Question

a block of mass 5 kg is being pulled along a board horizontally with a constant velocity; the coefficient of friction between the two surface is 0.25, then what is the frictional forces that opposes the motion​

Answer 1

The block is in equilibrium, so the net forces parallel and perpendicular to the surface are

∑ F[para] = F[normal] - mg = 0

∑ F[perp] = F[pull] - F[friction] = 0

where mg is the weight of the block. It follows that

F[normal] = mg = (5 kg) (9.8 m/s²) = 49 N

and so the frictional force has magnitude

F[friction] = 0.25 F[normal] = 12.25 N ≈ 12 N