Posted by **COFFEE** on Friday, June 29, 2007 at 11:00pm.

A force of 27N is required to maintain a spring stretched from its natural length of 12cm to a length of 15cm. How much work is done in stretching the spring from 15 to 25cm?

and this is what i did.. please check to see if i did it correctly.. thanks :)

f=27N

12cm to 15cm

f(x) = kx, k=spring constant

15-12cm = 3cm = 0.03cm

f(0.03) = 27N

0.03k = 27

k = 27/0.03 = 900

f(x) = 900x

w = integral of 0.03 to 0.13 for: 900x dx

w = 900*((x^2)/2)] from 0.03 to 0.13

w = 450[((0.13)^2)-((0.03)^2)]

w = 7.2J

Ok on k, 900 N/m

Ok on the next.

thanks :)

## Answer this Question

## Related Questions

- work force - a force of 10 pounds is required to stretch a spring of 4 inches ...
- Sorry, another math problem please - Hooke's Law asserts that the force required...
- Integral Calculus - A force of 4 pounds is required to hold a spring stretched 0...
- Calculus/Physics - A spring has a natural length of 10in. An 800lb force ...
- phy - A string of natural length L extends to a new length L' under tensile ...
- Calculus - Work of 1 joules is done in stretching a spring from its natural ...
- AP calc - A spring has a natural length of 18 cm. If a 22-N force is required to...
- Physics - Hooke's law states that it takes a force equal to kΔx is required...
- Physics - I think that for energy of a spring I should use (1/2)(k)(x^2) and (1/...
- physics help! - Two springs P and Q both obey Hooke’s law. They have spring ...