> Q. Starting from rest, a body slides down a $45^circ$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is – LIVE ANSWER TODAY

Q. Starting from rest, a body slides down a $45^circ$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is

In This post you will get answer of Q.
Starting from rest, a body slides down a $45^circ$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is

Solution:

The various forces acting on the body have been shown in the figure. The force on the body down the inclined plane in presence of friction $mu$ is

image

$F=mgsintheta -f=mgsintheta-mu N=ma$

or $a=gsintheta -mu gcostheta.$

Since block is at rest thus initial velocity u = 0

$therefore$ Time taken to slide down the plane

$t_1=sqrt{frac{2s}{a}}=sqrt{frac{2s}{g sin theta-mu g costheta}}$

In absence of friction time taken will be $t_2=sqrt{frac{2s}{g sin theta}}$

Given $:t_1=2t_2.$

$therefore t^2_1=4t^2_2$ or $frac{2s}{g(sin theta -mu cos theta)}={frac{2s times 4}{g (sin theta)}}$

or $sin theta=4sin theta-4 mu cos theta$ or $mu=frac{3}{4}tan theta =0.75$

Leave a Comment