question_answer 2)

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the (i) work done by the applied force in 10s, (ii) work done by the friction in 10 s, (iii) work done by the net force on the body in 10 s, and (iv) change in kinetic energy of the body in 10 s. Interpret your results.  

Answer:

Here   Force of friction,   Net force with which the body moves,   Acceleration,   Distance,   = 126 m (i) Work done by the applied force,                 (ii) Work done by the friction, . (iii) Work done by the net force, . (iv) Final velocity acquired by the body after 10 s,   Change in K.E. of the body   Thus the change in K.E. of the body is equal to the work done by the net force on the body.