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 ?

Explanation

Here, `F=5N,m=2kg,u=0,mu=0.2`
Froce of friction , `f=muR=mu mg=0.2xx2xx9.8`
`=3.92N`
Net force on the body, `F^(')=F-f=5-3.92=1.08N`
Acceleration of body `=(F^('))/(m)=(1.08)/(2)=0.54ms^(-2)`
Distance travelled in `5s` is
`s=ut+(1)/(2)at^(2)=0+(1)/(2)(0.54)xx5^(2)=6.75m`
(i) work done by applied force
`W_(1)Fxxs=5xx6.75J=33.75J`
(ii) work done by force of friction
`W_(2)=-fxxs=-3.92xx6.75=-26.46J`
(iii) work doen by net fogrce
`W_(3)=F^(')xxs=1.08xx6.75=7.29J`
(iv) velocity of body after `5s`
`upsilon=u+at=0+0.54xx5=2.7m//s`
Increase in `K.E.=(1)/(2)m upsilon^(2)-(1)/(2)m u^(2)`
`=(1)/(2)xx2(2.7)^(2)-0=7.29J`
Since increase in `K.E.` of the body `=` work done b the net force on the body, therefore work energy theorem stands verified.