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 $\dfrac{\sqrt3}{2}$. The force to be applied on the block so that the block will move down without acceleration is ______ N.
$(g = 10\ \text{m/s}^2)$.
Answer: (B) 12.5
Component of gravity down the incline:
$$mg\sin30^\circ = 5\times10\times\frac12 = 25\ \text{N}$$
Kinetic friction (acting up the incline while the block slides down):
$$\mu mg\cos30^\circ = \frac{\sqrt3}{2}\times50\times\frac{\sqrt3}{2} = 37.5\ \text{N}$$
Friction exceeds the gravity component, so an extra force $F$ must push the block down the incline. For zero acceleration:
$$F + 25 = 37.5 \Rightarrow F = 12.5\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics