Question

A spring of force constant $800N/m$ has an extension of $5cm$. Find work done in extending it from $5cm$ to $15cm$.

Answer 1

Step1: Given data

1. The spring constant of the spring is $K=$$800N/m$.
2. The initial extension of the spring is $x_1=5cm=0.05m.$
3. The final extension of the spring is $x_2=15cm=0.15m.$

Step2: Work done on a spring

1. The work done by a spring is the total amount of energy to extend the spring from one position to another.
2. The force exerted on a spring due to extension is proportional to the amount of extension, i.e, $F\propto x$, where, x is the amount of extension.
3. The work done by a spring is defined by the form, $U=\frac{1}{2}K\left({x_2}^2-{x_1}^2\right)$, where, K is the spring constant, $x_1$ is the initial, and $x_2$ is the final extension of the spring.

Step3: Diagram

Step4: Finding the work done

As we know, work done in spring due to extension is $U=\frac{1}{2}K\left({x_2}^2-{x_1}^2\right)$.

So,

$$\begin{gathered} U=\frac{1}{2}K\left({x_2}^2-{x_1}^2\right)=\frac{1}{2}\times 800\times \left[{\left(0.15\right)}^2-{\left(0.05\right)}^2\right] \\ orU=400\times \left[0.0225-0.0025\right] \\ orU=8joule. \end{gathered}$$

Therefore, find work done in extending the spring from $5cm$ to $15cm$ is $8joule.$