Correct Answer - B::C
Given `m=250gm=0.250kg
`u=40 cm/sec`
=0.4m/sec, mu=0, v=0`
here, `muR=ma`
(where a is deceleration)`
`a=((muR)/m)=((mumg)/m)=mug`
`=0.1xx9.8=098m/s^2`
`s=(v^2-u^2)/(2a)=0.082m`
Again work done against friction is given by
`-W=muRscostheta`
`=1xx2.5xx0.082xx1`
`rarr+0.02J`
`rarrW=-0.02J`