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:
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.
you are right up until where you set up the integral. so something must have happened after that
Originally Posted by fogel1497
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.