A child and sled with a combined mass of 48.0 kg slide down a frictionless hill that is 8.20 m high. If the sled starts from rest, what is its speed at the bottom of the hill?

Answer 1

PEg (the potential energy of gravity) = mgh

Plugging in the numbers, we have

PEg = (48.0 kg)(9.81 m/s^2)(8.20 m) Note: all numbers have 3 sig figs

PEg = 3861.216

PEg ≈ 3860 kg-m^2/s^2 = 3860 N-m (3 sig figs)

Law of Conservation of Energy states KE + PE == constant

At the top of the hill, the sled is at rest, so its KE = 0 N-m = 1/2 * m * v^2, where v = 0 m/s

At the bottom of the hill, PEg = 0 N-m (because PEg = mg * h, and h = 0 m)

So, KE (at bottom of hill) = PEg (at top of hill)

KE (at bottom of hill) ≈ 3860 N-m (ANSWER to 3 sig figs)