Force · 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.

Common Forces · question 1

Back to the question · AP Topic 2.7

Question: A 3 kg box is pushed parallel to a 20° incline, uphill at constant speed. The kinetic friction coefficient is 0.4. What push is required? Use g = 9.81 m/s².

Answer: 21.2 N uphill.

1. Use N = mg cos20° and fₖ = 0.4N. Why: The push is parallel to the ramp, so it does not alter the normal force.

2. Balance along the ramp: F = mg sin20° + 0.4mg cos20° ≈ 21.12 N. Why: Constant speed means zero acceleration; the choice rounds this to 21.2 N.

Common Forces · question 2

Back to the question · AP Topic 2.7

Question: If a man pulls on a (currently stationary) 10 kg block as shown, which of the following values for the applied force will slide the block along the ground without lifting it, if the relevant friction is μ s = 0.9 ?

Answer: 70 N and 80 N.

1. Use N = 98.1 − F sin40° and require F cos40° > 0.9N. Why: The upward pull lowers both normal force and the maximum static friction.

2. Solve F > 65.66 N, while contact requires F < 98.1/sin40° = 152.62 N. Why: 70 N and 80 N meet both tests; 50 N and 60 N do not.

Common Forces · question 3

Back to the question · AP Topic 2.5

Question: A construction worker raises the 50 kg bucket shown using a 10 kg counterweight. Treat the pulleys and ropes as ideal, with two vertical support segments at each movable pulley. What downward pull is required for steady lifting? Use g = 9.81 m/s².

Answer: 147.2 N downward.

1. Choose the bucket and both linked movable pulleys as one assembly. Its downward weight is 50g; four rope segments pull upward: two with tension Tpull and two with tension Tcounter. Why: Both movable pulleys are attached to the same load. Forces through their connecting bars are internal to the combined assembly; all four supporting rope segments are external forces.

2. For the 10 kg counterweight moving steadily, Tcounter = 10g = 10 × 9.81 = 98.1 N. Why: The counterweight has zero acceleration during steady lifting, so its upward rope tension balances its weight. An ideal rope has the same tension throughout that rope.

3. Balance the assembly vertically: 2Tpull + 2Tcounter − 50g = 0. Thus 2Tpull + 2(98.1 N) = 490.5 N. Why: The assembly also has zero acceleration. Each movable pulley is supported by two segments of its own rope; the two separate ropes need not have equal tensions.

4. Solve Tpull = (490.5 − 196.2)/2 N = 147.15 N. The worker must pull downward with 147.2 N to the offered precision. Why: The four upward support forces are 147.15 N, 147.15 N, 98.1 N, and 98.1 N, which sum to the 490.5 N weight. The worker’s pull equals the tension in the pulling rope.

Common Forces · question 4

Back to the question · AP Topic 2.5

Question: Three weights are tied together and suspended. Another 5 kg weight is added. If the rope can only support 500 N of tension, which length of rope snaps first?

Answer: Rope A.

1. Read the shown 1, 8 and 40 kg masses, plus the added 5 kg: A supports 54 kg, B 53 kg, C 45 kg and D 5 kg. Why: Each upper segment supports all loads below it.

2. At g = 9.81 m/s² these give 529.74, 519.93, 441.45 and 49.05 N, with A largest. Why: As the added load is applied, A reaches the 500 N limit first; the added full load would exceed B’s limit too if A could hold.

Common Forces · question 5

Back to the question · AP Topic 2.5

Question: A which points A, B, C, or D, are the tensions equal?

Answer: A, B and C.

1. Trace the uninterrupted light rope over the ideal pulley. Why: Its tension magnitude stays the same across those rope segments.

2. D belongs to the separate lower rope, illustrated with 49.05 N, rather than the continuous main rope at A, B and C. Why: Equal tension applies along one ideal rope, not automatically across separate ropes linked by pulleys.

Force Analysis · question 1

Back to the question · AP Topic 2.2

Question: Which free body diagrams correctly label all the forces on a block that is sliding to the left on a level floor while being pulled left?

Answer: Options 1 and 3.

1. List weight downward, normal upward, pull left, and kinetic friction right. Why: The block slides left relative to the floor.

2. Accept diagrams that use exactly those real forces and directions. Why: Friction opposes the sliding; it is not another leftward pull.

