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Step-by-Step Solution
Step 1: Identify the forces and motion involved
A body of mass m is projected up a rough inclined plane at an angle of 30^\circ to the horizontal. The frictional force opposes the direction of motion of the body. During ascent, friction acts down the plane, and during descent, it acts up the plane.
Step 2: Express the accelerations during ascent and descent
Let g be the acceleration due to gravity, and let \mu be the coefficient of friction. The component of gravitational acceleration down the incline is g \sin 30^\circ , and the normal force on the incline is mg \cos 30^\circ . Thus:
During ascent (going up), friction acts down the plane, so the net acceleration down the plane is
a_a = g \sin 30^\circ + \mu g \cos 30^\circ.
During descent (coming down), friction acts up the plane, so the net acceleration down the plane is
a_d = g \sin 30^\circ - \mu g \cos 30^\circ.
We also know \sin 30^\circ = \tfrac{1}{2} and \cos 30^\circ = \tfrac{\sqrt{3}}{2} .
Step 3: Write the time relation for ascent and descent
If s is the distance traveled along the plane (from the starting point to the highest point for ascent), the formula for the time taken under constant acceleration to cover distance s from rest is
t = \sqrt{\frac{2s}{a}}.
Given that the time of ascent, t_a , is half the time of descent, t_d , we have:
t_a = \frac{1}{2} \, t_d \quad \Longrightarrow \quad \sqrt{\frac{2s}{a_a}} = \frac{1}{2} \sqrt{\frac{2s}{a_d}}.
Step 4: Simplify the time relation to find a relationship between accelerations
Cancelling common factors on both sides and squaring, we get:
\frac{2s}{a_a} = \frac{1}{4} \, \frac{2s}{a_d} \quad \Longrightarrow \quad a_d = 4 \, a_a.
Step 5: Substitute expressions for a_a and a_d
Using
a_a = g \sin 30^\circ + \mu g \cos 30^\circ, \quad a_d = g \sin 30^\circ - \mu g \cos 30^\circ,
and the relation a_d = 4\,a_a , we have:
g \sin 30^\circ - \mu g \cos 30^\circ = 4 \Bigl(g \sin 30^\circ + \mu g \cos 30^\circ \Bigr).
Step 6: Solve for the coefficient of friction \mu
Substitute \sin 30^\circ = \tfrac{1}{2} and \cos 30^\circ = \tfrac{\sqrt{3}}{2} :
\frac{g}{2} - \mu \, g \, \frac{\sqrt{3}}{2} = 4 \left(\frac{g}{2} + \mu \, g \, \frac{\sqrt{3}}{2}\right).
Dividing through by g and simplifying:
\frac{1}{2} - \mu \, \frac{\sqrt{3}}{2} = 4 \left(\frac{1}{2} + \mu \, \frac{\sqrt{3}}{2}\right).
\frac{1}{2} - \frac{\sqrt{3}}{2}\mu = 2 + 2\sqrt{3}\,\mu.
Bring like terms together:
\frac{1}{2} - 2 = 2\sqrt{3}\,\mu + \frac{\sqrt{3}}{2}\mu,
- \frac{3}{2} = \mu \left(2\sqrt{3} + \frac{\sqrt{3}}{2}\right) = \mu \left(\frac{4\sqrt{3} + \sqrt{3}}{2}\right) = \mu \left(\frac{5\sqrt{3}}{2}\right).
Hence,
\mu = \frac{- \frac{3}{2}}{\frac{5\sqrt{3}}{2}} = - \frac{3}{2} \times \frac{2}{5\sqrt{3}} = - \frac{3}{5\sqrt{3}}.
Since we normally consider the magnitude of \mu (coefficient of friction is positive), we take
\mu = \frac{\sqrt{3}}{5}.
Step 7: Identify x from the given form of \mu
We are told that \mu = \frac{\sqrt{x}}{5} . From our result, \mu = \frac{\sqrt{3}}{5} . Therefore,
\sqrt{x} = \sqrt{3} \quad \Longrightarrow \quad x = 3.
Final Answer: x = 3
