Help! Troubleshooting Logic to Find Mass on Vertical Spring

AI Thread Summary
To find the mass that stretches a vertical spring by 6.0 cm with a spring constant of k=2.5 * 10^3 N/m, the correct approach is to use Hooke's law, F = kx. The force exerted by the mass (mg) equals the spring force, leading to the equation m * 9.8 = k * x. Substituting the values, m * 9.8 = 2500 * 0.06 results in a mass of approximately 15 kg. The initial method using work was incorrect for this problem.
superdave
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Okay,

I am messing something up here. I think it's my logic.

Given k=2.5 * 10^3 N/m

Find how much mass would have to be suspended from the vertical spring to move it 6.0 cm

I get W=1/2kx^2=.5(2500)*(.06m)^2=4.5J
I think that's right.

Then I do
W=F*d=m*a*d or m=W/a*d=4.5/(9.8m/s^2 / .06 m)=7.65 kg
But that's not the answer.

Help?
 
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Look up Hooke's law and the meaning of k.

Note: I assume the question is asking for the mass that would--if gently added to the spring, not dropped from a height--stretch the spring by the given amount.
 
Last edited:
Why are you using work, it would be much easier to use one of Hooke's laws;
F = - kx

Can you arrive at the correct answer using this?

[Edit] It seems Doc Al got there before me, damn he's so quick! :mad: [Edit]
 
Oh, so

F=kx
m*9.8=2500 * .06=150/9.8= 15 kg

I got it. Man, I sure went abot that the completely wrong way.

thanks
 
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