Distance Spring Stretches

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KindofSlow

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Mar 5, 2010
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Section: Conservative Forces and Potential Energy
Question: A spring with k=63 N/m hangs vertically next to a ruler. The end of the spring is next to the 15-cm mark on the ruler. If a 2.5-kg mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks.

My work:
I figure that the change in potential energy of the spring should equal the work done by gravity.
So \[\frac{1}{2}k{x^2} = mgx \]
\[\frac{1}{2}\times 63 \times x^2 = 2.5 \times 9.81 \times x\]
\[ 31.5 x^2 - 24.5x = 0 \]
Solving for x gives 78 cm, added to the starting point of 15, gives 93 cm.
Which is not even close to the book's answer of 54 cm.

If anyone could tell me where I went off track, I would appreciate it.

Thank you
 
KindofSlow said:
Section: Conservative Forces and Potential Energy
Question: A spring with k=63 N/m hangs vertically next to a ruler. The end of the spring is next to the 15-cm mark on the ruler. If a 2.5-kg mass is now attached to the end of the spring, where will the end of the spring line up with the ruler marks.

My work:
I figure that the change in potential energy of the spring should equal the work done by gravity.
So \[\frac{1}{2}k{x^2} = mgx \]
\[\frac{1}{2}\times 63 \times x^2 = 2.5 \times 9.81 \times x\]
\[ 31.5 x^2 - 24.5x = 0 \]
Solving for x gives 78 cm, added to the starting point of 15, gives 93 cm.
Which is not even close to the book's answer of 54 cm.

If anyone could tell me where I went off track, I would appreciate it.

Thank you
Click to expand...

You are assuming that the energy is conserved within the system - as the spring stretches isothermally. That is not correct.

The correct equation to use in this case is Hooke's law → F = kx

W = kx

2.5*9.81 = 63 * x

x = 0.389286 m
 
This is, in fact, a trick question. You are taught that energy considerations are usually a simpler way to solve a problem in mechanics. It is USUALLY the case. BUT NOT IN THIS CASE. That is because half the energy will be expended in a very hard to track way.

As Subhotosh Khan suggested .. half of the energy will go somewhere.
He suggests that after carefully attaching the weight, you just let it drop. Then the weight will oscillate up and down until the oscillations dampen out at half the original bounce.

It is lost to heat in spring and air resistance. If that where not true, then it would bounce up and down forever.
Forget Energy.
Use force equilibrium.
Force of Gravity = Force from Spring Tension (at equilibrium)
 
Last edited:
by the way

Once you understand this question thoroughly (and amused by it) you may be asking yourself the following question.
Problem: What if I slowly lower the weight into the final position with my hand and then release. Now virtually zero energy is lost to the air or heating the spring.

Answer: The weight will do work ON YOU as you lower it with support through delta x.
 
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