Work Problem w/ spring

**mmattson07** · September 18th 2009, 10:30 PM

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

**redsoxfan325** · September 18th 2009, 10:44 PM

This might be better suited for a physics forum. If you can set up the integral, I or someone else can help you, but other than that you might not get the help you're looking for.

**Grandad** · September 19th 2009, 12:16 AM

Hello mmattson07

Originally Posted by 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 $l$ and modulus $\lambda$ :

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

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

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

So, using the first of the two formulae above:

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

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

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

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

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

Grandad

**mmattson07** · September 19th 2009, 09:08 AM

We never finished the example we were doing in class so I was unsure on how to set up the integral...i know it has something to do with Hooke's Law. Thank you for the help though Grandad's method works and i follow it.

**Grandad** · September 19th 2009, 12:45 PM

Hello mmattson07

Originally Posted by mmattson07

We never finished the example we were doing in class so I was unsure on how to set up the integral...i know it has something to do with Hooke's Law. Thank you for the help though Grandad's method works and i follow it.

My 'method', as you describe it, is simply the result of applying Hooke's Law, which states that the tension in an elastic spring (or string) is proportional to the extension. In other words, with $T$ and $x$ defined as before:

$T = kx$ , for some constant $k$ .

Now if we define the modulus (or modulus of elasticity), $\lambda$ , to be the force required to double the length of the spring, then when $x = l, T = \lambda$ . So $k = \frac{\lambda}{l}$ , which in turn gives the first of the two formulae that I quoted, namely

$T = \frac{\lambda x}{l}$

The second formula is derived by integration, as follows:

Suppose the spring is extended from $x$ to $x+\delta x$ . Then the work done is given by

$\delta W \approx T\delta x$

$\Rightarrow \frac{dW}{dx}=T$

$\Rightarrow \int_0^W dW = \int_{x_1}^{x_2}Tdx$ , where $W$ is the work done extending the spring from $x_1$ to $x_2$ .

$=\int_{x_1}^{x_2}\frac{\lambda x}{l}dx$

$\Rightarrow W = \frac{\lambda({x_2}^2-{x_1}^2)}{2l}$

So there's nothing other than Hooke's Law here. However, it's much easier to learn and apply these two well-known formulae, than to derive the answers from first principles each time!

Grandad

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