Understanding the Spring Constant: Solving a Practical Problem

[louis2014]

(Moderator note: moved from technical forums, so no template)

Does anyone know how to answer this question?

A 1 meter spring lies horizontally on a table. You hang it vertically being held by one of its ends. Because of the mass of the spring itself it now extends to 1.1 meter. You hang a holder of unknown mass and now it's 1.2 meters long. You decide to hang a 500 gram mass on the holder and now, it's 1.3 meters long.
What is the spring constant of the spring?

I tried to use K = mgx, with m=0.5kg, g=9.81m/s2, and x=1.3-1.1m but the answer was not correct.
Thanks!

 

[.Scott]

The stretch is going to be proportional to the mass hanging from the bottom.

The mass of the spring affects the top of the spring more than the bottom. In fact, it has it's full effect on the top and no effect on the bottom. On average, its effect is the same as half the spring mass.

So that first 0.1M stretch is the same as if the spring was massless and you were hanging half the actual spring mass off the end of the spring. Then the unknown mass gives you another 0.1M. And the known mass of 500 grams gives you 0.1M.

So right away, you should be able to compute the unknown masses.

 

[louis2014]

The stretch is going to be proportional to the mass hanging from the bottom.

The mass of the spring affects the top of the spring more than the bottom. In fact, it has it's full effect on the top and no effect on the bottom. On average, its effect is the same as half the spring mass.

So that first 0.1M stretch is the same as if the spring was massless and you were hanging half the actual spring mass off the end of the spring. Then the unknown mass gives you another 0.1M. And the known mass of 500 grams gives you 0.1M.

So right away, you should be able to compute the unknown masses.

Oh I see, I was reading the question as the 500g mass replaced the original mass when it really was added to it. Thanks for your help!