# A 0.150 block moving vertically upward collides with a light vertical spring and compresses it...

## Question:

A 0.150 block moving vertically upward collides with a light vertical spring and compresses it 4.50 before coming to rest. If the spring constant is 54.0 , what was the initial speed of the block?

## Law of conservation of energy:

The law states that the energy cannot be generated or destroyed and the energy is always the same inside the system it could be converted from one form to another. It is used to convert mechanical energy into kinetic energy.

Given Data

• The mass of the block is {eq}m = 0.150\;{\rm{g}} {/eq}.
• The compression in the spring is {eq}C = 4.5\;{\rm{cm}} = 0.045\;{\rm{m}} {/eq}.
• The spring constant is {eq}k = 54\;{\rm{N/m}} {/eq}.

The expression for spring energy is given by.

{eq}{E_s} = \dfrac{1}{2}k{C^2} {/eq}

The expression of the kinetic energy is given by.

{eq}K.E = \dfrac{1}{2}m{v^2} {/eq}

Apply conservation of energy then the equation of energy is given by.

{eq}\begin{align*} \dfrac{1}{2}k{C^2}& = \dfrac{1}{2}m{v^2}\\ {v^2}& = \dfrac{{{C^2}k}}{m}\\ v& = C\sqrt {\dfrac{k}{m}} \end{align*} {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} v &= \left( {0.045} \right)\sqrt {\dfrac{{\left( {54} \right)}}{{\left( {0.150} \right)}}} \\ v &= 0.853\;{\rm{m/s}} \end{align*} {/eq}

Thus, the speed of the block is {eq}0.853\;{\rm{m/s}} {/eq}. 