Work Done by Spring Homework: -1.2J, -0.52J, -0.26J, 0.52J, 1.2J

AI Thread Summary
The discussion focuses on calculating the work done by a spring when a 3.0 kg mass is hung from it and then moved upward. The initial extension of the spring is 9.0 cm, and the block is moved an additional 12 cm. Participants emphasize the need to apply Hooke's law and work equations correctly, noting that the spring is not at its natural equilibrium due to the mass. There is a suggestion to calculate the spring constant and convert measurements from centimeters to meters for accuracy. The conversation highlights the importance of understanding potential energy in the system when determining the work done by the spring.
Spartan Erik
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Homework Statement



An ideal spring is hung vertically from the ceiling. When a 3.0 kg mass hangs at rest from it the spring is extended 9.0 cm from its relaxed length. An upward external force is then applied to the block to move it upward a distance of 12 cm. While the block is being moved by the force, the work done by the spring is:

-1.2J, -0.52J, -0.26J, 0.52J, 1.2J

Homework Equations



Hooke's law: F(spring) = -kd
W(spring) = 1/2kx(initial)^2 - 1/2kx(final)^2
Wapplied = -Wspring (assuming stationary before and after displacement)

The Attempt at a Solution



I think I can plug in 9 as x(initial) and -3 as x(final), giving me:
W(spring) = 1/2k81 - 1/2k9 = 40.5k-4.5k = 36K
so W(spring) = 36K

Not sure what else to do yet
 
Last edited:
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You need to work out the spring constant and converting from cm to meters might help. Also the spring is not at its natural equilibrium point when the mass is hung. It already has some potential energy.
 
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