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

Sol:

Case: 1

When there is no friction:

From newton’s second law, Force = mass * acceleration.

$mg sin \Theta = ma_{1}$ $a_{1} = g sin \Theta$

As the body starts from rest, the initial velocity (u) = 0 m/s.

Let, body travels ‘x’ distance in $t_{1} time.$

From equation of motion : x = ut +$\frac{1}{2} at^{2}$ $\Rightarrow x =\frac{1}{2} a_{1}t_{1}^{2}$ $\Rightarrow x = \sqrt{\frac{2x}{a_{1}}} = \sqrt{\frac{2x}{g sin \Theta }}$

Case: 2

When there is friction:

From newton’s second law, Force = mass * acceleration.

$\Rightarrow mg sin\Theta – \mu mg cos \Theta = ma_{2}$ $\Rightarrow a_{2} = g sin \Theta – \mu g cos \Theta$

As the body starts from rest, the initial velocity (u) = 0 m/s.

Let, body travels ‘x’ distance in $t_{2} time.$

From equation of motion: $x = ut + \frac{1}{2} at^{2}$ $x = ut + \frac{1}{2} at^{2}$ $x = \frac{1}{2} a_{2}t_{2}^{2}$ $\Rightarrow t_{2} = \sqrt{\frac{2x}{a_{1}}} =\sqrt{ \frac{2x}{g sin\Theta – \mu g cos\theta}}$

It is given that t_{2} = 2t_{1}[/latex] $\sqrt{\frac{2x}{g sin \theta- \mu g cos\theta}} = 2\sqrt{\frac{2x}{g {sin\theta}}}$

Now, substituting θ = 45 in the above equation, we get,

$\sqrt{\frac{1}{1 – \mu }} = 2$ $\mu = \frac{3}{4} = 0.75$

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