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:
Then I did F=KX (hooke's law) to solve for the k constant of the spring
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.
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, when its stretched 10 inches, we see that
That should be the desired result.
Does this make sense?