How can I calculate the compression of a support column under a heavier weight?

[kkmonte]

Homework Statement

A support column is compressed 2.46 x 10^ -4m under a weight of 5.42 X 10^5N. how much is the column compressed under a weight of 4.8 X10^6N?

Homework Equations

F=KX

The Attempt at a Solution

By using that equation, I get 542000 = .000246K



Solve for K, and I get a huge number, 2,203,252,032.52



Then by using that number as my new K, and solving for the compressed distance with the new weight, I get 0.0022 m.



Is that the right answer?



Similarly, I figured I could just make a ratio proportion of:



.000246....X

------------- = ----------------

542,000....4,800,000



Which also gives me an answer of 0.0022 m (2.2 X 10^ -3). Is this correct?

 

[PhanthomJay]

Yes, either method gives you the correct solution, as long as the column is still behaving elastically in accordance with Hooke's law.

 

[kkmonte]

Thanks Jay, when I first did this problem, I was looking at the answer of 2.2e-3 and shaking my head going how can this thing compress less when there is a heavier weight on it? However it wasn't after I was typing it in the computer that I realized that the first compression is 2.46e-4, and I went, duh. =)

Thanks for letting me know I'm correct.
Ken