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

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SUMMARY

The discussion focuses on calculating the work done by an ideal spring when a 3.0 kg mass is suspended from it and subsequently moved upward. Using Hooke's law and the work-energy principle, the work done by the spring is derived from the formula W(spring) = 1/2kx(initial)^2 - 1/2kx(final)^2. The initial extension of the spring is 9.0 cm, and the final position after an upward force is applied is -3.0 cm, leading to a calculated work done of 36K, where K is the spring constant. The importance of determining the spring constant and converting units from centimeters to meters is emphasized for accurate calculations.

PREREQUISITES
  • Understanding of Hooke's Law and spring mechanics
  • Familiarity with work-energy principles in physics
  • Ability to perform unit conversions, specifically from centimeters to meters
  • Knowledge of potential energy concepts related to springs
NEXT STEPS
  • Calculate the spring constant using the mass and extension data
  • Explore the implications of potential energy in spring systems
  • Learn about the applications of Hooke's Law in real-world scenarios
  • Investigate the effects of varying spring constants on work done
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Students studying physics, particularly those focusing on mechanics and spring dynamics, as well as educators looking for practical examples of Hooke's Law and work-energy calculations.

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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