Work done stretching a spring

Hello mmattson07 Quote:

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**mmattson07** A spring has a natural length of 28 cm. If a 28 N force is required to keep it stretched to a length of 31 cm, how much work W is required to stretch it from 28 cm to 38 cm?

W= ___ J

I know Work= Force * Distance but i dont know how to go about solving this i know it will be an integration of some sort

Two formulae you need to know about an elastic spring with natural length $\displaystyle l$ and modulus $\displaystyle \lambda$:

- When the extension in the spring is $\displaystyle x$, the tension, $\displaystyle T$, is given by $\displaystyle T=\frac{\lambda x}{l}$

- When the spring is stretched from an extension $\displaystyle x_1$ to an extension $\displaystyle x_2$, the work done, $\displaystyle W$, is given by $\displaystyle W=\frac{\lambda({x_2}^2-{x_1}^2)}{2l}$

So here, $\displaystyle l = 0.28$ m. And when $\displaystyle x = 0.03$ (because the total length is $\displaystyle 0.31$ m), $\displaystyle T = 28$.

So, using the first of the two formulae above:

$\displaystyle 28 = \frac{0.03\lambda}{0.28}$

$\displaystyle \Rightarrow \lambda = \frac{28^2}{3}$

And we need to find $\displaystyle W$ when $\displaystyle x_1=0$ and $\displaystyle x_2 = 0.10$

So $\displaystyle W = \frac{28^2(0.1^2-0^2)}{3\times 2 \times 0.28}$

$\displaystyle = \frac{28}{6}=4.67$ Joules

Grandad