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. $
\left(g=10 \mathrm{~m} / \mathrm{s}^2\right) .$

Professor PC Thomas Classes · Answer

The downward component of the block's weight along the incline is $W_{\parallel} = mg \sin(30^\circ) = 5 	imes 10 	imes 0.5 = 25$ N. Since the block is moving down, the kinetic friction acts upwards along the incline. The frictional force is $f_k = \mu N = \mu mg \cos(30^\circ) = \frac{\sqrt{3}}{2} 	imes 5 	imes 10 	imes \frac{\sqrt{3}}{2} = 50 	imes \frac{3}{4} = 37.5$ N. Because friction is greater than the downward weight component ($37.5 > 25$), the block would decelerate. To keep it moving at a constant speed (zero acceleration), an additional downward force $F$ is needed: $F + W_{\parallel} = f_k \implies F + 25 = 37.5 \implies F = 12.5$ N.