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is more work required for this spring? |
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| Jan11-06, 10:20 PM | #1 |
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is more work required for this spring?
A spring with a force constant of 50 N/m is stretched continuously from x0=0 to x2=20 cm. Is more work required to stretch the spring from x1=10 cm to x2=20 cm than from x0 to x1? Justify your answer mathematically.
I thought the work required is the same because the displacements are equal for both work equations... but I think I'm wrong. |
| Jan11-06, 10:25 PM | #2 |
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work = 1/2 kx^2
so then the difference would be greater between 20cm and 10cm w= EPEf-EPEi = 0.5(k)400-0.5(k)100 = 150k w0->1 = 0.5(k)(100) - 0.5k(0) = 50k |
| Jan11-06, 10:27 PM | #3 |
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but wouldn't it be:
work1 = 1/2 k (10cm)^2 work2 = 1/2 k (20cm - 10cm)^2=work1 ? |
| Jan11-06, 10:57 PM | #4 |
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is more work required for this spring?
no, the work is the difference between energies.
[tex] EPE_i + W = EPE_f [/tex] [tex] W = EPE_f - EPE_i [/tex] [tex] W = 0.5kx_f^2 - 0.5kx_i^2 [/tex] |
| Jan11-06, 11:08 PM | #5 |
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I remember now...
what does EPE mean? |
| Jan11-06, 11:18 PM | #6 |
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elastic potential energy
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