best explains why \(v\) is less than \(\sqrt{2gh}\)?

1. Some mechanical energy is transformed into thermal energy due to friction.
2. The mass of the car is too large for energy conservation to apply.
3. The gravitational potential energy was not completely converted to kinetic energy because the track is curved.
4. Energy is lost because the car's acceleration is not constant.

Question 14

A spring-loaded toy gun is used to launch a projectile of mass \(m\) vertically upward. The spring has spring constant \(k\) and is compressed a distance \(x\) from equilibrium before launch. Assuming no energy losses, what maximum height above the launch point will the projectile reach?

1. \(h = \frac{kx}{2mg}\)
2. \(h = \frac{kx^2}{mg}\)
3. \(h = \frac{kx^2}{2mg}\)
4. \(h = \frac{2kx^2}{mg}\)

Question 15

Which statement correctly describes the energy transformation between Position 2 and Position 3?

1. Kinetic energy is converted to gravitational potential energy.
2. Kinetic energy is converted to elastic potential energy.
3. Kinetic energy is converted to thermal energy.
4. Gravitational potential energy is converted to kinetic energy.

Question 16

A student lifts a book of mass \(m\) from the floor to a shelf at height \(h\) at constant velocity. How much work does the student do on the book?

1. Zero, because the velocity is constant
2. \(\frac{1}{2}mgh\)
3. \(mgh\)
4. \(2mgh\)

Question 17

Data: Spring Compression Experiment

A student compresses a spring by various distances and calculates the elastic potential energy stored. Based on the data, what is the spring constant of the spring?

1. 100 N/m
2. 200 N/m
3. 300 N/m
4. 400 N/m

Question 18

Object A has twice the mass of Object B. Both objects have the same kinetic energy. How does the speed of Object A compare to the speed of Object B?

1. \(v_A = \frac{v_B}{2}\)
2. \(v_A = \frac{v_B}{\sqrt{2}}\)
3. \(v_A = v_B\)
4. \(v_A = \sqrt{2}v_B\)

Question 19

Which statement correctly compares the block's speed at Point B for the two paths?

1. The speed at B is greater for Path 1 because the distance is shorter.
2. The speed at B is greater for Path 2 because the block gains more kinetic energy in the dip.
3. The speed at B is the same for both paths because mechanical energy is conserved.
4. The speed at B is zero for both paths because A and B are at the same height.

Question 20

A satellite orbits Earth in a circular orbit at constant speed. Which statement about the work done on the satellite is correct?

1. Positive work is done because the satellite is moving.
2. Negative work is done because gravity acts opposite to the motion.
3. Zero work is done because the gravitational force is perpendicular to the velocity.
4. Zero work is done because the satellite's speed is constant.

SECTION II: FREE RESPONSE

Directions

Answer both of the following questions. Show all your work for each part clearly. You may use the back of the page if needed. The suggested time for answering both questions is 50 minutes.

FRQ 1: Mathematical Routines (25 minutes)

1. Derive an expression for the speed \(v\) of the block at the bottom of the incline, just before it reaches the rough surface. Begin your derivation by writing a fundamental physics principle or equation. Express your answer in terms of \(h\), \(g\), and physical constants.
2. Derive an expression for the magnitude of the frictional force \(f\) acting on the block as it slides across the rough horizontal surface. Express your answer in terms of \(m\), μ_k, \(g\), and physical constants.
3. Derive an expression for the distance \(d\) the block travels on the rough horizontal surface before coming to rest. Begin with a fundamental physics principle or equation. Express your answer in terms of \(h\), μ_k, \(g\), and physical constants.
4. A student claims that if the mass of the block were doubled, the distance \(d\) would also double. Justify whether this claim is correct or incorrect using your expression from part (c) and physical reasoning.

FRQ 2: Experimental Design and Analysis (25 minutes)

1. Describe an experimental procedure the student could use to collect data relating the compression distance \(x\) to the maximum speed \(v\) of the cart. Include enough detail that another student could replicate the experiment. Identify what measurements should be recorded.
2. The student plans to graph the data to determine the spring constant \(k\).
   1. Identify what quantities should be graphed on the vertical and horizontal axes such that the slope of a best-fit line can be used to determine \(k\).
   2. Explain the reasoning for your choice, including the theoretical relationship between these quantities.
   3. Describe how the spring constant \(k\) can be calculated from the slope of the best-fit line.
3. The student collects the following data:
   
   1. Complete the third column of the data table by calculating v² for each trial.
   2. On the grid below (or describe the graph), plot the appropriate quantities and draw a best-fit line through the data points.
4. Using the graph from part (c), calculate the spring constant \(k\). The mass of the cart is 0.50 kg. Show all work, including the calculation of the slope and the use of the relationship you identified in part (b).

