Answer to Question #246739 in Physics for Kev

Question

A 15 kg block slides on a horizontal surface with an initial speed of 12 m/s. The speed of 5.0m/s is attained after travelling a distance of 25m. (a) What is the change in kinetic energy on the box? (b) How much work was done by friction on the block?

Answer 1

Given:

$m=15\:\rm kg$

$v_1=12\:\rm m/s$

$v_2=5.0\:\rm m/s$

$d=25\:\rm m$

(a) the change in kinetic energy on the box

$$\Delta E_k=\frac{mv_2^2}{2}-\frac{mv_2^2}{2}\\ =\frac{15*5.0^2}{2}-\frac{15*12^2}{2}=-890\:\rm J$$

(b) the work done by friction on the block

$$W=\Delta E_k=-890\:\rm J$$