The Total solution for NCERT class 6-12
A body of mass 2 kg initially at rest movesunder the action of an applied horizontal force of 7 N on a table withcoefficient of kinetic friction = 0.1. Compute the(a) Work done by the appliedforce in 10 s(b) Work done by friction in 10 s(c) Work done by the net force onthe body in 10 s(d) Change in kinetic energy ofthe body in 10 s and interpret your results.
(a) We know that Uk =frictional force/normal reactionfrictional force = Uk xnormal reaction= 0.1 x 2 kg wt = 0.1 x 2 x 9.8 N = 1.96 Nnet effective force = (7 – 1.96) N = 5.04 Nacceleration = 5.04/2 ms-2 =2.52 ms-2distance, s=1/2x 2.52 x 10 x 10 = 126 mwork done by applied force = 7 x 126 J = 882 J(b) Work done by friction = 1.96 x 126 = -246.96 J(c) Work done by net force = 5.04 x 126 = 635.04 J(d) Change in the kinetic energy of the body= work done by the net force in 10 seconds = 635.04 J (This is in accordancewith work-energy theorem).