Repeat all parts of above problem for `F=40N`

Explanation

(a). And (b). We have already calculated that, requirement of friction for pure rolling is
`f=(3)/(7)F`
for `F=40Nimpliesf=(120)/(7)N`
Now, this value of f is more than `mu_(S)N`, so backward slip `(altRalpha)` will take place because `f` is positive so kinetic friction or 10 N will act in forward direction.
(c).
The actual forces acting on the sphere are as under:
`a=(F_("net"))/(m)=(40+10)/(5)=10m//s^(2)`
`ane(a)/(R)`
but `alpha=(tau_("net"))/(I)=(FR-fR)/((2)/(5)mR^(2))`
`=(2.5(F-f))/(mR)=(2.5(40-10))/((5)(1))`
`=15rad//s^(2)`