Energy · quiz worked solutions

Tried the quiz first? Good. Here’s how to work through each answer. Follow the link beside a question to jump back to its original spot. Questions marked Extra exploration go beyond this AP Physics 1 course; Book controls questions explain the interface.

Introduction to Energy · question 1

Back to the question · AP Topic 3.3

Question: At the bottom of a moving pendulum’s swing, take gravitational potential energy to be zero. Which statement correctly describes its kinetic energy and the ratio K/U?

Answer: K > 0, U = 0, and K/U is undefined.

1. Use the stated zero of gravitational potential energy at the bottom. Why: The potential reference is a choice, not a claim that total energy vanishes.

2. A moving bob has positive kinetic energy, but division by U = 0 is undefined. Why: Do not turn an undefined ratio into zero or a finite number.

Introduction to Energy · question 2

Back to the question · AP Topic 3.2

Question: You push a car forward 1 mm with a constant 500 N force parallel to its motion, then release it. Its brakes subsequently return it to its original position. How much work did your pushing force do on the car during the forward motion?

Answer: 0.5 J.

1. Convert 1 mm to 0.001 m. Why: The pushing displacement must use metres.

2. Find W_push = (500)(0.001) = 0.5 J. Why: Later brake work does not erase the work done by the earlier separate pushing force.

Introduction to Energy · question 3

Back to the question · AP Topic 3.4

Question: A refrigerator requires a larger pulling force to move at the same constant speed on its sliding feet than on wheels. Over the same horizontal distance, why is more mechanical work required on its sliding feet?

Answer: More applied work becomes thermal energy through sliding friction.

1. At constant speed, ΔK = 0. Why: A larger applied force does not imply a kinetic-energy increase here.

2. Over equal distance, larger pulling force does more work against resistance. Why: Sliding dissipates more mechanical energy than the wheel model specified in the question.

Introduction to Energy · question 4

Back to the question · AP Topic 3.2

Question: A 50 kg object is pulled two laps around a track at a constant speed of 1 m/s. Model the pulling force as a 15 N tangential force directed along the object’s motion, balanced by resistance. If one lap is 0.5 km, how much work does the pulling force do over two laps?

Answer: 15,000 J of positive work by the pulling force.

1. Find the total path length: two laps cover 2 × 0.5 km = 1.0 km = 1000 m. Why: The force follows the direction of travel at each point, so its work accumulates along the entire path even though the final displacement after two laps is zero.

2. The stipulated 15 N pulling force is tangential and points along each displacement. Its work is Wpull = F × path length = (15 N)(1000 m) = 15,000 J. Why: The angle between the pull and every small displacement is zero, making every contribution positive. The force’s direction changes with the track; it is not a constant vector around a closed loop.

3. At constant speed, kinetic energy is unchanged. Resistance does −15,000 J, canceling the pulling force’s work; any ideal track-guiding force perpendicular to motion does zero work. Why: The work–energy theorem relates the change in kinetic energy to net work by all forces. The question asks for the pulling force’s work alone, which can be positive when net work is zero.

Introduction to Energy · question 5

Back to the question · AP Topic 3.2

Question: A man pulls a wagon by exerting a 30 N force on the handle, at an angle of 42 ∘ . What is the work done by the man if he pulls the wagon for 15 m ?

Answer: 334 J.

1. Take the horizontal part of the 30 N pull: 30 cos42° ≈ 22.29 N. Why: The wagon’s displacement is horizontal.

2. Find W = (30)(15)cos42° ≈ 334.42 J. Why: The vertical component contributes no work to the stated horizontal motion.

Kinetic Energy · question 1

Back to the question · AP Topic 3.1

Question: A baseball and a bowling ball are thrown at the same velocity. Which one has a larger kinetic energy?

Answer: The bowling ball.

1. Use K = ½mv² and hold v fixed. Why: Both balls are thrown at the same velocity.

2. The larger mass gives the larger kinetic energy. Why: K grows linearly with mass at fixed speed.

Kinetic Energy · question 2

Back to the question · AP Topic 3.1

Question: If the velocity of an object was to change from 5 m / s to 10 m / s , what is the ratio of the initial to final kinetic energy?

Answer: K_i/K_f = 1/4.

1. Keep the mass constant and write K_i/K_f = v_i²/v_f². Why: The common ½m cancels because the mass stays unchanged.

2. Calculate 5²/10² = 25/100 = 1/4. Why: Doubling speed quadruples kinetic energy.

Kinetic Energy · question 3

Back to the question · AP Topic 3.1

Question: You are given that the kinetic energy of two objects have the same value of 20 J , and that the velocity of the two objects are 3 m / s and 6 m / s , respectively. Find the difference between the masses.

Answer: 3.33 kg.

1. Use m = 2K/v²: m₁ = 40/9 = 4.444 kg and m₂ = 40/36 = 1.111 kg. Why: The slower object needs more mass to have the same energy.

2. Subtract 4.444 − 1.111 = 3.333 kg. Why: This is a positive difference between the masses.

Kinetic Energy · question 4

Back to the question · AP Topic 3.1

Question: When a running boy increases his speed by 2 m / s , his kinetic energy becomes three times the original value. What is his original speed (in m/s)?

