JEE Main 2026 · 28 January · morning · Single correct · +4 / −1

A block of mass 5 kg is moving on an inclined plane which makes an angle of 30° with the horizontal. Friction coefficient between the block and inclined plane surface is √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).

  1. (a) 25
  2. (b) 12.5
  3. (c) 15
  4. (d) 7.5
Answer and solution

Answer (b) 12.5

Solution

Physics sir · 4 steps

Gravity's pull along the slope is weaker than friction here, so you must push the block down to help it.

  1. Forces along incline. Gravity pulls the block down the slope with mg sin 30° = 5 × 10 × 1/2 = 25 N. The normal force is N = mg cos 30° = 50 × √3/2 = 25√3 N.
  2. Friction up the slope. The block is moving down, so kinetic friction acts up the slope. Its size is μ N = √3/2 × 25√3 = 37.5 N, which is more than the 25 N pulling it down.
  3. Zero net force. No acceleration means the forces along the slope must balance. Since friction wins, the applied force F must act down the slope to make up the difference.

    F + mg sin θ = μ mg cos θ ⇒ F = 37.5 - 25 = 12.5 N

  4. So. A force of 12.5 N down the incline keeps the block sliding at constant velocity, which is option (b).

The idea to keep. On an incline, first compare mg sin θ with μ mg cos θ. Friction always opposes the motion, and the applied force only has to make up the difference between the two.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Friction

Asked about 40 times in JEE Main since 2020. This question is from 2026.