An ideal spring is hung vertically from the ceiling. When a 2.0 kgmass hangs at

Question

An ideal spring is hung vertically from the ceiling. When a 2.0 kgmass hangs at rest from it the spring is extended 6.0 cm from itsrelaxed length. A downward external force is now applied to themass to extend the spring an additional 10 cm. While the spring isbeing extended by the force, the work done by the spring is:

a) -3.6 J
b) -3.3 J
c) -3.4e-5 J
d) 3.3 J
e) 3.6 J

I know that Ws = -Us I am havingdifficulty calculating k. When I use Equilibrium and Newton's lawsI get k=333.3N/m. When I use Ws = Wg I get k=666.6N/m I know they can not both be true - what am I missing? (Ithink 333.3N/m is right and that makes A correct - am I right?)

Explanation / Answer

2 * 9.81 = 0.06k --> k = 327 Nm--(1) 327 * 0.06^2 / 2. 327 * 0.16^2 / 2 The additional work W done on the spring is: W = 327(0.16^2 - 0.06^2) / 2 Substituting for k from (1): W = 327(0.16^2 - 0.06^2) / 2 = 3.6 J.

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