A$50g$ice cube can slide up and down a frictionless $30°$slope. At the bottom, a spring with spring constant $25N/m$is compressed $10cm$and used to launch the ice cube up the slope. How high does it go above its starting point?

Ice cube mass $=50g$

Energy stored in the spring k $=25N/m$

distance x $=10cm$

Mass potential energy at height h is $mgh$. Energy stored in the spring with spring constant k compressed by a distance x is $\frac{1}{2}k{x}^2$.A quantitative property that can be transmitted from an object so that it can do work.

At the bottom of the slope, energy stored in the spring is,

$$\begin{gathered} E=\frac{1}{2}k{x}^2 \\ E=\frac{1}{2}\left(25\right){\left(0.1\right)}^2 \\ E=0.125J \end{gathered}$$

When the ice cube reaches the top of the slope, at a height of h above the bottom, the energy will be transformed into PE,

$$\begin{gathered} E=mgh \\ h=\frac{E}{mg} \\ h=\frac{0.125}{0.05\times 9.81} \\ h=0.255m \end{gathered}$$

There is a conservation of energy. The ice cube's potential energy is converted to spring energy, which is then converted to PE. Since no friction exists between the slope and the ice cube, their energies are equal.