# A 0.56kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from...

## Question:

A 0.56kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.13 to 0.26m (relative to its unstrained length), the speed of the sphere decreases from 5.45 to 4.90m/s. What is the spring constant of the spring? (The sphere hangs from the bottom of the spring.)

## Conservation of energy:

As we know energy can only be transferred from one body to another. So, the energy of the system remains the same but gets transferred into a different type. In the question, we used the concept of conservation of energy to find the spring constant.

## Answer and Explanation:

Given Data:

Mass of metal sphere {eq}m = 0.56\;{\rm{kg}} {/eq}

Spring stretches from {eq}0.13\;{\rm{m}}\;{\rm{to}}\;{\rm{0}}{\rm{.26}}\;{\rm{m}} {/eq}

Speed of spring {eq}5.45\;{\rm{m/s}}\;{\rm{to}}\;4.90\;{\rm{m/s}} {/eq}

By applying, conservation of energy in the system.

{eq}{\rm{Increase}}\;{\rm{in}}\;{\rm{spring}}\;{\rm{energy}} = \;{\rm{Decrease}}\;{\rm{in}}\;{\rm{kinetic}}\;{\rm{energy}} + {\rm{Decrease}}\;{\rm{in}}\;{\rm{gravitational}}\;{\rm{energy}} {/eq}

So, mathematically it can be written as:

{eq}\dfrac{1}{2}k\left[ {{{\left( {0.26} \right)}^2} - {{\left( {0.13} \right)}^2}} \right] = \dfrac{1}{2}m\left[ {{{\left( {5.45} \right)}^2} - {{\left( {4.90} \right)}^2}} \right] + mg\left( {0.25 - 0.13} \right) {/eq}

On further solving, we get:

{eq}\begin{align*} \dfrac{1}{2}k\left[ {{{\left( {0.26} \right)}^2} - {{\left( {0.13} \right)}^2}} \right] &= \dfrac{1}{2}m\left[ {{{\left( {5.45} \right)}^2} - {{\left( {4.90} \right)}^2}} \right] + mg\left( {0.25 - 0.13} \right)\\ \dfrac{1}{2}k\left[ {{{\left( {0.26} \right)}^2} - {{\left( {0.13} \right)}^2}} \right] &= \dfrac{1}{2} \times 0.56 \times \left[ {{{\left( {5.45} \right)}^2} - {{\left( {4.90} \right)}^2}} \right] + 0.56 \times 10 \times \left( {0.25 - 0.13} \right)\\ 0.025k &= 1.593 + 0.672\\ k &= \dfrac{{2.265}}{{0.025}}\\ k &= 90.6\;{\rm{N/m}} \end{align*} {/eq}

Thus, the spring constant of spring is {eq}108.4\;{\rm{N/m}} {/eq}.

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from ICSE Environmental Science: Study Guide & Syllabus

Chapter 1 / Lesson 6