JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2026
- Medium
- MCQ
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.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 12.5
Step-by-step Solution
Detailed explanation

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