JEE Main Physics question

PhysicsLaws of MotionJEE Main

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 = 9.8 m/s², then the acceleration of the block is

Answer: (D)

Solution

Kinetic friction = μmg = 0.4×5×9.8 = 19.6 N. Net force = 24 − 19.6 = 4.4 N. Acceleration = 4.4/5 = 0.88 m/s².

Question ID VK-153463