Question

A sled weighing 100 lb. reaches the foot of a hill with a speed of 40 ft/s. The coefficient of kinetic friction between the sled and the horizontal surface of the ice at the foot of the hill is 0.030. How far will the sled travel on ice?

Answer 1

$m = 100Ib \space\space\space\space\space\space\space\space\space v = 40ft/s \space\space\space\space\space\space\space\space\space g = 32.2ft/s^2 \space\space\space\space\space\space\space\space\space \mu = 0.03$

The initial kinetic energy at the bottom equal to the work done against friction, after traveling a horizontal distance X, where it stops.

$\large\frac{mv^2}{2}$ = $\mu mgX \to X = \large\frac{v^2}{2\mu g}$ $=\large\frac{40*40}{2*0.03*32.2}$ $=82.82ft$