A 0.5 kg roller-coaster cart starts from rest at the top of a frictionless track of height h = 8.0 m. At the bottom of the track, it enters a rough horizontal section of length 10 m with a frictional coefficient that brings it to rest. What was the average frictional force exerted on the cart (take g = 10 m/s^2)?

Answer and explanation

Correct answer: D. 4.0 N

  1. Step 1: Step 1: Initial gravitational potential energy E_p = mgh = 0.5 kg 10 m/s^2 8.0 m = 40 J. On the frictionless track, all potential energy transforms into kinetic energy. Step 2. Step 2: On the rough section, work done by friction dissipates all kinetic energy: W_friction = F_friction d => 40 J = F_friction 10 m => F_friction = 4.0 N. Answer: 4.0 N. Why not the others: Conservation of energy links potential energy directly to braking work: mgh = F*d.

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