Question

A 12-kg box is released from the top of an incline that is 5.0m long and makes an angle of 40° to the horizontal. A 60-N friction force impedes the motion of the box. (a) What will be the acceleration of the box? (b) How long will it take to reach the bottom of the incline?

Answer 1

(a)

$mg\sin40°-60=ma\to a=g\sin40°-60/m=9.8\cdot\sin40°-60/12=1.3\ (m/s^2)$

(b)

$l=at^2/2\to t=\sqrt{2l/a}=\sqrt{2\cdot5/1.3}=2.77\ (s)$