# A spring with a natural length of 8 cm measures 12 cm after the 150-N weight is attached. Find...

## Question:

A spring with a natural length of 8 *cm* measures 12 *cm* after the 150-*N* weight is attached. Find the work required to stretch the spring 6 *cm* from its natural length.

## Spring:

Spring is a mechanical component that is generally used in the on-road vehicle to absorb different-different kinds of shocks. The total force needed to stretch the spring is equivalent to the product of the spring constant of the spring and the deformed length of the spring.

## Answer and Explanation: 1

**Given data**

- The natural length of the spring is {eq}{{l}_{1}}=8\ \text{cm}=8\times {{10}^{-2}}\ \text{m} {/eq}

- The measured length after applying force is {eq}{{l}_{2}}=12\ \text{cm}=12\times {{10}^{-2}}\ \text{m} {/eq}

- The total amount of force applied to the spring is {eq}F=150\ \text{N} {/eq}

- The stretched length at unknown work done is {eq}{{l}_{3}}=6\ \text{cm}=6\times {{10}^{-2}}\ \text{m} {/eq}

The deformed length of the spring from its original length is calculated as,

{eq}x={{l}_{2}}-{{l}_{1}}\Rightarrow x=\left( 12\times {{10}^{-2}}\ \text{m} \right)-\left( 8\times {{10}^{-2}}\ \text{m} \right)\Rightarrow x=4\times {{10}^{-2}}\ \text{m} {/eq}

By using the following relation, the spring constant of the spring is calculated as,

{eq}F=kx\Rightarrow \left( 150\ \text{N} \right)=k\left( 4\times {{10}^{-2}}\ \text{m} \right)\Rightarrow k=\dfrac{\left( 150\ \text{N} \right)}{\left( 4\times {{10}^{-2}}\ \text{m} \right)}\Rightarrow k=3750\ \text{N/m} {/eq}

By using the following integral, the work done needed to stretch the spring 6 cm from its natural length is calculated as,

{eq}W=\int\limits_{0}^{{{l}_{3}}}{Fdx}\Rightarrow W=\int\limits_{0}^{6\times {{10}^{-2}}\ \text{m}}{kxdx}\Rightarrow W=k\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{6\times {{10}^{-2}}\ \text{m}}\Rightarrow W=\left( 3750\ \text{N/m} \right)\left[ \dfrac{{{\left( 6\times {{10}^{-2}}\ \text{m} \right)}^{2}}}{2}-\dfrac{{{\left( 0 \right)}^{2}}}{2} \right]\Rightarrow W=6.75\ \text{N}\cdot \text{m} {/eq}

Thus, the work done to stretch the spring 6 cm from its natural length is 6.75 N meter.

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Chapter 4 / Lesson 19After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.