A uniform solid cylinder of mass m and radius R is released from rest on a rough plane inclined at angle θ to the horizontal, with sin θ=0.6. It…

For pure rolling of a solid cylinder, $I=\tfrac12 mR^2$, so $a=\dfrac{g\sin\theta}{1+I/mR^2}=\dfrac{0.6\,g}{1.5}=0.4\,g$, which makes (A) correct. Newton's law along the plane gives $mg\sin\theta-f=ma$, hence $f=0.6mg-0.4mg=0.2\,mg$, so (B) is correct. The normal force is $mg\cos\theta=0.8\,mg$, so the minimum friction coefficient is $f/N=0.2/0.8=0.25$, not $0.5$; (C) is wrong. Rolling kinetic energy splits in the ratio of $\tfrac12 mv^2:\tfrac12 I\omega^2=1:\tfrac12$, so the rotational part is $\tfrac{1/2}{3/2}=\tfrac13$ of the total energy, which equals $mgh$ because static friction does no work; (D) is correct. Answer: A, B, D. Note that friction here is static and does no work, which is why mechanical energy is conserved even though friction is essential to the motion.