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Found in: Page 202

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# What is the spring constant of a spring that stores ${\mathbf{25}}{\mathbf{\text{â€‰J}}}$of elastic potential energy when compressed by ${\mathbf{7}}{\mathbf{.5}}{\mathbf{\text{â€‰cm}}}$?

Spring constant is,$k=8.9Ã—{10}^{3}\text{â€‰N/m}$ .

See the step by step solution

## Step 1: Given

Elastic potential energy$U=25\text{â€‰J}$

Compressed length$x=7.5\text{â€‰cm}=7.5Ã—{10}^{âˆ’2}\text{â€‰m}$ .

## Step 2: To understand the concept

The problem is based on the concept of elastic potential energy. It is energy stored as a result of applying a force to deform an elastic object. By using the concept of elastic potential energy, we can find the spring constant.

Formula:

${\mathbit{U}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbit{k}}{{\mathbit{x}}}^{{\mathbf{2}}}$

## Step 3: Calculate the spring constant of the spring

Elastic potential energy can be written as

$\begin{array}{c}U=\frac{1}{2}k{x}^{2}\\ k=\frac{2U}{{x}^{2}}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}k=\frac{2Ã—25\text{â€‰J}}{{\left(7.5Ã—{10}^{âˆ’2}\text{â€‰m}\right)}^{2}}\\ k=8.9Ã—{10}^{3}\text{â€‰N/m}\end{array}$

Hence the spring constant is, $k=8.9Ã—{10}^{3}\text{â€‰N/m}$.

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