Very simple - mass on a spring

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  • #1

Homework Statement



The elastic constant of a spring holding an object in equilibrium is 600 N/m. Find the extension of the spring with respect to its initial length if the object has a mass of 480 g.


Homework Equations



hooke's law: F = -ky


The Attempt at a Solution



I know that the answer is supposed to be 0.0125m, but I'm not sure as to how to arrive at this solution. I'm also a bit unclear as to how everything fits together (for instance, does initial length refer to the equilibrium position?), as I'm new to this concept. Could anyone please give a simple, straight-forward walk-through? Thanks!
 

Answers and Replies

  • #2
ehild
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The spring is vertical and the object hangs on it. The spring holds the object and prevents it from falling down. So what force exerts the spring on the object of mass m?

ehild
 
  • #3
Does the spring exert a force of -ky on the mass? At equilibrium shouldn't this be equal to mg? How does this relate to the answer to the problem? If I set them equal and solve for y, I don't get the proper answer, so it seems that I'm missing something.
 
  • #4
ehild
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The magnitude of the spring force is kΔL and it acts upward. ΔL is the extension of the spring. Gravity is a downward force, mg. The forces have to cancel: mg=kΔL. What result did you get?


ehild
 
  • #5
ΔL = (mg)/k = (0.48*9.8)/600 = 0.00784 m

but this doesn't match up with the given solution of 0.0125 m :confused:

Is 0.00784 the equilibrium position, and I am supposed to find the max distance past that point, perhaps? The question seems a bit vague to me, so I don't have a good grasp on what exactly I am looking for and how the equilibrium position fits in.
 
  • #6
ehild
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I think your answer is correct and the given solution is wrong. I can not understand the problem in a different way: The spring holds the object in equilibrium, given the force constant and the mass of the object.
It happens quite often that wrong solutions are given. Do not worry about it.

ehild
 

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