Question

A spring of force constant 800 n/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is

Answer 1

check the attachment

Answer 2

Answer:

A spring of force constant 800 n/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is $8\ J$

Explanation:

• Elastic potential energy arises as a result of the deformation of the elastic objects such as the stretching of the spring.

• In the spring, work done is stored as elastic potential energy

• The expression of work done is given by,

        $\mathrm{W}=\frac{1}{2} \times \mathrm{k}\left(\mathrm{x}_{2}^{2}-\mathrm{x}_{1}^{2}\right)$       ...(1)

• Where k is the spring constant $x_1$ and $x_2$ is the original and extended length of the spring.

Given,

$\begin{array}{l}\mathrm{x}_{1}=5 \mathrm{~cm} = 0.05\ m \\\mathrm{x}_{2}=15 \mathrm{~cm} = 0.15\ m\end{array}$

$\mathrm{k}=800 \ N/m$

Substitute these values into equation(1)

$\mathrm{W}=\frac{1}{2} \times 800\left(0.15^{2}-0.05^{2}\right)=\frac{1}{2} \times 800\times 0.02 = 8 \ J$