The questions are on the work energy theorem test. Take it under real conditions before reading the mark scheme.
Quick answer: Work done by a variable force is the integral W = ∫F dx, the area under the force–position graph. This mark scheme gives the model answers and the marks for the 8 questions of the AP Physics C test on work energy theorem, from conservative force to engine power, with the equivalent answers to accept.
Model answers and marking guidance. Method marks are given for correct setting-up of integrals and derivatives even when the arithmetic is wrong. Do not penalise a unit slip twice, and apply error carried forward where later parts use an earlier incorrect number.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 3
Question 1 Conservative force (2 marks)
a. A force whose work between two points is independent of the path taken. b. F = −dU/dx.Marking: 1 mark each.
Question 2 Variable force (3 marks)
a. W = ∫₀² (2.0x + 3.0x²) dx = x² + x³ = 4.0 + 8.0 = 12 J. b. v = √(2 × 12/1.5) = √16 = 4.0 m/s.Marking: 1 mark for the integral set-up, 1 for 12 J, 1 for the speed.
Question 3 Energy diagram (4 marks)
a. The minimum is at x = 3.0 m. b. 2.0(x − 3.0)² = 8.0 gives x − 3.0 = ±2.0, so x = 1.0 m and 5.0 m. c. Kmax = 8.0 J, so v = √(2 × 8.0/0.50) = 5.7 m/s. d. F = −dU/dx = −4.0(x − 3.0) = −4.0 N at x = 4.0 m (towards the minimum).Marking: 1 mark per part.
Question 4 Equilibrium (3 marks)
a. dU/dx = 6.0x − 6.0x² = 0, so x = 0 and x = 1.0 m. b. d²U/dx² = 6.0 − 12x: at x = 0 it is +6.0, so stable; at x = 1.0 m it is −6.0, so unstable.Marking: 1 mark for positions, 1 mark for each classification with the second derivative.
Question 5 Spring launcher (2 marks)
a. ½kx² = ½ × 300 × (0.12)² = 2.2 J. b. v = √(2 × 2.16/0.20) = 4.6 m/s.Marking: 1 mark each; accept 2.16 J carried forward.
Question 6 Leaving the Moon (2 marks)
a. W = GMm/R = (6.67 × 10⁻¹¹ × 7.35 × 10²²)/(1.74 × 10⁶) = 2.8 × 10⁶ J. b. v = √(2 × 2.82 × 10⁶) = 2.4 × 10³ m/s.Marking: 1 mark each.
Question 7 Turning points (2 marks)
Since E = K + U and K ≥ 0, a position with U > E would need negative kinetic energy, which is impossible. At U = E the kinetic energy is zero, so the particle is momentarily at rest and reverses direction.Marking: 1 mark for K cannot be negative; 1 mark for the stop and reversal.
Question 8 Engine power (2 marks)
a. P = Fv. b. F = P/v = 15 000/25 = 6.0 × 10² N.Marking: 1 mark each.
