Question

A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and g = 9.8m / s2 , then the acceleration of the block

Answer 1

0.88 m/s2

24-(0.4*9.8*5)=4.4

Acceleration= 4.4/5=0.88

Answer 2

The answer to the given problem is as following

Given from the question,

Mass of block = 5kg

F = 24N

μ (friction) = 0.4 (force)

g (acceleration due to gravity)= 9.8 m/s²

friction, f = μmg = 0.4×5×9.8 = 19.6N

net force =F-f = 24-19.6 = 4.4N

acceleration = 4.4/5 = 0.88m/s²