The problems are on the work energy theorem worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Work done by a variable force is the integral W = ∫F dx, the area under the force–position graph. These worked answers correct the 20 problems of the AP Physics C worksheet on work energy theorem, from constant force work to force changing in time: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the integral or derivative, the substitution and the result with its unit. For sign questions the positive direction is stated. Use g = 9.81 m/s² unless the problem gives another value, and keep two or three significant figures in final answers.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 3
Problem 1 Constant force work
1. W = Fd cos 0° = 40 × 5.0 = 2.0 × 10² J.2. W = Fd cos 60° = 40 × 5.0 × 0.50 = 1.0 × 10² J.
Problem 2 Units of work
1. N × m = N·m, which is the definition of the joule.2. N = kg·m/s², so J = kg·m²/s².
Problem 3 Stretching a spring
W = ½kx² = ½ × 200 × (0.30)² = 9.0 J.Problem 4 Work from rest
W = ½mv², so v = √(2W/m) = √(240/3.0) = √80 = 8.9 m/s.Problem 5 Force from a spring potential
1. dU/dx = kx, so F = −kx.2. F = −50 × 0.20 = −10 N (towards the equilibrium position).
Problem 6 Conservative force
1. A force whose work between two points is independent of the path, so a potential energy can be defined.2. No. Friction’s work depends on the path length, and the energy is converted to thermal energy.
Problem 7 Linear force
1. W = ∫₀^2.5 4.0x dx = 2.0x² = 2.0 × 6.25 = 12.5 J.2. F at 2.5 m is 10 N, so area = ½ × 2.5 × 10 = 12.5 J, in agreement.
Problem 8 Force with an offset
1. W = ∫₀⁴ (5.0 + 2.0x) dx = 5.0x + x² = 20 + 16 = 36 J.2. v = √(2W/m) = √(72/1.5) = √48 = 6.9 m/s.
Problem 9 Piecewise force graph
1. Triangle: ½ × 2 × 6 = 6 J.2. Rectangle: 3 × 6 = 18 J.
3. The last triangle is ½ × 1 × 6 = 3 J, so the total is 6 + 18 + 3 = 27 J.
Problem 10 Force from a cubic potential
1. F = −dU/dx = −6.0x² + 6.0 newtons.2. F(2.0) = −24 + 6.0 = −18 N.
3. F = 0 at x = ±1.0 m. d²U/dx² = 12x, which is +12 at x = 1.0 (stable) and −12 at x = −1.0 (unstable).
Problem 11 Motorbike power
1. P = Fv = 600 × 20 = 1.2 × 10⁴ W.2. W = Pt = 12 000 × 30 = 3.6 × 10⁵ J.
Problem 12 Gravity from potential
1. F = −dU/dr = −GMm/r² (attractive).2. F = 3.99 × 10¹⁴/(6.37 × 10⁶)² = 3.99 × 10¹⁴/4.058 × 10¹³ = 9.8 N (magnitude).
Problem 13 Average force mistake
1. The average of the end values is only correct when F is linear in x. F = 3.0x² is curved.2. W = ∫₀² 3.0x² dx = x³ = 8.0 J.
Problem 14 Reading an energy diagram
1. U = E gives 2.0x² = 8.0, so x = ±2.0 m.2. Maximum K = E − Umin = 8.0 J, so v = √(2 × 8.0/0.50) = √32 = 5.7 m/s.
3. At x = 0, the bottom of the well, where U is smallest.
Problem 15 Lifting out of the Earth’s field
1. W = ∫ GMm/r² dr from R to 2R = GMm(1/R − 1/2R) = GMm/(2R) = 3.99 × 10¹⁴/(2 × 6.37 × 10⁶) = 3.1 × 10⁷ J.2. mgR = 1.0 × 9.8 × 6.37 × 10⁶ = 6.2 × 10⁷ J, about twice as large, because g falls with height.
Problem 16 Escape speed
1. W = ∫ from R to ∞ of GMm/r² dr = GMm/R = 3.99 × 10¹⁴/6.37 × 10⁶ = 6.26 × 10⁷ J per kg.2. ½v² = GM/R, so v = √(2GM/R) = √(1.25 × 10⁸) = 1.1 × 10⁴ m/s.
Problem 17 Molecular potential
1. dU/dr = U₀[−12a¹²r⁻¹³ + 12a⁶r⁻⁷], which is 0 at r = a. U(a) = U₀(1 − 2) = −1.0 × 10⁻²¹ J.2. d²U/dr² = U₀[156a¹²r⁻¹⁴ − 84a⁶r⁻⁸] = 72U₀/a² at r = a, so k = 72 × 1.0 × 10⁻²¹/(9.0 × 10⁻²⁰) = 0.80 N/m.
Problem 18 Sliding in a double well
1. F = −dU/dx = −4x³ + 8x = −0.5 + 4.0 = +3.5 N, towards +x.2. E = U(0.50) = 0.0625 − 1.0 = −0.9375 J. U(√2) = 4 − 8 = −4.0 J, so K = 3.0625 J and v = √(2 × 3.0625/1.0) = 2.5 m/s.
Problem 19 Lifting a chain
1. λ = 3.0/2.0 = 1.5 kg/m.2. dW = (λ dy) g y.
3. W = ∫₀^L λ g y dy = ½λgL² = ½mgL = 0.5 × 3.0 × 9.81 × 2.0 = 29 J.
Problem 20 Force changing in time
1. v = ∫(6.0t/2.0) dt = 1.5t² m/s.2. P = Fv = 6.0t × 1.5t² = 9.0t³ W.
3. W = ∫₀² 9.0t³ dt = (9.0/4)(16) = 36 J. K = ½ × 2.0 × (1.5 × 4)² = 36 J, so the work–energy theorem is confirmed.
