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A spring with a force constant of 4.2 x 10^4 N/m is initially at its equilibrium length.

(a) How much work must you do to stretch the spring 0.052 m?

Answer in Joules:

(b) How much work must you do to compress it 0.052 m?

Answer in Joules:
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Added Tue, 04 Aug '15
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1) first of all *same amount* of work is done either to stretch or compress the same spring a given amount as long as the deformation is in the "elastic range" of the spring. Obvously the force direction is opposite between a stretch and compression.

SPE = 1/2kx� {where k = 4.2e4, x = 0.052}
SPE = (0.5)(4.2e4)(5.2e-2)^2= 56.784 = 56.784 J ANS ANS
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