Question

A body of mass $m$ slides down a smooth inclined plane having an inclination of ${45}^o$ with the horizontal. It takes $4$ sec to reach the bottom. If the body is placed on a similar plane having coefficient of friction $0.5$. What is the time taken for it to reach bottom?

• A. $4\sqrt{2}sec$
• B. $4sec$
• C. $3\sqrt{2}sec$
• D. $2sec$

Answer 1

The correct option is A $4\sqrt{2}sec$

Given: a body of mass m slides down a smooth inclined plane having an inclination of ${45}^o$ with the horizontal. It takes 4 sec to reach the bottom. If the body is placed on a similar plane having coefficient of friction 0.5.

To find the time taken by the body to reach bottom

Solution:

First we need to find the distance

$d=ut+\frac{1}{2}g{t}^2⟹d=\frac{1}{2}\times g\sin {45}^{\circ }\times t^2⟹d=\frac{1}{2}\times g\sin {45}^{\circ }\times 4\times 4⟹d=8g\sin {45}^{\circ }$

Now for friction path we have acceleration

$a=g\sin \theta -\mu g\cos \theta$

So,

$d=ut+\frac{1}{2}g{t}^2⟹8g\sin {45}^{\circ }=\frac{1}{2}\times (g\sin \theta -\mu g\cos \theta )t^2$

Here $\theta ={45}^{\circ }$ and given $\mu =0.5=\frac{1}{2}$

So the above equation becomes,

$8\times \frac{1}{\sqrt{2}}=\frac{1}{2}(\sin {45}^{\circ }-\mu \cos {45}^{\circ })t^2⟹\frac{8}{\sqrt{2}}=\frac{1}{2}(\frac{1}{\sqrt{2}}-\frac{1}{2}\times \frac{1}{\sqrt{2}})t^2⟹\frac{8}{\sqrt{2}}=\frac{1}{2}\left(\frac{2-1}{2\sqrt{2}}\right)t^2⟹t^2=\frac{8\times 4\sqrt{2}}{\sqrt{2}}⟹t=4\sqrt{2}sec$

Hence the time taken by the body to reach bottom is $4\sqrt{2}sec$