A body of mass 2 kg initially kg intially at rest moves under the action of an appliedhorizontal force of 7 N on a tablewith coefficientof kinetic friction = 0.1Computethe (a) work done by the applied force in 10 s. (b) work done by frictionin 10 s . work done by the net forceon the bodyin 10 s. (d) Change in kinetic energy of the body in 10 s and interpret your details .

A body of mass 2 kg initially kg intially at rest moves under the acti

1.	A body of mass 2 kg initially kg intially at rest moves under the action of an appliedhorizontal force of 7 N on a tablewith coefficientof kinetic friction = 0.1Computethe (a) work done by the applied force in 10 s. (b) work done by frictionin 10 s . work done by the net forceon the bodyin 10 s. (d) Change in kinetic energy of the body in 10 s and interpret your details .
Answer» Solution :Frictional force , `f = muN =mu mg = 0.1 xx 2xx 9.8 = 1.96 N ` Neteffective force ` F = f = 7 - 1.96 = 5.04 N ` Acceleration of body under the NET force , `a = F/m =(5.04)/2 = 2.52 ms^(-2)` Distance covered in 10 s , `S =0+1/2 at^(2) [ :. u = 0 ] ` ` = 1/2 xx 2.52 xx10^(2)` ` = 126 m ` (a) work done by applied force , `W_(1)=(F)(s)` ` = 7 xx 126 = 882 J ` (b) Wrok done by friction , `W_(2) =-f(S)` ` = - 1.96 xx126 = - 246.96 = - 247 J ` (c ) Work done by net force= `5.04 xx126 = 635 J ` (d) CHANGE in the kinetic energy of the body. `DeltaK = DeltaW_(3)` = 635 J Interpretation : Work done by net force on the bod equal to the change in kinetic energy .