Question

A body takes double the time to slide down a rough inclined plane then an identical smooth inclined plane of angle 45^(0) .Then coefficient of friction is​

Answer 1

The coefficient of friction is 3/4

Explanation:

If the angle of inclination of inclined plane is $\theta$ and the coefficient of friction of inclined plane is $\mu$

Then

net acceleration of the body in the downward direction on a rough inclined plane

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

Here, $\theta=45^\circ$

Therefore,

$a=g\sin 45^\circ-\mu g\cos 45^\circ$

$\implies a=\frac{g}{\sqrt{2}}-\frac{\mu g}{\sqrt{2}}$

Similarly, acceleration on a smooth inclined plane ($\mu=0$

$a'=\frac{g}{\sqrt{2}}$

If the body covers a distance s on inclined plane, time taken

$s=0\times t+\frac{1}{2}at^2$

$\implies t=\sqrt{\frac{2s}{a}}$

For rough inclined plane

$t=\sqrt{\frac{2s}{(1-\mu)g/\sqrt{2}}}$

$\implies t=\sqrt{\frac{2\sqrt{2}s}{(1-\mu)g}}$

Similarly, for smooth inclined plane

$t'=\sqrt{\frac{2s}{g/\sqrt{2}}}$

$\implies t'=\sqrt{\frac{2\sqrt{2}s}{g}}$

Given t = 2t'

Thus

$\sqrt{\frac{2\sqrt{2}s}{(1-\mu)g}}=2\sqrt{\frac{2\sqrt{2}s}{g}}$

$\implies \frac{2\sqrt{2}s}{(1-\mu)g}=4\times \frac{2\sqrt{2}s}{g}$

$\implies \frac{1}{1-\mu}=4$

$\implies 1-\mu=\frac{1}{4}$

$\implies \mu=1-\frac{1}{4}$

$\implies \mu=\frac{3}{4}$

Hope this answer is helpful.

Know More:

Q: a block of mass m takes time t to slide down on a smooth inclined plane of angle of inclination θ and height h. if same block slide down on a rough inclined plane of same angle of inclination and same height and takes time n times of initial value, then coefficient friction between block and inclined plane is :

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