A uniform solid sphere of mass m and radius R is released from rest on a rough fixed incline of angle θ with sin θ=0.6 and rolls down without…

For rolling without slipping $a=R\alpha$ and $I=\tfrac25mR^2$. Along the incline $mg\sin\theta-f=ma$ and torque about the centre gives $fR=I\alpha$, so $f=\tfrac25ma$. Hence $a=\tfrac{5}{7}g\sin\theta=\tfrac57(10)(0.6)=30/7\ \mathrm{m\,s^{-2}}$, so (A) is correct. Then $f=\tfrac25m\cdot a=\tfrac27mg\sin\theta$, directed up the incline because it must supply the torque that spins the sphere, so (B) is correct. The normal force is $mg\cos\theta$, so $\mu_{min}=f/N=\tfrac27\tan\theta=\tfrac27\times\tfrac34=\tfrac{3}{14}$, not $2/7$; (C) is wrong (it forgets the $\tan\theta$ factor). Finally the rotational and translational energies are $\tfrac12I\omega^2=\tfrac15mv^2$ and $\tfrac12mv^2$, whose ratio is $2/5$, not $2/7$ ($2/7$ is the rotational fraction of the total energy); (D) is wrong. Answer: A and B.