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Old Mar 13th 2011, 01:07 PM   #1
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Compression of a Spring

This problem is not clicking with me.

A 1.50 box moves back and forth on a horizontal frictionless surface between two different springs, as shown in the accompanying figure. The box is initially pressed against the stronger spring, compressing it 4.00 , and then is released from rest.

By how much will the box compress the weaker spring?

The only equation I have is W = k (change in x)^2 / 2 and W = (change in x)(F) / 2

What am I missing?

What is the maximum speed the box will reach?

thanks in advance!
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Old Mar 13th 2011, 01:50 PM   #2
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Originally Posted by pre pt marc View Post
This problem is not clicking with me.

A 1.50 box moves back and forth on a horizontal frictionless surface between two different springs, as shown in the accompanying figure. The box is initially pressed against the stronger spring, compressing it 4.00 , and then is released from rest.

By how much will the box compress the weaker spring?

The only equation I have is W = k (change in x)^2 / 2 and W = (change in x)(F) / 2

What am I missing?

What is the maximum speed the box will reach?

thanks in advance!
Since there is no diagram I have to ask a few questions. What are the spring constants? Are both springs attached to the box? Or is the box sliding between two disconnected springs? If they are attached to the box how far is the other spring stretched initiallY?

-Dan
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Old Mar 13th 2011, 03:33 PM   #3
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reply to spring constants

Stronger spring - k = 32 N /cm

Weaker spring - k = 16 N /cm

The box is not attached to either spring. My apologies for not supplying all the pertinent info.

thanks!
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Old Mar 13th 2011, 06:03 PM   #4
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Originally Posted by pre pt marc View Post
Stronger spring - k = 32 N /cm

Weaker spring - k = 16 N /cm

The box is not attached to either spring. My apologies for not supplying all the pertinent info.

thanks!
No problem.

The spring with the larger constant (K = 32 N/cm) has an initial potential energy of (1/2)Kx^2 where x = 4.0 cm, so S = (1/2)(3200 N/m)(0.04 m)^2 = 2.56 J. When the box has moved awy from the stronger spring all of the spring's potential energy has been converted to kinetic energy (see the second part of the problem) and then it hits the second spring and all that kinetic energy gets converted to spring potential energy. So for part I we have that 2.56 J = (1/2)ky^2, where k is the spring constant for the second spring (k = 16 N/cm) with a spring displacement y. For the second part all of the spring potential energy was converted to kinetic energy so we have 2.56 J = (1/2)mv^2.

Note carefully what I have done with the units for the first spring constant. It was reported in N/cm, but I converted it to N/m when using the equation.

-Dan
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Old Mar 13th 2011, 06:25 PM   #5
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Max speed of the box

Given that the weaker spring compresses 5.66 cm and the stronger spring compresses 4.00 cm, we have a total of 9.66 cm covered by the box moving back and forth.

How do I utilize this info to calculate max speed when i lack the variable of time?

Thank you kindly!
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Old Mar 13th 2011, 07:16 PM   #6
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Originally Posted by pre pt marc View Post
Given that the weaker spring compresses 5.66 cm and the stronger spring compresses 4.00 cm, we have a total of 9.66 cm covered by the box moving back and forth.

How do I utilize this info to calculate max speed when i lack the variable of time?

Thank you kindly!
The sum of the spring displacements tells you nothing since (I presume) the springs are not compressed at the same time.

As I said, when the box is between springs all the energy of the box is all in its kinetic energy. So 2.56 J is kinetic energy thus the maximum speed will be given by the equation 2.56 J = (1/2)mv^2.

-Dan
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Old Sep 1st 2011, 07:51 PM   #7
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Two Springs and a Box

Hi All,

I liked this problem but I was busy. But now I've worked it. See what you think of this solution.

Two Springs and a Box

I have to work at my site - I don't know Latex well enough to use the entry box.

Good Luck, Gymbeaux
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