Posted by **deb** on Thursday, March 11, 2010 at 9:47am.

A force of 10 pounds stretches a spring 2 inches. Find the work done in stretching this spring 3 inches beyond its natural length

- calculus/physics -
**FredR**, Thursday, March 11, 2010 at 6:49pm
Assuming an ideal spring (linear) and using Hooke's law, the force generated by the spring is:

F = -kx (1)

where x is the distance the spring is stretched from the rest position. So, substituting the info into (1) you get:

-10 = -k(2)

or

k = 5 lb/in

work is the 1st integral of force with respect to x, which would be:

W = (1/2)kx^2

= (1/2)5(3)^2

= 22.5 in-lb

## Answer This Question

## Related Questions

- Calculus/Physics - A spring has a natural length of 10in. An 800lb force ...
- Pre-calculus - Hooke's Law states that the distance a spring will stretch beyond...
- work force - a force of 10 pounds is required to stretch a spring of 4 inches ...
- Math - Hooke's Law states that the distance a spring will stretch beyond its ...
- Calculus - A force of 6 pounds is required to hold a spring stretched 0.2 feet ...
- Integral Calculus - A force of 4 pounds is required to hold a spring stretched 0...
- math - The length a spring stretches is directly proportional to the force ...
- algebra - The length a spring stretches is directly proportional to the force ...
- Calculus - Work of 1 joules is done in stretching a spring from its natural ...
- Algebra 1 - The amount a spring will stretch S is proportional to the force (or ...

More Related Questions