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.

The time taken by the boy to reach the height h is

Options

A. (a) `sqrt((2h)/(g))`
B. (b) `sqrt((6h)/(g))`
C. (c) `2sqrt(h/g)`
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`