Question

A body takes just twice the time as long to slide plane inclined at ${30}^o$ to the horizontal as if the plane frictionless.The coefficient of friction between and the plane is

Answer 1

Angle $\theta ={30}^0$

Time $t_2=2t_1$

We know that,

In absence of friction

The acceleration of the block

$a_1=g\sin {30}^0$

$a_1=\frac{g}{2}$

Now, in presence of friction

The acceleration of the block

$a_2=g(\sin {30}^0-{\mu }_k\cos {30}^0)$

$a_2=\frac{g}{2}(1-\sqrt{3}{\mu }_k)$

Now, $t_2=2t_1$

We know that,

The square of time of slide is inversely proportional to the acceleration

So,

$\frac{t_2^2}{t_1^2}=\frac{a_1}{a_2}$

$\frac{4t_1^2}{t_1^2}=\frac{1}{(1-\sqrt{3}{\mu }_k)}$

$4-4\sqrt{3}{\mu }_k=1$

${\mu }_k=\frac{\sqrt{3}}{4}$

Hence, the coefficient of friction is $\frac{\sqrt{3}}{4}$