A block of mass 1kg is pushed up a surface inclined to horizontal at an angle of `30^(@)` by a force of 10N parallel to the inclined surface `[figure]`. The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate
(a) work done against gravity
(b) work done against force of friction
(c) increase in potential energy
(d) increase in kinetic energy
(e) work done by applied force.

Explanation

Consider the adjacent digraam the block is pushed up by applyin a force F.

Normal reaction(N) and frictionless force(f) is shown
Given mass=m=1kg, `theta=30^(@)`
`F=10N,mu,=0.1` and S=distance moved by the block along th inclined plane=10m
(a) Work done against gravity=Increase in PE of the block
`=mgxx"Vertical distance travelled"`
`=mgxxs(sin theta)=mgxxs(sin theta)=(mgs)sintheta`
`=1xx10xx10xxsin 30^(@)=50J " " (thereforeg le 10m//s^(2))`
(b) Work done against friction
`wf=fxxs=muNxxs=mu mg cos theta thetaxx5`
`=0.1xx1xx10xxcos 30^(@)xx10`
`=1xx10xx10xxsin30^(@)xx10`
(c ) Increase in PE=mgh`mg(s sin theta)`
`=1xx10xx10xxsin 30^(@)`
`100xx(1)/(2)=50J`
(d) By work energy theorem, we know that work done by all the foreces=change in KE
`(W)=DeltaK`
`Deltak=W_(g)+W_(f)+W_(f)`
`=-mgh-fs+FS`
`-50-8.66+10xx10`
`=50-8.66=41.34J`
(e) Work done by applied force, F=FS
`=(10)(10)=100J`
(a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved. Ball 3 is having negligible friction hence, there is no loss of energy.
(b) Ball 1 acquire rotational energy, ball 2 loses energy by friction, They cannot cross at C. Ball 3 can cross over.
(c ) Ball 1,2 turn back before C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in "wrong" sense when it reaches B. It cannot roll back to A, because of kinetic friction.