Question

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.

Answer 1

Here, `m=2kg, u=0, F=7N, mu=0.1, W=?, t=10s`
Acceleration produced by applied force `, a_(1)=F//m=7//2=3.5m//s^(2)`
Force of friction, `f=muR=mumg=0.1xx2xx9.8=1.96N`
Retardation produced by friction, `a_(2)=(-f)/(m)=-(1.96)/(2)=-0.98ms^(-2)`
Net acceleration with which body moves, `a=a_(1)+a_(2)=3.5-0.98=2.52ms^(-2)`
Distance moved by the body in 10second, from `s=ut+(1)/(2)at^(2)=0+(1)/(2)xx2.52xx(10)^(2)=126m.`
(a) Work done by the applied for `=Fxxs`
`w_(1)=7xx126=882J`
(b) Work done by the force of friction `W_(2)=-fxxs=-1.96xx126=-246.9J`
(c) Work done by the net force `W_(3)=` Net force `xx` distance `=(F-f)s=(7-1.96)126=635J`
(d) From `v=u+at`
`v=0+2.52xx10=25.2ms^(-1) :.` Final K.E. `=(1)/(2)mv^(2)=(1)/(2)xx2xx(25.2)^(2)=635J` .
Initial K.E.`=(1)/(2)m u^(2)=0 :.` Change in K.E. `=635-0=635J`
This shows that change in K.E. of the body is equal to work fone by the net force on the body.