A 6.25 kg solid ball with a radius of 0.185 m rolls without slipping at 3.55 m/s. What is its total kinetic energy?

Answer 1

It consists of two parts:
(a) The translational KE,
(1/2) M V^2,
which is due to the motion of the center of mass at velocity V = 3.55 m/s, and
(b) The rotational KE,
(1/2)*I*w^2
where w is the angular velocity, V/R, and I is the moment of inertia, which for a solid sphere is (2/5)M*R^2

Put then all together and you get
KE(total) = (1/2 + 1/5) M V^2
= (7/10)*M V^2