Laws of Motion

JEE Main level, single correct, medium

130 of 210

Question 130 of 210

A block of mass 5 kg is moving on an inclined plane which makes an angle of 30∘30^{\circ} with the horizontal. Friction coefficient between the block and inclined plane surface is 32\frac{\sqrt{3}}{2}. The force to be applied on the block so that the block will move down without acceleration is ____\_\_\_\_ N.
(g=10 m/s2)\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right).

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Answer: (C) 12.5

For motion without acceleration, net force = 0.
Component of weight along incline: mgsin⁡(30°)=5×10×0.5=25mg\sin(30°) = 5 \times 10 \times 0.5 = 25 N (down).
Normal force: N=mgcos⁡(30°)=50×32=253N = mg\cos(30°) = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} N.
Friction force opposing downward motion:
f=μN=32×253=37.5f = \mu N = \frac{\sqrt{3}}{2} \times 25\sqrt{3} = 37.5 N (up).
For equilibrium along the incline, applying force F down the incline:
mgsin⁡θ+F=fmg\sin\theta + F = f, so 25+F=37.525 + F = 37.5, giving F=12.5F = 12.5 N.

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