A spring of negligible mass and spring constant k is attached to the wall on one end, and the other end is attached to the center of mass axis of a solid sphere of mass m and radius r. The spring is stretched a distance A from equilibrium and let go. The sphere rolls freely on a smooth surface. What is the linear kinetic energy when the sphere is at the equilibrium position?

Question

Quizplus · Accepted Answer

When the spring is released from a distance A from equilibrium, its potential energy is given as:
$U_s = \frac{1}{2}kA^2$
When the sphere reaches the equilibrium position, the spring is unstretched and hence its potential energy is zero. At this point, the total mechanical energy of the system is equal to the kinetic energy of the sphere. The total mechanical energy of the system is given as:
$E_{tot} = U_s + K$
where $K$ is the kinetic energy of the sphere. When the sphere reaches the equilibrium position, all the potential energy of the spring is transferred to the kinetic energy of the sphere. Hence:
$E_{tot} = \frac{1}{2}mv_{CM}^2$
where $v_{CM}$ is the velocity of the center of mass of the sphere.
The potential energy of the spring at the equilibrium position is zero. Hence, the total mechanical energy of the system is equal to the kinetic energy of the sphere at the equilibrium position, which is given by:
$E_{tot} = \frac{1}{2}mv_{CM}^2 = \frac{1}{2}kA^2$
Solving for $v_{CM}$, we get:
$v_{CM} = \sqrt{\frac{kA^2}{m}}$
The linear kinetic energy of the sphere is given by:
$K = \frac{1}{2}mv_{CM}^2 = \frac{1}{2}kA^2$
Substituting the given values, we get:
$K = \frac{1}{2}(m)(\frac{kA^2}{m}) = \frac{1}{2}kA^2$
Therefore, the correct answer is B.

A Spring of Negligible Mass and Spring Constant K Is