Force Analysis · question 2

Back to the question · AP Topic 2.7

Question: What is the 10 kg block’s acceleration if μ k = 0.3 ? Assume it is sliding to the left along a level plane.

Answer: 0.057 m/s² left.

1. Find N = (10)(9.81) = 98.1 N, so fₖ = 0.3(98.1) = 29.43 N right. Why: There is no vertical acceleration on the level floor.

2. Use the pictured 30 N leftward pull: a = (30 − 29.43)/10 = 0.057 m/s² left. Why: The force imbalance is small; dividing force by mass matters.

Force Analysis · question 3

Back to the question · Extra exploration

Question: A 5 kg box falls at terminal speed 20 m/s through air of density 1.275 kg/m³. Its cross-sectional area is 4 m². Find its drag coefficient. Use g = 9.81 m/s².

Answer: C_D ≈ 0.048.

1. At terminal speed, set upward drag equal to mg = 49.05 N. Why: Terminal speed means zero vertical acceleration.

2. Solve C_D = 2mg/(ρAv²) = 98.1/[(1.275)(4)(20²)] ≈ 0.0481. Why: This uses the supplied quadratic-drag model rather than an AP-required drag calculation.

Force Analysis · question 4

Back to the question · Extra exploration

Question: What has a higher terminal velocity, a 5 kg ball of diameter 1 m , or a 5 kg cube with a side length of 1 m falling face-down? Assume both have the same C D .

Answer: The ball.

1. Compare projected areas: A_ball = π(0.5)² = 0.785 m²; A_cube = 1 m². Why: The masses, fluid and drag coefficients are the same.

2. Use v_t = √(2mg/(ρC_DA)). Why: The smaller projected area needs a larger speed for drag to balance weight.

Force Analysis · question 5

Back to the question · AP Topic 2.7

Question: A 10 kg block slides left on a level floor. A pull acts at 60° above the leftward horizontal with magnitude F = mg/sin 60° ≈ 113.28 N. Find its horizontal acceleration magnitude for μₖ = 0.4 and μₖ = 0.3, at the threshold where the normal force is zero. Use g = 9.81 m/s².

Answer: 5.66 m/s² for both coefficients, leftward.

1. Calculate N = mg − F sin60° = 0. Why: The question places the block exactly at the loss-of-contact threshold.

2. Then fₖ = μₖN = 0 for either coefficient, and a = F cos60°/m = g cot60° ≈ 5.664 m/s². Why: With zero normal force, changing μ cannot change the friction force.

Spring Force · question 1

Back to the question · AP Topic 2.8

Question: This block is attached to a spring. What state is the spring in, given the direction of F → s ?

Answer: Compression.

1. Read the rightward spring-force arrow as pointing away from the left anchor and toward the block’s relaxed-length equilibrium position. Why: An ideal spring exerts a restoring force toward its relaxed length, opposite the block’s displacement from that position.

2. The spring pushes the block rightward to increase its current length, so it is compressed. With right positive, Fspring = −kx and Fspring > 0 imply x < 0 relative to relaxed-length equilibrium. Why: Compression places the block closer to the anchor than at relaxed length. Extension would place it farther right and make the spring pull left toward the anchor.

Spring Force · question 2

Back to the question · AP Topic 2.8

Question: If each blue arrow represents an external horizontal applied force on the block, which diagram shows a force that can compress the spring?

Answer: Option 2.

1. Identify where the spring is anchored and which way shortens it. Why: Compression reduces the end-to-end separation.

2. Choose the applied arrow that drives the block toward the anchor. Why: The applied force and the restoring spring force then point oppositely.

Spring Force · question 3

Back to the question · AP Topic 2.8

Question: Given the figure below, label the forces acting on the apple.

Answer: Target 1: weight; target 2: spring force.

1. Assign Earth’s force to the downward arrow. Why: Weight acts toward Earth.

2. Assign the upward arrow to the stretched spring. Why: At rest it balances the weight, but the two forces are not a third-law pair.

Spring Force · question 4

Back to the question · AP Topic 2.8

Question: An ideal spring of relaxed length 5.0 cm is held in static compression at length 0.10 cm by an applied force of 25 N. Neglect other forces along the spring. Find the spring constant.

Answer: 510 N/m.

1. Convert lengths: 5.0 cm − 0.10 cm = 4.90 cm = 0.0490 m. Why: Hooke’s law uses change from relaxed length, not final length.

