# A 50 g ice cube can slide without friction up and down a 32 degrees slope. The ice cube is...

## Question:

A {eq}50 \ g {/eq} ice cube can slide without friction up and down a {eq}32 {/eq} degrees slope. The ice cube is pressed against the spring at the bottom of the slope, compressing the spring {eq}10.16 \ cm {/eq}. The spring constant is {eq}22 \ N/m {/eq}.

When the ice cube is released, what distance will it travel up the slope before reversing direction?

## Energy:

Energy is the ability to perform work, and its S.i unit is joule. From energy equations, Equation of motion can be derived. Energy is a scalar quantity. In this problem, the concept of conservation of energy is used to determine upward traveling distance.

Given

Mass of the ice cube {eq}m = 50\, {\rm{g}}\, {\rm {= 0}}{\rm{.05}}\, {\rm{Kg}} {/eq}

Slope {eq}\theta = 32^\circ {/eq}

Compression in spring up to {eq}x = 10.16\, {\rm{cm}}\, {\rm {= 0}}{\rm{.1016}}\, {\rm{m}} {/eq}

Spring constant {eq}K = 22\, {\rm{N/m}} {/eq}

From conservation of energy, potential energy of the spring is same as potential energy of the ice cube, so

Potential energy of the spring is

{eq}P.E = \dfrac{1}{2}k{x^2}.........\left (1 \right) {/eq}

At height h potential energy of the ice cube is

{eq}P.E = mgh {/eq}

Here {eq}h = l\sin \theta {/eq} so above expression becomes

{eq}P.E = mgl\sin \theta ...........\left (2 \right) {/eq}

Now from equation 1 and 2

{eq}\begin{align*} \dfrac{1}{2}k{x^2} &= mgl\sin \theta \\ 0.5 \times 22 \times {0.1016^2} &= 0.05 \times 9.81 \times l \times \sin 32^\circ \\ 0.1135 &= 0.259l\\ l &= 0.43\,{\rm{m}}\\ {\rm{l}} &= 43\, {\rm{cm}} \end{align*} {/eq}

Hence {eq}43\, {\rm{cm}} {/eq} ice will travel up the slope before reversing the direction.