Integrals and work

**fogel1497** · Sep 14th 2008, 06:53 PM

Question: If the work required to stretch a spring 1 ft beyond its natural length is 9 ft-lb, how much work W is needed to stretch it 10 in. beyond its natural length?

So my teacher today gave us some physics equations and said to do this homework. Im a little stuck on this, usually they give us the force required to stretch a spring, not the work. Here is what i tried:

W=FD
so
9=F*1ft
F=9

Then I did F=KX (hooke's law) to solve for the k constant of the spring
so
9 = K * 1 ft
K = 9

So I now know that F = 9X (equation for force of a spring when you know the k constant of that spring)
And then i said the:
INTEGRAL of: 9x from 0-10/12 = Total work needed to stretch the spring 10inches.
I get 7.25, but this answer is wrong.

**Jhevon** · Sep 14th 2008, 07:05 PM

Originally Posted by fogel1497

Question: If the work required to stretch a spring 1 ft beyond its natural length is 9 ft-lb, how much work W is needed to stretch it 10 in. beyond its natural length?

So my teacher today gave us some physics equations and said to do this homework. Im a little stuck on this, usually they give us the force required to stretch a spring, not the work. Here is what i tried:

W=FD
so
9=F*1ft
F=9

Then I did F=KX (hooke's law) to solve for the k constant of the spring
so
9 = K * 1 ft
K = 9

So I now know that F = 9X (equation for force of a spring when you know the k constant of that spring)
And then i said the:
INTEGRAL of: 9x from 0-10/12 = Total work needed to stretch the spring 10inches.
I get 7.25, but this answer is wrong.

you are right up until where you set up the integral. so something must have happened after that

the work in ft-lb is given by:

$W = \int_0^{10/12}9x~dx = \frac 92x^2 \bigg|_0^{10/12}$

**fogel1497** · Sep 14th 2008, 07:22 PM

Your answer comes out to be 3.125, which my online homework tells me is an incorrect value.

**Chris L T521** · Sep 14th 2008, 07:38 PM

Originally Posted by fogel1497

Question: If the work required to stretch a spring 1 ft beyond its natural length is 9 ft-lb, how much work W is needed to stretch it 10 in. beyond its natural length?

So my teacher today gave us some physics equations and said to do this homework. Im a little stuck on this, usually they give us the force required to stretch a spring, not the work. Here is what i tried:

W=FD
so
9=F*1ft
F=9

Then I did F=KX (hooke's law) to solve for the k constant of the spring
so
9 = K * 1 ft
K = 9

So I now know that F = 9X (equation for force of a spring when you know the k constant of that spring)
And then i said the:
INTEGRAL of: 9x from 0-10/12 = Total work needed to stretch the spring 10inches.
I get 7.25, but this answer is wrong.

Here's a different approach:

$W=\int_a^b F(x)\,dx$

We are told that the distance stretched is 1 ft, and the work done is 9 ft-lbs.

So we see that $9=\int_0^1 kx\,dx$ , where $k$ is the spring constant.

Thus, $9=\left.\left[\frac{1}{2}kx^2\right]\right|_0^1\implies 9=\tfrac{1}{2}k\implies \color{red}\boxed{k=18}$

Thus, when its stretched 10 inches, we see that $W=18\int_0^{\frac{10}{12}}x\,dx=9\left.\bigg[x^2\bigg]\right|_0^{\tfrac{10}{12}}=9\left(\frac{10}{12}\ri ght)^2=\color{red}\boxed{6.25~ft-lb}$

That should be the desired result.

Does this make sense?

--Chris

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