Mechanics problem involving work, spring and a-jack-in-the-box

  • #1

Homework Statement


A jack-in-the-box is made using a spring of natural length 0.2m and modulus 100N and a "jack" of mass 0.5kg. When the lid is closed, the spring is compressed to a length of 0.1m. Assuming the spring to be vertical throughout, calculate the maximum distance that the "jack" will rise when the lid is suddenly raised.


Homework Equations


E.P.E= λx^2/l*1/2 and P.E=mgh where λ=modulus, l=natural length x= extension, m=mass and h= distance.


The Attempt at a Solution


I have really attempted to solve the problem using the principle of conversion of energy but really failed to get the exact answer as I expected. Could you guys tell me if there is a much more appropriate way of solving it.
 

Answers and Replies

  • #2
lewando
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That is the appropriate way. Please show what you did.
 
  • #3
Okay, this is how I went about this problem.
y= maximum distance
extension, x= y-0.1
All the work done by the spring through it's compression will be converted into potential energy when the jack reaches it's maximum point. So, according to condition
E.P.E=P.E
or, 100*(y-0.1)^2/2*0.2= 0.5*9.8*y
But, I just simply know that the expression for the extension that I wrote here is absolutely wrong. Plz enlightren me on this or throw some subtle hints that might prove to be helpful.
 
  • #4
lewando
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I am confused with what you are trying to do with "extension, x=y-0.1" You know from the problem statement how much the spring is being compressed. Use that to determine KEspring. If the mass, sitting on the compressed spring, is at y=0, then the Δy is your h in mgh (PEmass,max). Δy or h is what you are looking for.
 
  • #5
Initially i also tried to solve it simply by plugging in the given info in this problem. However, it didn't work out and gave me an answer of 0.511 which is wrong. That's why I tried to do in several methods. It seemed easy to me at the first glance because h is the only unknown quantity that needs to be determined.
 
  • #6
lewando
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If energy is being conserved, then think about how the initial energy stored in the spring is transformed as the jack rises to h. If you know PEspring, what can you say about PEmass,max?
 
  • #7
I am sorry that I haven't been able to ask you questions regarding this problem as this has really been a hectic week for me. Anyway, back to the problem if I know P.E of the spring can I solve for h in the simplest way by plugging in the values or do I have to write a different expression for h. I am asking this again because I have really tried to reach the exact solution of this prob. Any subtle hint on your part will be helpful.
 
  • #8
lewando
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What did you get for the PE of the compressed spring?
 
  • #9
P.E of the compressed spring= 2.5 Joules
 
  • #10
lewando
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Does not look correct. How did you arrive at that value?
 
  • #11
Given that natural length, l= 0.2, λ=100N and hence, x= 0.2-0.1=0.1
P.E= 0.1^2*100/2*0.2=2.5 could you point out my mistake???
 
  • #12
lewando
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Sorry, no mistake. Its been a while and I needed to get my bearings (I thought you were using k = 100N/m...)

So as you tried to do before, you equated this to mgh and got h = .511m. That's what I got. Do you know the actual answer? With sig figs, its more like 0.5m.
 
  • #13
This is actually a problem from my M3 book and the actual answer given in this book is 0.180metres.
 
  • #14
lewando
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The problem is--incorrectly thinking that the head is not physically attached to the spring and is just being "launched". Being attached, the spring will initially push the jack for .1m and then begin to pull back on the jack. Sorry I missed that till now.
 
Last edited:
  • #15
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Given that natural length, l= 0.2, λ=100N and hence, x= 0.2-0.1=0.1
P.E= 0.1^2*100/2*0.2=2.5 could you point out my mistake???
What is the meaning of the 0.2 in the expression of the PE?
And check your data again. Is it 0.5 kg or 0.05 kg?
 
  • #16
yes, I have checked the data clearly and 0.2m here is the natural length of the spring and the problem clearly states that the mass is 0.5 kg and the spring is compresseed to 0.1m when the P.E is stored in it which is why it's compression is, x=0.1m
 
  • #17
lewando
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So, moving forward with the "tight-head jack" model--you need to construct an energy equation. You have correctly determined the initial energy in the compressed spring. Now where will the energy be when the head is at its maximum h?
 
  • #18
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Now I see that you gave the "modulus" in N. What is this supposed to be?
The spring constant is in N/m. Young/s modulus is in N/m^2.
Is this the spring constant multiplied by the initial length?
 
  • #19
lewando
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λ, from a form of Hooke's Law:

T = tension
L = natural length
x = extension

T = (λ/L)x

gives a λ with units of N.
 
  • #20
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Thank you for clarification.
Then you get 0.18 m for the case with the jack attached to the spring.
 
  • #21
Nasu, Could you please explain to me how you got 0.18m using T=λ/lx because from hooke's law we know that it is actually T=λx/l
 
  • #22
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Nasu, Could you please explain to me how you got 0.18m using T=λ/lx because from hooke's law we know that it is actually T=λx/l
When did I say that I used T=λ/lx?
 
  • #23
Okay, It wasn't you. I think it was lewando. I think I know what the equation of this system will look like when the spring with the jack attached to it is released. Please confirm if it is okay.

λx^2/2l= mgh+ λx/l
 
  • #24
lewando
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You are on the right track, but your 3rd term, λx/l needs some work. It should involve h.
 
  • #25
So, there is going to be h instead of x in the equation of tension. Thank you very much once again for your help lewando
 

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