Question

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

Answer 1

Hint Use the formula of the acceleration and calculate it for both the frictionless condition and the frictional condition. Find the relation between the acceleration and the time taken and substitute the known values in it, to find the coefficient of the friction.

Useful formula
The formula for the acceleration of the body is
${a_{}} = g\left( {\sin {\theta ^ \circ } - {\mu _k}\cos {\theta ^ \circ }} \right)$
Where is the acceleration of the body, $g$ is the acceleration due to gravity , ${\mu _k}$ is the coefficient of the friction and $\theta $ is the angle of the surface with the horizontal.

Complete step by step solution
It is given that the
Time taken to slide the plane, ${t_2} = 2{t_1}$
The angle that the plane forms with the horizontal, $\theta = {30^ \circ }$
It is known that when there is no friction between the surfaces of sliding, then the acceleration of the block is equal to $g\sin \theta $ .
${a_1} = g\sin {30^ \circ }$
By simplifying the above equation, we get
${a_1} = \dfrac{g}{2}$
In the presence of the frictional force, the acceleration of the block is calculated as follows.
${a_2} = g\left( {\sin {{30}^ \circ } - {\mu _k}\cos {{30}^ \circ }} \right)$
By further simplification,
${a_2} = \dfrac{g}{2}\left( {1 - \sqrt 3 {\mu _k}} \right)$
It is known that the square of time is inversely proportional to that of the acceleration,
$\dfrac{{t_2^2}}{{t_1^2}} = \dfrac{{{a_1}}}{{{a_2}}}$
Substituting that ${t_2} = 2{t_1}$ and the value of the acceleration in the above equation,
$\dfrac{{4t_1^2}}{{t_1^2}} = \dfrac{1}{{\left( {1 - \sqrt 3 {\mu _k}} \right)}}$
By simplifying the above equation,
${\mu _k} = \dfrac{{\sqrt 3 }}{4}$

Hence the coefficient of the friction is obtained as $\dfrac{{\sqrt 3 }}{4}$ .

Note Remember that the friction helps the plane to stand at a particular place by making it ${30^ \circ }$ inclined with the horizontal. In the frictionless condition, it falls down soon. The relation between the time and the acceleration is calculated from the formula ${\text{acceleration}} = \dfrac{{{\text{distance}}}}{{{\text{tim}}{{\text{e}}^2}}}$.