A College Board-style free-response question on Unit 3: Work, Energy, and Power. Work through each part, then reveal the model answer to see exactly what earns each point.
The ramp is frictionless, so mechanical energy is conserved. Take the bottom of the ramp as the zero of gravitational potential energy.
Initial state: K_i = 0 (released from rest), U_i = mgh = 2 · 10 · 5 = 100 J
Final state: K_f = ½mv², U_f = 0
Conservation of mechanical energy:
K_i + U_i = K_f + U_f
0 + 100 = ½(2)v² + 0
100 = v² → v = √100 = 10 m/s
Answer: v = 10 m/s at the bottom of the ramp.
On the rough surface, friction is the only horizontal force. All of the block's KE is dissipated as the block slides to a stop.
Initial KE on horizontal surface (from part A): K = ½(2)(10)² = 100 J
Friction force: The block is on a horizontal surface, so F_N = mg = 2 · 10 = 20 N. Then:
F_friction = μ_k · F_N = 0.20 · 20 = 4 N
Energy dissipated by friction over distance d:
E_dissipated = F_friction · d = 4d
Block comes to rest when all KE is gone:
F_friction · d = K_initial
4d = 100 → d = 25 m
Answer: The block travels 25 m on the rough surface before stopping.
The student is incorrect. The distance the block travels is INDEPENDENT of mass.
The student is right that more friction acts on a heavier block (F_friction = μ_k · mg increases with m). But they're missing that the block also arrives at the bottom with more kinetic energy.
Re-derive symbolically:
From part A: ½mv² = mgh → v² = 2gh (mass cancels)
From part B: μ_k(mg)d = ½mv² → μ_k · g · d = ½v²
Solving for d:
d = v²/(2·μ_k·g) = 2gh/(2·μ_k·g) = h/μ_k
Mass cancels everywhere! d = h/μ_k = 5/0.20 = 25 m — same answer regardless of mass.
Numerical check for m = 4 kg:
At bottom: KE = mgh = 4·10·5 = 200 J. F_friction = 0.20·40 = 8 N. d = 200/8 = 25 m. Same.
Physically: When you double the mass, you double the KE AND you double the friction force — both effects scale the same way, so the stopping distance is unchanged.