Answer to Question #290782 in Physics for Nard

Question

A block slides from rest from the top of an inclined plane 10 m long which is inclined 37° with

the horizontal. If the coefficient of kinetic friction is 0.30, determine how long it will take the

block to reach the bottom of the plane.

Answer 1

Given:

$l=10\rm \: m$

$\theta=37^\circ$

$\mu_k=0.30$

The Newton's second law says

$$a=\frac{F_{\rm net}}{m}=\frac{mg\sin\theta-\mu_k mg\cos\theta}{m}$$

So, the acceleration of the block

$$a=g(\sin\theta-\mu_k \cos\theta)$$

Hence, time of motion

$$t=\sqrt{2l/a}=\sqrt{2l/g(\sin\theta-\mu_k \cos\theta)}$$

$$t=\sqrt{2*10/9.8(\sin37^\circ-0.30\cos 37^\circ)}=2.37\:\rm s$$