Answer: 1 + √3 ≈ 2.73 m/s.

1. Write (v + 2)² = 3v² for unchanged mass. Why: Tripling kinetic energy multiplies squared speed by three.

2. Use v + 2 = √3v and solve v = 2/(√3 − 1) = 1 + √3. Why: Speed is positive, so the positive square-root branch applies.

Kinetic Energy · question 5

Back to the question · AP Topic 3.2

Question: A 1 kg particle initially moves at 2 m/s along +x. A constant net force of 4 N acts along +x through a displacement of 3 m. Find v_f/v_i.

Answer: v_f/v_i = √7.

1. Find K_i = ½(1)(2²) = 2 J and W = (4)(3) = 12 J. Why: The specified force is the net force along the displacement.

2. Then K_f = 14 J and v_f = √28 m/s, so v_f/2 = √7. Why: Energy ratios are squared speed ratios; the old math choice represents the square root.

Potential Energy · question 1

Back to the question · AP Topic 3.4

Question: What velocity does a ball need to make it to the top of a 50 m high hill?

Answer: 31.3 m/s.

1. Use the minimum-speed condition v_top = 0 and negligible losses. Why: All starting kinetic energy becomes gravitational potential energy.

2. Set ½mv² = mg(50), giving v = √(2·9.81·50) ≈ 31.32 m/s. Why: Mass cancels; this is the ideal point-particle hill model.

Potential Energy · question 2

Back to the question · AP Topic 3.4

Question: If a snowboarder is approaching a vertical ramp at 15 m / s , and the ramp is 10 m high, how high above the ramp will he get?

Answer: 1.5 m above the ramp.

1. Find the maximum vertical rise v₀²/(2g) = 15²/(2·9.81) ≈ 11.468 m. Why: A vertical ramp directs the remaining velocity upward in the lossless model.

2. Subtract the ramp’s 10 m height to get 1.468 m. Why: The question asks height above the ramp, not above the starting ground.

Potential Energy · question 3

Back to the question · AP Topic 3.4

Question: If two balls are dropped from different heights, one from 10 m and the other from 15 m , how much faster will the 15 m ball be going right before it hits the ground than the 10 m high ball right before it hit the ground?

Answer: 3.2 m/s faster.

1. Find v_15 = √(2·9.81·15) ≈ 17.155 m/s and v_10 = √(2·9.81·10) ≈ 14.007 m/s. Why: Each drop begins from rest with no losses.

2. Subtract the two speeds to get 3.148 m/s, about 3.2 m/s. Why: Subtracting heights inside one square root would answer a different question.

Potential Energy · question 4

Back to the question · AP Topic 3.4

Question: If a ski jumper starts 20 m above the tip of the ramp, how fast will he leave the ramp?

Answer: 19.8 m/s.

1. Treat the jumper as starting from rest and neglect losses. Why: The initial gravitational energy supplies the speed.

2. Set mg(20) = ½mv², giving v = √(392.4) ≈ 19.809 m/s. Why: The ramp’s shape does not affect this energy result.

Potential Energy · question 5

Back to the question · AP Topic 3.4

Question: If a skateboarder is going 20 m / s , and he approaches a 30 ∘ ramp that ends 2 m above the ground, how high above the ground will he go?

Answer: 6.6 m above the ground.

1. First find launch speed at the 2 m ramp top: v_launch² = 20² − 2g(2) = 360.76 m²/s². Why: The climb uses part of the initial kinetic energy.

2. Only the vertical component rises further: Δh = v_launch² sin²30°/(2g) ≈ 4.597 m. Why: The horizontal kinetic energy remains at the projectile peak.

3. Add the 2 m launch height to get 6.597 m. Why: The peak height is measured from the ground.

Potential Energy · question 1

Back to the question · AP Topic 3.4

Question: A 1 kg ball is launched horizontally on a frictionless surface by an ideal spring with k=100 N/m initially compressed 0.5 m. Find the ball’s speed when the spring returns to its unloaded length. Ignore spin.

Answer: 5.0 m/s.

1. Find spring energy ½kx² = ½(100)(0.5²) = 12.5 J. Why: The model excludes losses, gravity changes and spin.

2. Set ½(1)v² = 12.5, giving v = 5.0 m/s. Why: At the unloaded length the spring no longer stores elastic energy.

Potential Energy · question 2

Back to the question · AP Topic 3.4

Question: A ball is released 5 m above an unloaded vertical spring with k=50 N/m. It compresses the spring 1 m before stopping instantaneously. Neglect losses. Find the ball’s mass. Use g=9.81 m/s².

Answer: 0.425 kg.

1. The ball drops 5 m before contact and another 1 m while compressing, so Δh = 6 m. Why: Gravity keeps doing work during compression.

2. Set mg(6) = ½(50)(1²), giving m = 25/(6·9.81) ≈ 0.4247 kg. Why: Ignoring the extra one metre would overestimate the mass.

