In the figure shown, a solid sphere of mass `'m'` and radius `r` is released from a height `6r` to slide down a smooth surface. A plank of same mass `'m'` touches the horizontal portion of the surface at the ground. The co-efficient of friction between the plank and the sphere is `mu` and that between the plank and the ground is `mu//4`. Find the work done by the friction force between the plank and the ground till the sphere starts pure rolling on the plank. Neglect the height of the plank.

Explanation

The sphere will reach the plank with speed `v_(0)=sqrt(10gr)` (in horizontal direction)
`FBD` of the sphere when it moves on the plank.
`N_(1)=mg`…..(`i`)
`veca_(sph)=-mu ghati`
(as `f_(1)=muN_(1)` kinetic friction)
`rArr vecv_(sph(t))=(v_(0)-mu "gt")hati` .......(`ii`)
`vecalpha_(sph)=mu mgr(-hatk)`
`rArr vecomega_(sph(t))=-(mu mgr)/(I_(cm))thatk=(-5mug)/(2r)thati`......(`iii`)
`FBD` of plank,
`N_(2)=mg+N_(1)=2mg` ........(`iv`)
`f_(2)=(mu)/(4)N_(2)=(mu)/(4) xx 2mg = (mumg)/(2)` .....(`v`)
`veca_(palpha)=((f_(1)-f_(2))hati)/(m)=(mug)/(2)hati`
`rArrvecv_(Palpha)(t)=(mug)/(2)thati`
For pure cooling,
`vecv_(sph)+vecomega_(sph)xxvecr=vecv_(palpha`
(Here` vecr` is the position vector of the lower most point of the sphere with respect to the centre of the sphere).
`rArr(v_(0)-mu"gt")hati-(5mu"gt")/(2)hati=(mug)/(2)thati`
`rArrt=(v_(0))/(4mug)`
Distance moved by the plank during this time is `s=(1)/(2)(mug)/(2)xx(v_(0)^(2))/(16(mgu)^(2))=(v_(0)^(2))/(64mug)`
`:. W_(f2)=-(mumg)/(2)xx(v_(0)^(2))/(64mug)=-(mv_(0)^(2))/(128)`