JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
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.
- A 25
- B 12.5
- C 7.5
- D 15
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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