Two 30 kg blocks rest on a massless belt which passes over a fixed pulley and is attached to a 40 kg block. If coefficient of friction between the belt and the table as well as between the belt and the blocks B and C is `mu` and the system is released from rest from the position shown, the speed with which the block B falls off the belt is

Options

A. `2sqrt2(m)/(s)` if `mu=0.2`
B. `sqrt2(m)/(s)` if `mu=0.2`
C. `2(m)/(s)` if `mu=0.5`
D. `2.5(m)/(s)` if `mu=0.5`

Answer

A, C

Explanation

Maximum possible acceleration of block will be `=mug`
(i) When `mu=0.2` maximum acceleration possible so that blocks does not slip is `2(m)/(s^2)`. Now acceleration of system assuming blocks don't slip is
As `3.8gt2` hence the assumption is wrong and blocks slip on the belt.
Maximum force applicable on block B is `mumg`
Acceleration of block`=2(m)/(s^2)`
Hence velocity of B when it strikes pulley is `V^2=0^2+2xx2xx2`
`V^2=8`
`V=2sqrt2(m)/(s)`
Hence (a).
(ii) When `mu=0.5` maximum acceleration possible so the block dows not slip is `5(m)/(s^2)`
Now acceleration of system assuming blocks don't slip is
`a=(40g-2mumg)/(100)=1(m)/(s^2)`
`mu=0.5`
as `1lt5` hence, blocks do not slip and acceleration of block B is `1(m)/(s^2)`
hence velocity of B when it strikes pulley is
`V^2=0^2+2.1.2`
`V^2=4`
`V=2(m)/(s)`
Hence (c).