Question

A body of mass m is launched up on a rough in dines plane making the angle $45^\circ $ with the horizontal. If time of ascent is $\dfrac{1}{3}$ of the time of descent, the frictional coefficient between the plane and body is

Answer 1

Hint:In the question, we are provided with a relation between the time of ascent and the descent. For that firstly put both the formulas and then by solving the relation by putting the given values. The formula of ascent is ${t_a} = \sqrt {\dfrac{{2s}}{{{a_a}}}} $ and the descent is ${t_d} = \sqrt {\dfrac{{2s}}{{{a_d}}}} $.

Complete step by step answer:
In the question, we are given that the time of ascent is $\dfrac{1}{3}$ to the time of descent. The time of ascent is defined as the time taken by the body thrown up to reach the maximum height and the time of descent is the time taken by a freely falling to touch the ground.The formula for the time of ascent is ${t_a} = \sqrt {\dfrac{{2s}}{{{a_a}}}} $ and ${a_a}$ is the acceleration for the ascent.

Writing the equation for the acceleration of the ascent,
$m{a_a} = mg\sin \theta + \mu g\cos \theta $
Similarly, the time of descent ${t_d} = \sqrt {\dfrac{{2s}}{{{a_d}}}} $ and s is the distance and ${a_d}$ is the acceleration of the descent.
Equation for the descent is $m{a_d} = mg\sin \theta - \mu g\cos \theta $ here $\mu $ is the frictional coefficient.
According to the question,
${t_a} = \dfrac{1}{3}{t_d}$

Putting the values,
$\sqrt {\dfrac{{2s}}{{g\sin \theta + \mu g\cos \theta }}} = \dfrac{1}{3}\sqrt {\dfrac{{2s}}{{g\sin \theta - \mu g\cos \theta }}} $
Squaring and solving, while $\theta = 45^\circ $(given)
$9g\sin \theta - 9\mu g\cos \theta = g\sin \theta + \mu g\cos \theta \\
\Rightarrow 8g\sin \theta = 10\mu g\cos \theta \\
\Rightarrow \mu = \dfrac{8}{{10}}\tan 45^\circ \\
 \therefore\mu = \dfrac{4}{5} \\ $
Hence, the frictional coefficient between the plane and the body is $\dfrac{4}{5}$.

Note:The time of flight is taken to be addition of the time of ascent and the time of descent. The formulas used in the question should be remembered carefully. We can also learn the derivation for both the time which is the time of ascent and the time of descent.