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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

A force of 5 N acts on a 15 kg body initially at rest. The work done by the force during the first second of motion of the body is

Answer: (B)

Solution

a = F/m = 5/15 = 1/3 m/s². Distance in the first second, s = ½at² = ½(1/3)(1)² = 1/6 m. Work done, W = Fs = 5 × 1/6 = 5/6 J.

Question ID VK-153855