2. Find k = F/(compression magnitude) = 25/0.0490 ≈ 510.2 N/m. Why: Static balance makes the applied force equal to the spring-force magnitude.

Spring Force · question 5

Back to the question · AP Topic 2.8

Question: A block with mass 200 g is attached to a spring with stiffness of 75 N / m . The spring is stretched by 7 cm . Determine the acceleration and direction of the spring block system. Ignore friction.

Answer: 20.6 m/s² uphill.

1. Convert m = 0.200 kg and x = 0.070 m; find Fₛ = kx = 5.25 N uphill. Why: The stretched spring pulls toward its uphill anchor.

2. Subtract mg sin35° ≈ 1.125 N and divide by 0.200 kg to get 20.62 m/s² uphill. Why: Only the gravity component parallel to the ramp opposes the spring.

FFR - Work · question 1

Back to the question · AP Topic 3.2

Question: A box weighs 500 N. Compare the applied work needed to raise it steadily by 5 m vertically with the work needed to push it steadily up a frictionless 30° ramp to the same height. The push is parallel to the ramp. How much less work is needed using the ramp?

Answer: 0 J less; both require 2500 J.

1. For vertical lifting, W = (500 N)(5 m) = 2500 J. Why: Steady lifting balances weight.

2. On the ramp, F = 500 sin30° = 250 N and distance = 5/sin30° = 10 m, so W = 2500 J. Why: The ramp trades smaller force for a longer distance.

FFR - Work · question 2

Back to the question · AP Topic 3.2

Question: If a 57 g tennis ball falls 10 m , with air providing a resistive force of 0.25 N , what is the net work exerted on the ball?

Answer: 3.09 J.

1. Calculate gravity’s work (0.057)(9.81)(10) = 5.5917 J. Why: Weight and downward displacement point together.

2. Subtract the air’s work magnitude (0.25)(10) = 2.50 J to get W_net = 3.0917 J. Why: Resistance points opposite the displacement.

FFR - Work · question 3

Back to the question · AP Topic 3.2

Question: What force exerts more work, a 20 N force exerted at 45 ∘ from the displacement, or a 15 N force exerted in line with the displacement? Assume both forces are exerted on the same object, so the distance traveled is equal.

Answer: The 15 N force.

1. Compare components along the same displacement: 20 cos45° = 14.14 N and 15 cos0° = 15 N. Why: Only the parallel component contributes to work.

2. Multiply each by the common distance. Why: The larger parallel component produces more work, despite the smaller total force.

FFR - Work · question 4

Back to the question · AP Topic 3.2

Question: If a man pulls on a block with a rope that is at a 60 ∘ angle to the block as it slides along the ground, how much work does the man do in sliding the block 20 m ? The man is pulling with a 100 N force.

Answer: 1000 J.

1. Resolve the pull along the horizontal path: F_parallel = 100 cos60° = 50 N. Why: The vertical component has no horizontal work contribution.

2. Find W = (50)(20) = 1000 J. Why: Work uses the displacement component along the force.

FFR - Work · question 5

Back to the question · AP Topic 3.2

Question: If a block has 4 forces acting on it, as shown, and if moves horizontally 10m, rank the forces in order of increasing work done. The forces are: F 1 = 10 N , F 2 = 15 N , F 3 = 10 N , F 4 = 5 N .

Answer: Increasing work: 4, 1, 3, 2.

1. Use W = Fₓ(10 m): W₄ = −50 J, W₁ = 0, W₃ = 100 J. Why: Force 4 opposes the path and force 1 is perpendicular to it.

2. Find W₂ = (15 cos25°)(10) ≈ 135.95 J and sort the signed values. Why: A negative work ranks below zero, and the sloping force still has a large forward component.

FFR - Work · question 1

Back to the question · AP Topic 3.3

Question: What is the work done by gravity on a 10 kg box, if it is pulled up 10 m and then dropped back to its original position?

Answer: 0 J.

1. Use gravity’s work −mgΔy over the complete trip. Why: Gravity is conservative near Earth in this model.

2. The final height equals the initial height, so Δy = 0. Why: The negative upward work and positive downward work cancel.

FFR - Work · question 2

Back to the question · AP Topic 3.2

Question: A 15 kg box slides along the 7 m, 2 m, 3 m and 8 m legs shown on a level floor. A horizontal pull keeps it moving, and μₖ = 0.24. Find the work done by friction. Use g = 9.81 m/s².

