The problems are on the friction and inclined planes worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Newton's second law states that the resultant force equals mass times acceleration, ΣF = ma. These worked answers correct the 20 problems of the Physics 11-12 worksheet on friction and inclined planes, from static and kinetic to crate on a truck: each solution shows the formula, the substitution and the final answer with its unit.
Each answer starts from the resultant-force equation and shows the substitution and the result in bold. Small differences in the last digit are acceptable when your rounding differs. Next, check that your free-body diagram matches the forces used in the method.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB A.2
Problem 1 Static and kinetic
1. Moreover, static friction, which equals the push.2. In addition, the maximum static friction is usually larger. μs for static and μk for kinetic friction.
Problem 2 Net force and acceleration
1. a = F ÷ m = 3000 ÷ 1200 = 2.5 m/s².2. F = ma = 1200 × 4.0 = 4800 N.
Problem 3 Weight and reaction
1. W = 12 × 9.81 = 118 N.2. The box is in equilibrium, so N = 118 N upwards.
Problem 4 Sliding crate
1. N = 25 × 9.81 = 245 N.2. Then, the friction is f = 0.30 × 245.25 = 73.6 N.
Problem 5 Pulled box
1. 60 − 22 = 38 N to the right.2. Then, the acceleration is a = 38 ÷ 8.0 = 4.75 m/s².
Problem 6 Weight components
1. mg sin θ = 4.0 × 9.81 × sin 25° = 39.24 × 0.4226 = 16.6 N.2. Next, the second component is mg cos θ = 39.24 × 0.9063 = 35.6 N.
Problem 7 Pushing a crate
1. f = 0.35 × 40 × 9.81 = 137 N.2. Then, the net force is 180 − 137.3 = 43 N.
3. Finally, the acceleration is a = 42.7 ÷ 40 = 1.07 m/s².
Problem 8 Moving a bookcase
1. fmax = 0.45 × 15 × 9.81 = 66 N.2. Next, the kinetic friction is f = 0.32 × 147.15 = 47 N.
3. Finally, the acceleration is a = (66.2 − 47.1) ÷ 15 = 1.3 m/s².
Problem 9 Skidding car
1. a = μg = 0.80 × 9.81 = 7.85 m/s².2. Then, the stopping distance is s = v² ÷ 2a = 625 ÷ 15.7 = 39.8 m.
3. a = 3.92 m/s², so s = 625 ÷ 7.848 = 79.6 m, twice as far.
Problem 10 Frictionless sledge
1. a = g sin θ = 9.81 × 0.2588 = 2.54 m/s².2. Then, the speed is v = √(2as) = √(2 × 2.539 × 20) = 10.1 m/s.
Problem 11 Ramp for a crate
1. mg sin 20° = 117.7 × 0.3420 = 40.3 N.2. Next, the friction is f = 0.15 × 117.7 × cos 20° = 0.15 × 110.6 = 16.6 N.
3. Finally, the acceleration is a = (40.3 − 16.6) ÷ 12 = 1.97 m/s².
Problem 12 Angle of slipping
1. At the limit, mg sin θ = μs N = μs mg cos θ, so μs = sin θ ÷ cos θ = tan θ.2. μs = tan 27° = 0.51.
Problem 13 Two blocks pulled
1. f = 0.20 × 5.0 × 9.81 = 9.81 N.2. Then, the acceleration is a = (25 − 9.81) ÷ 5.0 = 3.04 m/s².
3. For the 2.0 kg block: T − 0.20 × 2.0 × 9.81 = 2.0 × 3.038, so T = 6.08 + 3.92 = 10.0 N.
Problem 14 Lift scale reading
1. N − mg = ma, so N = 70 × (9.81 + 1.2) = 771 N.2. If the acceleration is downwards, N = 70 × (9.81 − 1.2) = 603 N.
3. At constant speed, N = 70 × 9.81 = 687 N.
Problem 15 Spot the slope error
1. She used N = mg. However, on a slope N = mg cos θ.2. f = 0.25 × 78.48 × cos 35° = 16.1 N. mg sin 35° = 45.0 N, so a = (45.0 − 16.1) ÷ 8.0 = 3.6 m/s².
Problem 16 Trolley and hanging mass
1. f = 0.10 × 2.0 × 9.81 = 1.96 N.2. 0.50 × 9.81 − 1.962 = 2.5 × a, so a = 2.943 ÷ 2.5 = 1.18 m/s².
3. T = 0.50 × (9.81 − 1.177) = 4.32 N. Then, check on the trolley: T − 1.962 = 2.0 × 1.177 gives 4.32 N.
Problem 17 Pushing up a ramp
1. F = mg(sin θ + μ cos θ) = 196.2 × (0.4226 + 0.2719) = 136 N.2. a = (180 − 136.3) ÷ 20 = 2.19 m/s² up the ramp.
Problem 18 Friction coefficient from data
1. s = ½at², so a = 3.0 ÷ 1.21 = 2.48 m/s².2. The mass cancels because ma = mg sin θ − μ mg cos θ.
3. μ = (sin 22° − a ÷ g) ÷ cos 22° = (0.3746 − 0.2527) ÷ 0.9272 = 0.13.
Problem 19 Car and caravan
1. a = (3500 − 400 − 300) ÷ 2000 = 1.40 m/s².2. For the caravan: T − 300 = 800 × 1.40, so T = 1420 N.
Problem 20 Crate on a truck
1. Also, friction provides the acceleration: a = μs g = 0.40 × 9.81 = 3.9 m/s².2. The crate slides, so a = μk g = 0.30 × 9.81 = 2.9 m/s².
3. Relative acceleration = 4.5 − 2.943 = 1.557 m/s², so s = ½ × 1.557 × 4.0 = 3.1 m.
