Net pulling for on the system is
`F_("net")=2g-lg=20-10=N`
Toal mass being pulled `m=(+2)=3kg`
Therfore, acceleration of the system will be
`a=(F_("net"))/(m)=(10)/(3)m//s^2`
Displacement of both the blocks in `1 s` is
`S=1/2at^2=1/2((10)/(1))(1)^2=(5)/(3)m `
Free bosy diagram of 2 kg block is shown.(b).
Using `sumF=ma,` we get
`20-T=2_a=2(10/3)` or `T=20-20/3=40/3N`
`:.` work done by sting (tension) on `1 kg` block in `1 s` is
`W_2=(T)(S)cos 0^@)`
=`(40/3)(5/3)(1)=200/9J`
similarly, work done by string on 2 kg block in 1 s will be
`W_2=(T)(S)(cos 180^@)`
`=(40/3)(5/3)(1)=200/9J`.
