A body of mass 'm' is launched up on a rough inclined plane making an angle of 30 ° with the horizontal. The coefficient of friction between the body and plane is √(x)/5 if the time of ascent is half of the time of descent. The value of x is __________. Numerical answer type. Work it out, then open the answer.

Answer: 3. Up: au = g(sin θ + μcos θ). Down: ad = g(sin θ - μcos θ). Same distance, so t ∝ 1/√(a). tu = td/2 ⇒ au = 4ad sin θ + μcos θ = 4sin θ - 4μcos θ ⇒ μ = 3/5tan θ μ = 3/5 · 1/√(3) = √(3)/5 So x = 3.

A body of mass 'm' is launched up on a rough inclined plane making an… | JEE Main 2021 (20 Jul, Shift 2)

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Friction on an Incline

A body of mass 'm' is launched up on a rough inclined plane making an angle of 30 ${}^\circ$ with the horizontal. The coefficient of friction between the body and plane is $\frac{\sqrt{x}}{5}$ if the time of ascent is half of the time of descent. The value of x is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Up: $a_u = g(\sin\theta + \mu\cos\theta)$ . Down: $a_d = g(\sin\theta - \mu\cos\theta)$ . Same distance, so $t \propto \frac{1}{\sqrt{a}}$ .
$t_u = \frac{t_d}{2} \Rightarrow a_u = 4a_d$
$\sin\theta + \mu\cos\theta = 4\sin\theta - 4\mu\cos\theta \Rightarrow \mu = \frac{3}{5}\tan\theta$
$\mu = \frac{3}{5} \cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{5}$
So $x = 3$ .

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