Friction and inclines worksheet answers for AP Physics 1 (free PDF download)

Problem 1 Static or kinetic

1. Static (the book does not slide).
2. Kinetic (surfaces slide).
3. Static (no slipping at the contact point).
4. Kinetic once the sofa is moving.

Problem 2 Friction on a crate

1. N = mg = 20 × 9.81 = 196 N.
2. f_k = 0.30 × 196.2 = 59 N.

Problem 3 Stretching a spring

1. F = kx = 250 × 0.12 = 30 N.
2. x = 50 ÷ 250 = 0.20 m.

Problem 4 Weight components

mg = 5.0 × 9.81 = 49.05 N.
1. mg sin 40° = 49.05 × 0.643 = 31.5 N.
2. mg cos 40° = 49.05 × 0.766 = 37.6 N.

Problem 5 Harder to start than to keep going

1. Static friction can rise to a maximum μ_sN, which must be exceeded to start the trolley. Once it slides the friction falls to the smaller value μ_kN, so less force is needed.
2. μ_s is greater than μ_k for most surfaces.

Problem 6 Pushing a box

N = 12 × 9.81 = 117.7 N.
1. f_s,max = 0.45 × 117.7 = 53 N.
2. 40 N is less than 53 N, so the box stays still and the friction is 40 N.
3. f_k = 0.35 × 117.7 = 41.2 N; a = (60 − 41.2) ÷ 12 = 1.6 m/s².

Problem 7 Reading a friction graph

1. 30 N.
2. 24 N.
3. N = 60 N, so μ_s = 30 ÷ 60 = 0.50 and μ_k = 24 ÷ 60 = 0.40.

Problem 8 Sliding down a ramp

1. N = mg cos 25° = 98.1 × 0.906 = 88.9 N.
2. f = 0.15 × 88.9 = 13.3 N.
3. a = (mg sin 25° − f) ÷ m = (41.46 − 13.34) ÷ 10 = 2.8 m/s².

Problem 9 Braking distance

1. a = μ_kg = 0.80 × 9.81 = 7.85 m/s².
2. d = v² ÷ 2a = 625 ÷ 15.7 = 39.8 m.
3. a = 3.92 m/s², d = 625 ÷ 7.85 = 79.6 m.

Problem 10 Atwood machine

1. a = (6.0 − 4.0) × 9.81 ÷ 10 = 1.96 m/s².
2. T = 4.0 × (9.81 + 1.96) = 47.1 N.

Problem 11 Spring with a hanging mass

1. mg = kx, so k = 0.30 × 9.81 ÷ 0.060 = 49 N/m.
2. x = 0.50 × 9.81 ÷ 49.05 = 0.10 m (10 cm).

Problem 12 Table and hanging mass

1. a = m_hangg ÷ (m_total) = 2.0 × 9.81 ÷ 5.0 = 3.9 m/s².
2. T = 3.0 × 3.92 = 11.8 N.

Problem 13 Table with friction

1. f = 0.20 × 3.0 × 9.81 = 5.9 N.
2. a = (19.62 − 5.89) ÷ 5.0 = 2.7 m/s².
3. T = 3.0 × 2.75 + 5.89 = 14.1 N.

Problem 14 Spot the error

1. Friction on a slope is μN with N = mg cos θ, not mg; the student ignored the cosine.
2. f = 0.30 × 8.0 × 9.81 × cos 30° = 20.4 N.

Problem 15 When does it start to slide?

1. At the point of slipping mg sin θ = μ_s mg cos θ; the mg cancels, so tan θ = μ_s.
2. μ_s = tan 28° = 0.53.
3. θ = tan⁻¹(0.60) = 31°.

Problem 16 Pushing a crate uphill

1. Weight mg down; normal force N perpendicular to the slope; push F up the slope; kinetic friction down the slope (opposing the motion).
2. F = mg sin 30° + μ_k mg cos 30° = 98.1 + 42.5 = 141 N.
3. a = g (sin 30° − 0.25 cos 30°) = 9.81 × (0.500 − 0.2165) = 2.8 m/s².

Problem 17 Block on a slope and hanging mass

1. System: 4.0 × 9.81 − 5.0 × 9.81 × sin 30° = 39.24 − 24.53 = 14.72 N = 9.0a, so a = 1.6 m/s².
2. T = 4.0 × (9.81 − 1.635) = 32.7 N.
3. 2.0 × 9.81 = 19.6 N is less than 24.5 N, so the 5.0 kg block slides down the slope and lifts the hanging mass.

Problem 18 Spring pulling a block

1. kx = μ_smg, so x = 0.50 × 4.905 ÷ 80 = 0.031 m (3.1 cm).
2. x = 0.30 × 4.905 ÷ 80 = 0.018 m (1.8 cm).
3. The friction falls from μ_sN to μ_kN, so a smaller spring force is enough to balance it.

Problem 19 Friction data

1. F ÷ N = 0.39, 0.39, 0.39, 0.39 (1.9 ÷ 4.9 = 0.388, 3.8 ÷ 9.8 = 0.388, 5.8 ÷ 14.7 = 0.395, 7.7 ÷ 19.6 = 0.393).
2. Gradient = (7.7 − 1.9) ÷ (19.6 − 4.9) = 0.39; it is the kinetic coefficient μ_k.
3. With zero normal force there is no contact force pressing the surfaces together, so the friction must be zero.

Problem 20 Design a test for μ_s

Place the block on the board and raise one end slowly until the block just begins to slide. Measure the angle θ with the protractor (or measure the height h and length l of the board and use sin θ = h ÷ l). At that point mg sin θ equals the maximum static friction μ_s mg cos θ, so μ_s = tan θ. Sources of error are the reading of the angle at the instant of slipping and a changing surface (dust or wear). Reduce the first by raising the board slowly, repeating at least five times and averaging; reduce the second by cleaning the surface and using fresh sections of the board.