Two unequal masses of 1 kg and 2 kg are attached at the two ends of a light inextensible string passing over a smooth pulley as shown in figure. If the system is released from rest, find the work done by string on both the blocks in 1 s. (Take g = 10 `ms^(-2)`)

Explanation

Net pulling force on the system is `F_("net")=2g-1g=20-10 = 10 N`
Total mass being pulled
`m=(1+2)=3kg`

Therefore, acceleration of the system will be
`a=(F_("net"))/(m)=(10)/(3)ms^(-2)`
Displacement of both the blocks in 1 s is
`s=(1)/(2)at^(2)=(1)/(2)((10)/(3))(1)^(2)=(5)/(3)m`
Free body diagram of 2 kg block is hown in fig .(ii).
Using `sum F=ma`, we get
`20-T=2a=2((10)/(3))`
or` " " T=20-(20)/(3)=(40)/(3) N`
`:.` Work done by string (tension) on 1 kg block in 1 s is
`W_(1)=(T)(s) cos0^(@)`
`=((40)/(3))((5)/(3))(l)=(200)/(9) J`
Similarl, work done by string on 2 kg block in 1 s will be
`W_(2)=(T)(s)(cos 180^(@))`
`=((40)/(3))((5)/(2))(-1)=-(200)/(9) J`