Multiple problems dealing with springs and Hooke's Law

  • Thread starter TmrK
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  • #1
TmrK
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Homework Statement


A vertical ideal spring is mounted on the floor and has a spring constant of 132 N/m. A 0.80-kg block is placed on the spring in two different ways. (a) In one case, the block is placed on the spring and not released until it rests stationary on the spring in its equilibrium position. Determine the amount (magnitude only) by which the spring is compressed. (b) In a second situation, the block is released from rest immediately after being placed on the spring and falls downward until it comes to a momentary halt. Determine the amount (magnitude only) by which the spring is now compressed.

Homework Equations


Fapplied=kx


The Attempt at a Solution


First part is solved with an answer of 0.0594m. Second part has me lost.

EDIT: Solved problem.
 
Last edited:

Answers and Replies

  • #2
technician
1,506
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You have the first part with no trouble. The mass has a weight of 0.8 x 9.81 =7.84N and you have used the spring constant value correctly to calculate the extension, 0.059m.
You now need to think about ENERGY. When the weight is PLACED on the spring the force on the spring increases uniformly from 0N up to 7.84N.
(I hope that you can picture the straight line graph of F against extension)
The work done on the spring (elastic PE stored) = AVERAGE FORCE x distance.
The average force is 0.5F and so the energy stored is 0.5F x extension (area under the straight line graph?... you should be familiar with this equation)
When the weight is dropped (released) the force on the spring does not start at 0 and increase to 7.84N uniformly. It is 7.84N from the start so the PE given up by the weight is not 0.5F x ext, it is F x ext, but at this point the energy stored in the spring is only 0.5F x ext..... there is some extra energy that needs accounting from.... think KE
Can you take it from there?
 

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