Q.

A body is pushed up on a rough inclined plane making an angle 30⁰ to the horizontal. If its time of ascent on the plane is half the time of its descent, find coefficient of friction between the body and the inclined plane

Moderate

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$\frac{\sqrt{3}}{5}$

$\sqrt{\frac{3}{5}}$

$\frac{3}{\sqrt{5}}$

$\frac{3}{5}$

answer is A.

(Detailed Solution Below)

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Detailed Solution

Given:
t_a =1/2t_d

Let's assume, the length of inclined plane is s.

s = 1/2at2

⇒t=√(2s/a)

So from the given condition,

√(2s/aa) = 1/2 √(2s/ad) .........(1)

Here, friction force, f=μmgcosθ

a_a = gsinθ + μgcosθ

a_a = g/2 + √3/2 μg (∵θ=30^∘)

a_d = gsinθ − μgcosθ

a_d = g/2 − √3/2μg

using the above values of aa and ad, and putting in equation (1), we will get

μ = √3/5

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