Potential Energy · question 3

Back to the question · AP Topic 3.4

Question: Treat the shown 5 kg wheel as a sliding point mass and neglect rotation. It starts from rest 10 m above a horizontal spring with k=200 N/m on a frictionless path. How far does it compress the spring?

Answer: 2.2 m.

1. The specified point-mass model begins with mg(10) = (5)(9.81)(10) = 490.5 J. Why: The prompt deliberately removes rotational energy.

2. Set ½(200)x² = 490.5, giving x = √4.905 ≈ 2.215 m. Why: Maximum compression has zero instantaneous kinetic energy.

Potential Energy · question 4

Back to the question · AP Topic 3.4

Question: A 60 kg diver stands on an ideal massless board modeled as a vertical spring with k=15 kN/m. Its unloaded surface is 5 m above the water and is initially depressed 0.2 m, so the diver starts 4.8 m above the water at rest. With no extra muscular work or losses, how high above the water does the diver rise?

Answer: 5.31 m above the water.

1. Start at 4.8 m with spring energy ½(15000)(0.2²) = 300 J. Why: The board’s depressed starting position is below its unloaded 5 m level.

2. Use mg h_max = mg(4.8) + 300, giving h_max = 4.8 + 300/(60·9.81) ≈ 5.310 m. Why: There is no extra muscular work in the stated ideal model.

Potential Energy · question 5

Back to the question · AP Topic 3.4

Question: If a penny was dropped from the top of the Empire State Building, will it pierce the awning at the base of the building? Assume the awning acts as a spring with a spring constant of 1200 N / m , and it will tear if it is moved more than 0.15 m from equilibrium. The Empire State Building is 450 m high, and a penny has a mass of 2.5 g .

Answer: No, in the ideal spring model.

1. The tear threshold stores ½(1200)(0.15²) = 13.5 J. Why: That is the maximum spring energy before the given tear displacement.

2. Gravity can supply only (0.0025)(9.81)(450 + 0.15) ≈ 11.04 J by that point. Why: This is below 13.5 J even with no air drag, so the compression stops before the tear threshold; a real awning needs a different damage model.

FFR - Power · question 1

Back to the question · AP Topic 3.5

Question: A 500 kg elevator is lifted 15 m in 1 min at a constant speed. How much power does the lifting mechanism exert on the elevator?

Answer: 1.23 kW.

1. At constant speed, the lift force balances mg and does work mgh. Why: The elevator gains gravitational energy without gaining kinetic energy.

2. Find P_avg = (500)(9.81)(15)/60 = 1226.25 W = 1.23 kW. Why: One minute must be converted to 60 seconds.

FFR - Power · question 2

Back to the question · AP Topic 3.2

Question: A 4 kg block is sliding along horizontal pavement, which has a μ k = 0.8 . If it is sliding at 4 m / s at x = 0 , how far will it slide before stopping?

Answer: 1.02 m.

1. Use friction work −μₖmgd to remove initial kinetic energy ½mv². Why: Only sliding friction does horizontal net work in this model.

2. Find d = v²/(2μₖg) = 16/(2·0.8·9.81) ≈ 1.019 m. Why: The mass cancels between energy and friction.

FFR - Power · question 3

Back to the question · AP Topic 3.5

Question: A hamster running in its wheel can run at about 50 cm / s . If that hamster has to exert a force of 0.25 N to maintain this velocity, how many hamsters are needed to power a 1 MW power plant?

Answer: 8,000,000 hamsters.

1. Convert the hamster’s speed 50 cm/s to 0.50 m/s, giving P = Fv = (0.25)(0.50) = 0.125 W. Why: Mechanical power is force along the velocity times speed.

2. Divide 1×10⁶ W by 0.125 W = 8×10⁶. Why: This whimsical model assumes perfect conversion and continuous output.

FFR - Power · question 4

Back to the question · AP Topic 3.5

Question: A roller coaster can ascend this 40 m high hill, with a 55 ∘ slope in 1 min at a constant speed. Neglecting friction, how much power does the lifting chain exert on the 1.6 Mg train?

Answer: 10.5 kW.

1. Convert 1.6 Mg to 1600 kg and use vertical gain 40 m. Why: Only the height change sets the lossless steady lifting work.

2. Find P = mgh/t = (1600)(9.81)(40)/60 = 10464 W. Why: The incline angle changes chain force and path length, but not their work product.

FFR - Power · question 5

Back to the question · AP Topic 3.5

Question: The pictured water lifter has five rising buckets, each containing 5 kg of water, all moving vertically at 1 m/s. Neglect empty-bucket mass, friction, and speed changes. Find the mechanical power required.

Answer: 245.3 W.

1. Five buckets contain total rising mass 5(5) = 25 kg. Why: All five masses gain height at the same instant.

2. Use P = mgv = (25)(9.81)(1) = 245.25 W. Why: This neglects bucket mass and resistance as the question states.

Model references

These original worked steps use the stated question models. Physics references: Mechanical energy, Power. AP topic labels follow the supplied-edition College Board CED, Effective Fall 2024, ©2026, pinned in the course audit.