Physics · Friction

JEE Main 2026 — 28 January, Morning Shift — Question 35

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.

Option A:
25
Option B:
12.5
Correct
Option C:
7.5
Option D:
15

Answer: B

Step-by-step solution

$\mathrm{mg} \sin 30^{\circ}=\mathrm{F}+\mu \mathrm{mg} \cos 30^{\circ}$ $\mathrm{F}=5 \times 10 \times \frac{1}{2}-\frac{\sqrt{3}}{2} \times 5 \times 10 \times \frac{\sqrt{3}}{2}$ $\mathrm{F}=25-\frac{75}{2}=25-37.5$ $\mathrm{F}=-12.5 \mathrm{~N}$ ∴ force will be downward on incline of magnitude 12.5 N

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Friction
Topic: Single Block Problems Involving Friction