A small solid sphere of mass `m` is released from point A. portion AB is sufficiently rough (to provide accelerated pure rolling) BC is smooth and after C the ball moves freely under gravity find gravitational potential energy (U), rotational kinetic energy `(K_(R))` ad translational kinetic energy `(K_(T))` at points A, B, C , D and E.

Explanation

AB, is sufficient rough BC is smooth and after C motion is under gravity. So, total mechanical energy `(U+K_(R)+K_(T))` is always constant. On AB:
`(K_(R))/(K_(T))=(2)/(5)` (for solid sphere)
as accelerated pure rolling is taking place.
After B, rotational kinetic enerrgy will becom constant, after C centre of the ball will follow a projectile motion.
`thereforev_(D)=v_(C)cos60^(@)=(v_(C))/(2)=` half of `v_(C)`
`therefore(K_(T))_(D)=(1)/(4)(K_(T))_(C)`
At point A
`U=3mgh`
`K_(R)=0impliesK_(T)=0`
`therefore` mechanical energy `E=3mgh=` constant
At point B
`U=0impliesE=3mgh`
`thereforeK=E=3mgh`
but `(K_(R))/(K_(T))=(2)/(5)`
`thereforeK_(R)=(2)/(7)K=(6)/(7)mgh`
`K_(T)=(5)/(7)K=(15)/(7)mgh`
At point C
`U=mgh`
`K_(R)=(K_(R))_(B)=(6)/(7)mgh`
`K_(T)=E-U-K_(R)`
`=3mgh-mgh-(6)/(7)mgh=(8)/(6)mgh`
At point D
`K_(R)=(K_(R))_(B)=(6)/(7)mgh`
`K_(T)=(1)/(4)(K_(T))_(C)=(2)/(7)mgh`
`U=E-K_(R)-K_(T)`
`=3mgh-(6)/(7)mgh-(2)/(7)mgh`
`=(13)/(7)mgh`
At point E
`U=0`
`K_(R)=(K_(R))_(B)=(6)/(7)mgh`
`K_(T)=E-K_(R)-U`
`=3mgh-(6)/(7)-0=(15)/(6)mgh`