A block of mass 5 kg is moving on an inclined plane which makes an angle of $30^{\circ}$ with the ho

Question

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.

Accepted Answer

$${mgsin30^{\circ} = F + \mu mgcos30^{\circ} }{F = 5 \times 10 \times \frac{1}{2} - \frac{\sqrt{3}}{2} \times 5 \times 10 \times \frac{\sqrt{3}}{2} }{F = 25 - \frac{75}{2} = 25 - 37.5 }{F = - 12.5\text{ }N }$$∴ force will be downward on incline of magnitude 12.5 N

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