Question

A body of mass 2 Kg, initially at rest, moves under the influence of an external force of magnitude 4 N on plane ground. What is the work done by the force and the change in kinetic energy in the first 10 seconds?

(a) 400 J, 400 J

(b) -400 J, 400 J

(c) 400 J, -400 J

(d) -400 J, -400 J

The question was asked in a national level competition.

This key question is from Work-Energy Theorem in chapter Work, Energy and Power of Physics – Class 11

Answer 1

The correct choice is (a) 400 J, 400 J

To explain I would say: By work-energy theorem, the total work done on a body is equal to the change in the total energy. K.E. = \(\Big(\frac{1}{2}\Big)\)mv^2. The acceleration of the body is 4/2 = 2 m/s^2. The final velocity after 10 s can be found out by the first equation of motion. The total change in the total energy is the total change in kinetic energy as the body is moving on plane ground and is equal to 400 J. Hence the work done is 400 J.