Question

A block is pushed with some velocity up a rough inclined plane. It stops after ascending few meters and then reverses its direction and returns back to point from where it started. If angle of inclination is ${37}^{\circ }$ and the time to climb up is half of the time to return back then coefficient of friction is:

• A. $\frac{9}{20}$
• B. $\frac{7}{5}$
• C. $\frac{7}{12}$
• D. $\frac{5}{7}$

Answer 1

The correct option is A $\frac{9}{20}$

Here for ascending, we can write the equation.

$S=vt-\frac{1}{2}g{t}^2$

After moving distance S it stops means v=0.

And acceleration =$-gsin\theta -\mu gcos\theta$

So, $S=\frac{-1}{2}\times (-gsin\theta -\mu gcos\theta )t_a^2$

$S=\frac{1}{2}\times (gsin\theta +\mu gcos\theta )t_a^2$

Now for descending, we can write as :-

$S=vt+\frac{1}{2}g{t}^2$

After moving distance S it stops again means v_1=0.

And acceleration =$gsin\theta -\mu gcos\theta$

So, $S=\frac{1}{2}\times (gsin\theta -\mu gcos\theta )t_d^2$

$S=\frac{1}{2}\times (gsin\theta -\mu gcos\theta )t_d^2$

It is given that $t_a=\frac{t_d}{2}$

As the distan covered is same so, taking the ratio we get,

$(\frac{t_a}{t_d}{)}^2=\frac{gsin\theta -\mu gcos\theta }{gsin\theta +\mu gcos\theta }$

We have $\theta ={37}^{\circ }andratiooftimeis2$

$4=\frac{sin{37}^{\circ }-\mu cos{37}^{\circ }}{sin{37}^{\circ }+\mu cos{37}^{\circ }}$

From here after calculation we get $\mu =\frac{9}{20}$