Thread: Integrals and work

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.

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:

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

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:



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

So we see that , where is the spring constant.

Thus,

Thus, when its stretched 10 inches, we see that

That should be the desired result.

Does this make sense?

--Chris