JEE Main 2026 — Jan 28, Shift 1 — Physics solution

JEE Main2026Jan 28, Shift 1Physics

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 $\frac{3}{2}$ . The force to be applied on the block so that the block will move down without acceleration is ______ N.

A
25
B
12.5
C
7.5
D
15

Solution

The block slides down at constant velocity. Friction opposes motion, so friction acts up the incline. Along the incline (taking down the slope as positive), the block needs a net force of zero. The friction force magnitude is $μ m g cos θ$ .

Compare gravity component along the slope to friction:

$m g sin 3 0^{\circ} = 5 \cdot 10 \cdot \frac{1}{2} = 25$ N (down).

$μ m g cos 3 0^{\circ} = \frac{3}{2} \cdot 5 \cdot 10 \cdot \frac{3}{2} = \frac{75}{2} = 37.5$ N (up).

Friction alone exceeds gravity along the slope, so without external help the block would not slide down — it would be held in place. To make it move down at constant velocity, an additional force is needed acting down the slope.

Balance: $m g sin θ + F = μ m g cos θ \Rightarrow F = 37.5 - 25 = 12.5$ N.

Option (2). (The required force has magnitude 12.5 N, directed down the slope.)

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