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Spring Conservation of Energy

  • #1
1. Problem Statement:
A vertical spring has one end attached to the ceiling and a 3kg bag attached to the other one. When the system is at rest, the spring is stretched by 40cm. 1) determine the spring constant. 2) Let the bag drop from a position in which the spring is not deformed. Using the conservation of energy law, find: how fast the bag is moving after it drops 40 cm, and how far down the bag will drop before starting to come back. Calculate and show the direction of the bag's acceleration when the bag is at the highest, lowest, and middle of the oscillation positions.

Homework Equations

:[/B]
K=mg/x
KE=1/2m(v^2)
GPE=mgh
EPE=1/2k(x^2)

3. Attempt at the Solution:
So I found a spring constant. I'm not sure if it's correct but I got, k=(3)(9.8)/(.4)= 73.5 N/m. I ran into problems with the rest though. For finding the velocity of the bag dropping 40cm, I did GPE=KE+EPE but I'm not sure if that's right. As for the farthest distance the bag would drop and the acceleration, I wasn't exactly sure how to approach them.

Update: To find the velocity at 40cm, should it actually only be GPE=KE because we're treating the 40cm as our equilibrium? And then to find the farthest distance the bag would drop would it be GPE=EPE to find our displacement? Or am I completely wrong?
 
Last edited:

Answers and Replies

  • #2
haruspex
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GPE=KE+EPE
GPE lost = KE gained + EPE gained, yes.
to find the farthest distance the bag would drop would it be GPE=EPE to find our displacement
Yes, but what is the GPE lost in this case?
 
  • #3
Yes, but what is the GPE lost in this case?
Would it be (3)(9.8)(d) because we don't know the distance?
 
  • #4
haruspex
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Would it be (3)(9.8)(d) because we don't know the distance?
Right.
 

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