A particular spring has a force constant of 2.5 X 103 N/m.

(a) How much work is done in stretching the initially-relaxed spring by 6.0 cm?

(b) How much more work is done in stretching the spring an additional 2.0 cm?

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- Jun 29th 2006, 09:42 AM #1

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- Jun 30th 2006, 04:43 AM #2Originally Posted by
**Candy**

$\displaystyle W = S - S_0 = (1/2)kx^2-(1/2)kx_0^2$

The other is to consider that the work done is the integral of the force over the infinitesimal displacements:

$\displaystyle W = \int_{x_0}^x dx \, kx$

which, of course, gives the same formula as above.

Either way you want to look at it:

a) The spring is initially relaxed, so it's at its equilibrium position x0 = 0 m.

The spring is stretched 0.060 m. So the final state of the spring has x = 0.060 m. (ALWAYS use m for this!) Thus:

$\displaystyle W = (1/2)kx^2-(1/2)kx_0^2$ = 4.5 J

b) We wish to find how much extra work is done, so use x0 = 0.060 m and x = 0.080 m.

$\displaystyle W = (1/2)kx^2-(1/2)kx_0^2$ = 3.5 J

-Dan