#### A smooth block is released at rest on a  45o incline and then slides a distance  . The time taken to slide  is  time as much to slide on rough incline than on a smooth incline. The coefficient of friction is Option 1) Option 2) Option 3) Option 4)

As we learnt in

Kinetic or Dynamic Friction -

kinetic friction

coefficient of kinetic friction

R = reaction

- wherein

depends on the nature of surface in contact.

Component of $\dpi{100} g$ down the plane $\dpi{100} =\; gsin\theta$

$\dpi{100} \therefore$      For smooth plane,

$\dpi{100} d=\frac{1}{2}(gsin\theta )t^{2}\; ............(i)$

For rough plane,

Frictional retardation up the plane $\dpi{100} =\; \mu _{k}(gcos\theta )$

$\dpi{100} \therefore \; \; \; d=\frac{1}{2}(gsin\theta -\mu _{k}gcos\theta )(nt)^{2}$

$\dpi{100} \therefore \; \; \; \frac{1}{2}(gsin\theta )t^{2}=\frac{1}{2}(gsin\theta -\mu _{k}gcos\theta )n^{2}t^{2}$

or     $\dpi{100} sin\theta =n^{2}(sin\theta -\mu _{k}cos\theta )$

Putting $\dpi{100} \theta =45^{\circ}$

$\dpi{100} or \; \; sin45^{\circ}=n^{2}(sin45^{\circ}-\mu _{k}cos45^{\circ})$

$\dpi{100} or\; \; \; \frac{1}{\sqrt{2}}=\frac{n^{2}}{\sqrt{2}}(1-\mu _{k})$

$\dpi{100} or\; \; \; \mu _{k}=1-\frac{1}{n^{2}}$

Correct option is 3.

Option 1)

This is an incorrect option.

Option 2)

This is an incorrect option.

Option 3)

This is the correct option.

Option 4)

This is an incorrect option.