1. ## Work Problem?

A force of 7 pounds is required to hold a spring stretched 6 in. beyond its natural length. How much work is done in stretching the spring from its natural length to 8 in. beyond its natural length? I know work=Integral(kx). I've tried all the way I know how..I know you change the bounds to feet (Bounds are 0 to 2/3 ft.) , because work is in Ft-lbs. I need the number for K.

Do I use W=F*D, and plug in the work of 7lbs?
Not really sure where to go, and I have a test on this tomorrow.

2. Originally Posted by Bracketology
A force of 7 pounds is required to hold a spring stretched 6 in. beyond its natural length. How much work is done in stretching the spring from its natural length to 8 in. beyond its natural length? I know work=Integral(kx). I've tried all the way I know how..I know you change the bounds to feet, because work is in Ft-lbs. I need the number for K.

Do I use W=F*D, and plug in the work of 7lbs?
Not really sure where to go, and I have a test on this tomorrow.
you begin by using Hooke's law: F = kx, F is the force needed to stretch the spring x feet from its natural length. k is the spring constant. it is the same k in the integral formula for work.

can you continue now?

3. Yes! Thanks. I didn't have that formula written down.