Answer: −706.3 J.

1. Find friction magnitude μₖmg = (0.24)(15)(9.81) = 35.316 N. Why: A horizontal pull leaves N = mg.

2. Use the full sliding path 7 + 2 + 3 + 8 = 20 m: W_f = −(35.316)(20) = −706.32 J. Why: Friction follows path length, not the net displacement.

FFR - Work · question 3

Back to the question · AP Topic 3.2

Question: If a 5 kg block falls 10 m onto a 20 ∘ ramp, and then slides down the ramp for 5 m , what is the work done by gravity on the box? What is the work done by friction? μ k = 0.5

Answer: Gravity: +574 J; friction: −115 J.

1. Find the total vertical drop 10 + 5 sin20° = 11.710 m, giving W_g = (5)(9.81)(11.710) ≈ 574.38 J. Why: Gravity depends on the change in height.

2. On the ramp, W_f = −(0.5)(5)(9.81)cos20°(5) ≈ −115.23 J. Why: Friction acts during the sliding leg, not the preceding free fall.

FFR - Work · question 4

Back to the question · AP Topic 3.2

Question: If a crane lifts a 100 kg beam 80 m up with a 1000 N force, and then drops it onto a 75 m high floor, how much work does the crane do on the beam?

Answer: 80,000 J.

1. Identify the interval during which the crane’s force acts: the upward 80 m lift. Why: The crane does no work after it releases the beam.

2. Calculate W_crane = (1000)(80) = 80,000 J. Why: The later fall changes gravity’s work, not the crane’s earlier work.

FFR - Work · question 5

Back to the question · AP Topic 3.3

Question: How much work does gravity do on a 60 kg diver who starts on a diving board 5 m above the water’s surface, jumps 0.5 m up and then dives to a depth of 2 m below the surface?

Answer: About 4120 J.

1. Measure the net height drop from +5 m to −2 m: 7 m. Why: The extra upward part of the dive cancels in gravity’s net work.

2. Find W_g = (60)(9.81)(7) = 4120.2 J. Why: Other forces in water do not change the work assigned specifically to gravity.

FFR - Work · question 1

Back to the question · AP Topic 3.2

Question: What is the total work done by the force in the graph below over the entire displacement?

Answer: 12.5 J.

1. Split the graph at x = 0, 3, 6, 9, 12, 13 and 15 m and use signed trapezoid areas. Why: Area under a force–position graph is work.

2. Add 4.5 + 9 + 0 − 7.5 + 0.5 + 6 = 12.5 J. Why: Areas below the axis subtract, and each N·m is a joule.

FFR - Work · question 2

Back to the question · AP Topic 3.2

Question: During what intervals is the force resisting the motion of the object it acts upon?

Answer: 7.5–12.4 m.

1. Treat the displayed increasing x as the direction of displacement. Why: A negative x component of force then resists that displacement.

2. Read the portion below the zero-force axis. Why: The two crossings bound the resisting interval, about 7.5 m to 12.4 m.

FFR - Work · question 3

Back to the question · AP Topic 3.2

Question: What is the total work done by the force during the interval 5m - 10m?

Answer: −13 J.

1. Read F(5) = −2 N, F(6) = −4 N and F(10) = −1 N. Why: These are the endpoints of the two straight graph segments in the requested interval.

2. Add ((−2 − 4)/2) × 1 + ((−4 − 1)/2) × 4 = −3 − 10 = −13 J. Why: Both trapezoids lie below zero and contribute negative work.

FFR - Work · question 4

Back to the question · AP Topic 3.2

Question: When is the force encouraging the motion of the body it acts upon?

Answer: 0–4 m and 11.33–15 m.

1. Take increasing x as the motion direction. Why: The force encourages motion wherever its x component is positive.

2. Select the graph portions above the axis. Why: The zero crossings near 4 m and 11.33 m separate positive and negative work contributions.

FFR - Work · question 5

Back to the question · AP Topic 3.2

Question: Which of the following intervals is the force doing the most work on the body?

Answer: 0–2 m.

1. Compare signed areas: 0–2 gives 4 J; 2–5 gives 2 J; 10–12 gives −0.5 J; 12–15 gives 1.5 J. Why: Work is area, not simply the highest force at one point.

2. Choose the largest positive area, 4 J. Why: Negative areas cannot exceed that positive work.

Model references

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