A block of mass 5 kg is moving on an inclined plane which makes an… | JEE Main 2026 (28 Jan, Shift 1)

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Friction on an Incline

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

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Answer: (b)

Weight component down the slope: $mg\sin 30^\circ = 50 \times 0.5 = 25\text{ N}$ .
Kinetic friction (up the slope while moving down): $\mu mg\cos 30^\circ = \dfrac{\sqrt{3}}{2} \times 50 \times \dfrac{\sqrt{3}}{2} = 37.5\text{ N}$ .
Zero acceleration: $F + 25 = 37.5 \Rightarrow F = 12.5\text{ N}$ (down the incline).
So the answer is (b) 12.5 N.

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