By mechanical energy conservation, kinetic energy converts to spring potential energy: $\frac{3}{4}mv^2 = \frac{1}{2}kx^2$. Substituting the values gives $x = 0.6\text{ m}$.

The acceleration of a rolling body depends on its moment of inertia ratio $I/mR^2$. The solid cylinder has a smaller moment of inertia ratio ($1/2$) compared to the hollow cylinder ($1$), giving it a higher acceleration and causing it to reach the bottom first.

Acceleration of the centre of mass is given by $a = frac{F}{m} + frac{tau}{I}R_{text{eff}}$. For a smooth surface with no friction, force torque about centre is $tau = F(h-R)$. Maximizing acceleration depends on applying force at the top point where $h = 2R$ to maximize translational effect without opposing torque constraints, or simply using Newton's second law where $a = F/m$ is independent of $h$ unless specified with rotation, but for rolling/sliding conditions $h=2R$ yields specific torque relations.

Using conservation of mechanical energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For a solid sphere ($I = \frac{2}{5}MR^2$), solving yields $v = \sqrt{\frac{10}{7}gh}$.

Acceleration of a rolling body is given by $$a = \frac{g \sin\theta}{1 + I/MR^2}$$. Since the solid cylinder has a smaller moment of inertia ratio than the hollow cylinder, its acceleration is greater, so it reaches the bottom first.

The acceleration on an inclined plane is inversely proportional to $1 + I/MR^2$. Solid sphere has the lowest moment of inertia coefficient ($2/5$), giving it maximum acceleration and shortest time to reach the bottom.