Atmospheric pressure and the spring constant

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SUMMARY

The discussion centers on calculating the work done by atmospheric pressure on a spring in a piston-cylinder system. The spring constant is 4300 N/m, and the piston radius is 0.017 m. The atmospheric pressure compresses the spring by 0.0214 m, resulting in a force of 91.995 N. While the initial calculation of work done was 1.969 J, it was clarified that due to the dynamics of the system, the effective work done by the atmosphere is approximately 2 J, with only about 1 J being stored in the spring.

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rasputin66
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Homework Statement


I already found A, but I keep getting wrong answers for B.

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Friction is absent. The spring constant of the spring is 4300 N/m. The piston has a negligible mass and a radius of 0.017 m. (a) When the air beneath the piston is completely pumped out, how much does the atmospheric pressure cause the spring to compress? (b) How much work does the atmospheric pressure do in compressing the spring?

k= 4300 N/m
r= 0.017 m
A= 0.00091 m^2
P= 101 kN/m
F= 91.995 N

Homework Equations


P= F/A
W= F d

The Attempt at a Solution



I found the answer to A which is 0.0214 m. So then I tried to use the Work equation and got 1.969 J. But this is wrong. I'm stuck! Please help!
 
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Please post your working.
 
OK, well the spring moved 0.0214 m and the F involved is 91.995 N.
W = F d
W = (91.995 N) (0.0214 m)= 1.968693 J
The homework program on the computer is telling me this is wrong. I don't know why. Do you think this is OK? Sometimes the program is mistaken.
 
That all seems correct, though my numbers are slightly different: .02133m and 1.956J.
But I wonder if the problem setter has made a mistake here. If you calculate the work done on the spring, kx2/2, you will only get half as much.
Imagine all the air being drawn out very quickly. At first, the force from the atmospheric pressure will be much more than the compressive force in the spring. Since all masses are taken to be zero, this will result in infinite acceleration! When that sort of thing happens to equations you have to relax your assumptions to reintroduce some sanity. In this case, allow the piston to have some small mass.
Now the piston will acquire KE and momentum. At the point where the spring has reached the equilibrium compression you calculated, 0.0213m, half of the work done by the atmosphere is in the form of KE of the piston. Note that this is true no matter how light we make the piston. In a real environment, the spring will then oscillate, losing energy to friction, asymptotically converging to the 0.0213 displacement.
In short, the work the atmosphere does will be about 2J, but the work then stored in the spring will be only about 1J.
 

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