`2gk` block is kept on `1kg` block as shown . The friction between `1kg` block and fixed surface is absent and the coefficient of friction between `2kg` block is `mu=0.1`. A constant horizontal force `F=4N` is applied on `1kg` block. If the work done by the friction on `1 kg` block in `2s` is `-X J`, then find `X`. Take `g=10m//s^(2)`.

Answer

8

Explanation

`a_(2kg)=(f)/(m)=(2N)/(2kg)=1m//s^(2)`

`F-f=ma`
`rArr 4-1=1xxa_(1kg)`
`rArr a_(1kg)=2m//s^(2)`
Distance travelled by `1kg ` in `t=2s`.
`S=(1)/(2)xxat^(2)=(1)/(2)xx2xx2^(2)=4m.`
Velocity of the `1kg` block after `t=2s, `
`v=a=2xx2m//s=4m//s`
`:.` work down by `F=F.S.=4xx4J=16J`
`KE` of `1kg` block `=(1)/(2)xxmxxv^(2)=(1)/(2)xx1xx4^(2)=8J`
Using work energy theorem
`W_(n e t)=DeltaKE`
`W_(F)+W_(f r iction)=DeltaKE`
`16+W_(f r i ction)=8`
`rArr W_(f r i ction)=-8J`
Ans. 8