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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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?
There are two (slightly) different ways to look at this problem. The first is to state that the work done on the spring is equal to the change in the spring potential energy:Quote:
Originally 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
