The questions are on the friction and inclines test. Take it under real conditions before reading the mark scheme.
Quick answer: Friction is the contact force that resists sliding. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics 1 test on friction and inclines, from kinetic friction to sled pulled at an angle, with the equivalent answers to accept.
The answers show the expected method and marks. Award follow-through marks when a correct method is applied to a wrong earlier value.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 2
Question 1 Kinetic friction (1 mark)
fk = μkN, where fk is the kinetic friction force (N), μk the coefficient of kinetic friction (no unit) and N the normal force (N).
Marking: 1 mark for the equation and symbols.
Question 2 Box on a floor (3 marks)
- N = 8.0 × 9.81 = 78.5 N; fs,max = 0.60 × 78.5 = 47 N.
- 40 N is below 47 N, so the box does not move and the friction force equals the push: 40 N.
Marking: (a) 1 for N, 1 for the product; (b) 1.
Question 3 Sliding on a slope (3 marks)
a = g (sin θ − μk cos θ) = 9.81 × (0.342 − 0.10 × 0.940) = 9.81 × 0.248 = 2.4 m/s² down the slope.
Marking: 1 for N = mg cos θ, 1 for the net force equation, 1 for the value.
Question 4 Atwood machine (4 marks)
- a = (5.0 − 3.0) × 9.81 ÷ 8.0 = 2.5 m/s².
- For the 3.0 kg mass T = 3.0 × (9.81 + 2.45) = 37 N.
Marking: (a) 1 for the system equation, 1 for the value; (b) 1 for the equation on one mass, 1 for the value.
Question 5 Spring (3 marks)
- F = kx = 120 × 0.080 = 9.6 N.
- mg = 9.6, so m = 9.6 ÷ 9.81 = 0.98 kg.
Marking: (a) 1; (b) 1 for mg = kx, 1 for the value.
Question 6 Why μ is not constant (2 marks)
The maximum static friction is larger than the kinetic friction (μs is larger than μk). A push must exceed μsN to start the motion, but once sliding only μkN has to be balanced to keep a constant speed.
Marking: 1 for the larger static value, 1 for the link to starting and sliding.
Question 7 Stopping on the road (3 marks)
a = μkg = 0.70 × 9.81 = 6.87 m/s²; d = v² ÷ 2a = 400 ÷ 13.7 = 29 m.
Marking: 1 for f = μmg, 1 for a, 1 for d.
Question 8 Sled pulled at an angle (1 mark)
The rope’s upward component reduces the normal force, N = mg − F sin θ, so f = μN is smaller.
Marking: 1 mark for the reduced normal force.
