A block of mass 5?kg is moving on an inclined plane which makes an angle of with the horizontal. The coefficient of friction between the block and the inclined plane surface is The force to be applied on the block so that the block will move down without

Question

A block of mass 5 kg is moving on an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and the 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.

$(g = 10 {m/s}^{2})$ [2026]

Smart Achievers · Accepted Answer

(1)

$m g \sin 30 ° = F + μ m g \cos 30 °$

$F = 5 \times 10 \times \frac{1}{2} - \frac{\sqrt{3}}{2} \times 5 \times 10 \times \frac{\sqrt{3}}{2}$

$F = 25 - \frac{75}{2} = 25 - 37.5$

$F = - 12.5 N$

$∴$ $Force will be downward on incline of magnitude 12.5 N$

Solution

1 12.5

2 25

3 7.5

4 15

Ans.
(1)

$m g \sin 30 ° = F + μ m g \cos 30 °$

$F = 5 \times 10 \times \frac{1}{2} - \frac{\sqrt{3}}{2} \times 5 \times 10 \times \frac{\sqrt{3}}{2}$

$F = 25 - \frac{75}{2} = 25 - 37.5$

$F = - 12.5 N$

$∴$ $Force will be downward on incline of magnitude 12.5 N$

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