Question

A 47.0 kg student starts from rest and slides 26.0 m down a hill where the average force of friction is 55.0 N. What is the final speed of the student at the bottom of the hill if the vertical drop was 17.0 m?

Answer 1

Given:

$m=47.0\:\rm kg$

$F_f=55.0\:\rm N$

$d=26.0\:\rm m$

$h=17.0\:\rm m$

The energy-work theorem says

$$\Delta E_k+\Delta E_p=W$$

We get

$$\frac{mv_f^2}{2}=mgh-Fd$$

$$v_f=\sqrt{\frac{2}{m}(mgh-Fd)}$$

$$v_f=\sqrt{\frac{2}{47.0}(47.0*9.81*17.0-55.0*26.0)}\\ =16.5\:\rm m/s$$