A horizontal force of `5N` acts on a body of mass `2kg` initally at rest. It starts moving on a table having coefficient of friction `=0.2` Calculate
(i) work doen by the applied force in `5s`
(ii) work done by force of friction in `5s`
(iii) work done by net force is `5s`
(iv) change in `K.E.` of the body in `5s`.
What do you conclude from this ?

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 . Calculate the (a) work done by applied force in 10s. (b) work done by friction in 10s. (c ) work done by the net force on the body in 10s. (d) change in K.E. of body in 10s, and interpret your result.

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 . Calculate the (a) work done by applied force in 10s. (b) work done by friction in 10s. (c ) work done by the net force on the body in 10s. (d) change in K.E. of body in 10s, and interpret your result.

A body of mass 2 kg initially kg intially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1 Compute the (a) work done by the applied force in 10 s. (b) work done by friction in 10 s . work done by the net force on the body in 10 s. (d) Change in kinetic energy of the body in 10 s and interpret your details .

A body of mass 2 kg initially at rest moves under the action of 0 an applied horizontal force of 7 N on a table with coefficient of kinetic friction= 0.1. Compute the (a) work done by the applied force in 10 s, .

A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the:- work done by the applied force in 10 s.

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 Work done by the applied force in 10s.