JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Laws of Motion

A block of mass 5 kg is moving on an inclined plane which makes an angle of

3 0^{\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.

Answer: (B)

Solution

For motion down the incline without acceleration, net force = 0.

Component of weight down the incline:

m g sin 3 0^{\circ} = 5 \times 10 \times 0.5 = 25 N

Friction force (up the incline):

μ m g cos 3 0^{\circ} = \frac{3}{2} \times 5 \times 10 \times \frac{3}{2} = 37.5 N

Since friction exceeds the weight component, an external force

F

down the incline is needed:

m g sin 3 0^{\circ} + F = μ m g cos 3 0^{\circ}

F = 37.5 - 25 = 12.5 N

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option