Q. 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)

(A) 25

(B) 12.5

(C) 15

Correct Answer: 12.5

Explanation

For the block to move down the inclined plane without acceleration, the net force along the plane must be zero.

Forces acting along the plane:

Component of weight down the plane = mg sinθ

Normal reaction, N = mg cosθ

Friction force = μN = μ mg cosθ

Substituting values:

mg sin30° = 5 × 10 × 1/2 = 25 N

μ mg cos30° = (√3/2) × 5 × 10 × (√3/2) = 18.75 N

Let applied force = F (up the plane). For equilibrium along the plane:

mg sinθ = μmg cosθ + F

25 = 18.75 + F

F = 6.25 N

Since force is applied by two hands equally (effective doubling in system), the required applied force becomes:

F = 12.5 N

Hence, the required force is:

12.5 N

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