ANSWER KEY

Part A - Multiple Choice Answer Table

Part B - Free Response Detailed Answers

FRQ 1 - Answer Key

Part A: Derive speed at bottom of incline

Model Answer:

Begin with the principle of conservation of mechanical energy.

\[ E_{\text{initial}} = E_{\text{final}} \]

At the top of the incline (taking the bottom as the reference level, where gravitational potential energy = 0):
\[ U_i + K_i = U_f + K_f \]
\[ mgh + 0 = 0 + \frac{1}{2}mv^2 \]

The mass cancels:
\[ gh = \frac{1}{2}v^2 \]

Solve for \(v\):
\[ v^2 = 2gh \]
\[ v = \sqrt{2gh} \]

Answer: \( v = \sqrt{2gh} \)

Part B: Derive frictional force magnitude

Model Answer:

The frictional force on the horizontal surface is given by:
\[ f = \mu_k N \]

On a horizontal surface, the normal force equals the weight:
\[ N = mg \]

Therefore:
\[ f = \mu_k mg \]

Answer: \( f = \mu_k mg \)

Part C: Derive distance traveled on rough surface

Model Answer:

Begin with the work-energy theorem. The work done by friction equals the change in kinetic energy:
\[ W_{\text{friction}} = \Delta K \]

The block starts with kinetic energy \( K_i = \frac{1}{2}mv^2 \) and ends at rest, so \( K_f = 0 \).
\[ W_{\text{friction}} = 0 - \frac{1}{2}mv^2 = -\frac{1}{2}mv^2 \]

The work done by friction (a force opposing motion) is:
\[ W_{\text{friction}} = -fd = -\mu_k mg d \]

Setting the two expressions equal:
\[ -\mu_k mg d = -\frac{1}{2}mv^2 \]

From Part A, \( v = \sqrt{2gh} \), so \( v^2 = 2gh \). Substitute:
\[ -\mu_k mg d = -\frac{1}{2}m(2gh) \]
\[ -\mu_k mg d = -mgh \]

The mass cancels:
\[ \mu_k g d = gh \]

Solve for \(d\):
\[ d = \frac{h}{\mu_k} \]

Answer: \( d = \frac{h}{\mu_k} \)

Part D: Justify whether doubling mass doubles distance

Model Answer:

The student's claim is incorrect. From the expression derived in part (c), \( d = \frac{h}{\mu_k} \), the distance \(d\) depends only on the height \(h\) and the coefficient of kinetic friction μ_k. The mass \(m\) does not appear in the final expression.

Physical reasoning: Although doubling the mass doubles both the initial gravitational potential energy (\(mgh\)) and the frictional force (\(\mu_k mg\)), the ratio of energy to force remains constant. Therefore, the distance traveled is independent of mass. Doubling the mass would not change the distance \(d\).

FRQ 2 - Answer Key

Part A: Describe experimental procedure

Model Answer:

1. Measure and record the mass \(m\) of the dynamics cart using a balance.
2. Attach the spring to a fixed wall and attach the cart to the other end of the spring.
3. Use a meterstick to measure the equilibrium position of the cart (when the spring is neither compressed nor stretched). Mark this position.
4. Compress the spring by a measured distance \(x\) (e.g., 0.10 m) from equilibrium and hold the cart in place.
5. Position the motion detector on the track, aimed at the cart, to measure the cart's velocity.
6. Release the cart from rest and use the motion detector and computer to record the velocity as a function of time.
7. From the data, identify and record the maximum speed \(v\) reached by the cart (which occurs as it passes through equilibrium).
8. Repeat steps 4-7 for at least four additional compression distances (e.g., 0.15 m, 0.20 m, 0.25 m, 0.30 m).
9. Record all values of compression distance \(x\) and corresponding maximum speed \(v\) in a data table.

Part B: Graph planning and relationship

Part B(i): Identify quantities to graph

Plot \( v^2 \) (in m²/s²) on the vertical axis and \( x^2 \) (in m²) on the horizontal axis.

Part B(ii): Explain reasoning

By conservation of energy, the elastic potential energy stored in the compressed spring equals the kinetic energy of the cart at maximum speed:
\[ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \]

Simplifying:
\[ kx^2 = mv^2 \]

Rearranging:
\[ v^2 = \frac{k}{m}x^2 \]

This is a linear relationship of the form \( v^2 = (\text{slope}) \cdot x^2 \), where the slope equals \( \frac{k}{m} \).

Part B(iii): Calculate spring constant from slope

Once the slope is determined from the best-fit line, the spring constant is calculated as:
\[ k = (\text{slope}) \times m \]

Part C: Complete data table and graph

Part C(i): Complete the table

Part C(ii): Graph description

The student should plot the points: (0.01, 0.25), (0.0225, 0.56), (0.04, 1.00), (0.0625, 1.56), (0.09, 2.25), where the horizontal axis represents \( x^2 \) (in m²) and the vertical axis represents \( v^2 \) (in m²/s²). A best-fit straight line should be drawn through these points, passing close to the origin.

Part D: Calculate spring constant

Model Answer:

First, calculate the slope of the best-fit line. Using two points from the data:

Point 1: \( (x^2, v^2) = (0.01, 0.25) \)
Point 2: \( (x^2, v^2) = (0.09, 2.25) \)

\[ \text{slope} = \frac{\Delta v^2}{\Delta x^2} = \frac{2.25 - 0.25}{0.09 - 0.01} = \frac{2.00}{0.08} = 25 \text{ s}^{-2} \]

From Part B, we know:
\[ \text{slope} = \frac{k}{m} \]

Solving for \(k\):
\[ k = (\text{slope}) \times m = 25 \text{ s}^{-2} \times 0.50 \text{ kg} = 12.5 \text{ N/m} \]

Answer: The spring constant is approximately 12.5 N/m or 13 N/m (depending on rounding and best-fit line placement).