A boy of mass m climbs up a conveyor belt with a constant acceleration. The speed of the belt is `v=sqrt(gh//6)` and the coefficient of friction between the body and conveyor belt is `mu=(5)/(3sqrt3)`. The boy starts from A and moves with the maximum possible acceleration till he reaches the highest point B.

Work done by the boy is

Options

A. (a) Equal to work done by boy
B. (b) `Equal to work done by the motor in running the conveyor belt
C. (c) Zero
D. (d) None of above

Answer

B

Explanation

`a=(F_(max)-mg sin theta)/(m)`
`=(mumgcos theta-mg sin theta)/(m)=g/3`

`AB=vt+1/2at^2`
`(h)/(sin 30^@)=sqrt((gh)/(6))t+1/2g/3t^2`
`g t^2+sqrt(6ght)-12h=0`
Solving, we get `t=sqrt(6h/g)`
`v_f=v_i+at=sqrt((gh)/(6))+g/3sqrt(6h/g)=3sqrt((gh)/(6))`
`DeltaKE=1/2m(v_f^2-v_i^2)=2/3mgh`
Distance moved by conveyour belt till the boy reaches B:
`s=v_it=sqrt((gh)/(6))sqrt((6h)/(g))=h`
w.r.t. conveyor belt, distance moved by the boy,
`2h-h=h`
Work done by gravity w.r.t. conveyor
`W_(mg)=-mgh sin 30^@=-(mgh)/(2)`
From the frame of belt: `v_1=0`, `v_f=at=2sqrt((gh)/(6))`
`W_(boy)=DeltaKE+DeltaPE`
`=1/2m(v_f^2-v_i^2)+mgh/2`
`=1/2m[4(gh)/(6)]+(mgh)/(2)=(5mgh)/(6)`
From the frame of ground:
`W_(boy)+W_f+W_(mg)=DeltaKE`
`W_f=2/3mgh+mgh-5/6mgh=5/6mgh`