Question

neglecting friction and the mass of the spring, what is the change in height ($\delta h$) of the sphere?
a 0 m
b $\infty$ m
c $\infty$ m
d 0 m

Explanation:

Step1: Apply conservation of mechanical energy

Since there is no friction and spring - mass is neglected, the total mechanical energy is conserved. The initial mechanical energy is the elastic potential energy of the spring $U_{s}=\frac{1}{2}kx^{2}$ (where $k$ is the spring constant and $x$ is the compression of the spring), and when the spring is released, this elastic potential energy is converted into gravitational potential energy $U_{g}=mgh$ of the sphere. Let the initial height of the sphere be $h_1$ and the final height be $h_2$, so $\Delta h=h_2 - h_1$.
The elastic potential energy of the spring is converted into gravitational potential energy of the sphere: $\frac{1}{2}kx^{2}=mg\Delta h$.
However, if we assume the spring is on a horizontal surface (from the context of the problem - no information about an inclined - plane or non - horizontal setup), and the spring is just pushing the sphere horizontally, there is no change in the vertical height of the sphere.

Step2: Determine the change in height

The force exerted by the spring is horizontal and there is no vertical force component (neglecting any non - existent vertical forces due to the setup). So, $\Delta h = 0$.

Answer:

A. 0 m