Posted by **Mischa** on Monday, April 23, 2007 at 3:46pm.

A force of 250 N is required to stretch a spring 5m from rest. Using Hooke's law, F=kx, how much work, in joules, is required to stretch the spring 7m from rest?

The energy required scales with x^2.

Multiply 250 J by (7/5)^2.

- calc/physics -
**derek**, Wednesday, December 5, 2012 at 4:05pm
350 J

## Answer This Question

## Related Questions

- Physics - Hooke's law states that it takes a force equal to kΔx is required...
- physics - The force required to stretch a Hooke’s-law spring varies from 0 N to ...
- Physics - The force required to stretch a Hooke’s-law spring varies from 0 N to ...
- Physics - The force required to stretch a Hooke’s-law spring varies from 0 N to ...
- physics - The force required to stretch a Hooke’s-law spring varies from 0 N to ...
- Physics - The force required to stretch a spring varies directly with the amount...
- Physics - When a 2.00-kg object is hung vertically on a certain light spring ...
- Physics - When a 3.00-kg object is hung vertically on a certain light spring ...
- physics - When a 2.60-kg object is hung vertically on a certain light spring ...
- science - It takes 3.38 J of work to stretch a Hooke’s-law spring 5.63 cm from ...

More Related Questions