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Spring compression - Conservation of Energy

  • Thread starter jtw2e
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Spring compression -- Conservation of Energy

1. Homework Statement
So far I'm only working on part A. I didn't know whether to create separate posts for each part I had trouble with or just put the whole thing in one post.

5501435672_1eab351652_b.jpg



2. Homework Equations
KE = 1/2mv^2
PE = mgh
Uelas = 1/2kx^2


3. The Attempt at a Solution

Narrowed full CoE equation down to:
K0 = Ugrav2 + Uelas2

As the example in our text, I set my datum at the point where the block just comes into contact with the springs but has not yet compressed them. My point 2 is the lowest point the block reaches.

Used quadratic equation to solve for height (mgh2)which should also equal x if I understand this correctly.

Got x = 0.130m or -0.332m BOTH were wrong.

Please help, I've been at this for 2 hours and I have a freaking long way to go (working ahead...have to start days ahead of the others just to keep up b/c I'm a moron).
 

PhanthomJay

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Re: Spring compression -- Conservation of Energy

Well, first you have to solve for the equivalent k of the 2 springs together. You didn't show your work, so what value did you use for k? This is a bit difficult, it requires some thought.

Then in your equation, on the left side, you neglected the U grav term.
 
27
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Re: Spring compression -- Conservation of Energy

Well, first you have to solve for the equivalent k of the 2 springs together. You didn't show your work, so what value did you use for k? This is a bit difficult, it requires some thought.

Then in your equation, on the left side, you neglected the U grav term.

The Ugrav term does not apply on the left side because y1 = 0 there.
There were 2 given values for k; I used them both, separately.

As for my work:

K0 = Ugrav2 + Uelas2

1/2(2.20kg)(2.05m/s)2 = (2.20kg)(9.81m/s2)(h) + 1/2(142N/m)(x)2 + 1/2(72N/m)(x)2

4.62275 J = 21.582N(h) + 71(x)2 + 36(x)2

Now since my datum at point 1 is just as the block begins to touch the spring, and point 2 is the lowest the block goes before the springs rebound, h = x.

Moved terms around and setup quadratic equation:

0 = 107(x)2 + 21.582(x) - 4.62275J

Solving for x = 0.130m or -0.332m But as I said, both of these were counted wrong. So I am completely lost.
 

PhanthomJay

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Re: Spring compression -- Conservation of Energy

You are neglecting the fact that the block is 1.3 m above the springs when it starts. It will be moving faster when it first hits the springs. So when you start at the beginning of the motion, the block has a U term associated with it (you can call it 0 but I don't recommend so).

Your spring Uelastic terms on the right side are not correct. You might want to check out this site to calculate an equivalent stiffness of 2 springs in series:

http://nienaber.disted.camosun.bc.ca/Physics154/TablesandInfo/SpringCombs.pdf
 
27
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Re: Spring compression -- Conservation of Energy

You are neglecting the fact that the block is 1.3 m above the springs when it starts. It will be moving faster when it first hits the springs. So when you start at the beginning of the motion, the block has a U term associated with it (you can call it 0 but I don't recommend so).

Your spring Uelastic terms on the right side are not correct. You might want to check out this site to calculate an equivalent stiffness of 2 springs in series:

http://nienaber.disted.camosun.bc.ca/Physics154/TablesandInfo/SpringCombs.pdf
Thank you, I see your point about the U term on the left...the KE will be higher because of a higher velocity once it reaches the datum. As far as the 2 springs in series, the professors did not mention any system with more than one spring, and neither does the text, so I thank you for your link.

Based on your link, the two springs though appear to be acting in parallel, not in series. In such a case it appears just treating the two as one spring should work, which is what I have done above. Yet it was wrong. Still completely lost (but at least I have a correct KE on the left side now...).
 
Last edited:
27
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Re: Spring compression -- Conservation of Energy

Solved parts A & B finally...

We were supposed to treat the 2 springs as one and just add their k values. I'm am now totally stuck on part C. Been at this for 8 hours... like a full time job. Getting nowhere fast.. Please help on Part C.
 

PhanthomJay

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Re: Spring compression -- Conservation of Energy

Solved parts A & B finally...

We were supposed to treat the 2 springs as one and just add their k values. I'm am now totally stuck on part C. Been at this for 8 hours... like a full time job. Getting nowhere fast.. Please help on Part C.
That's not right, the springs are on top of each other,and hence in series.....if they were next to each other, they'd be in parallel....maybe the pic is wrong....anyway, for part C, use your equation again....this time, what is the final energy when the block is at the top and has no speed...solve for the height.
 
27
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Re: Spring compression -- Conservation of Energy

That's not right, the springs are on top of each other,and hence in series.....if they were next to each other, they'd be in parallel....maybe the pic is wrong....anyway, for part C, use your equation again....this time, what is the final energy when the block is at the top and has no speed...solve for the height.

I wrote to the prof:
I don't see anything about springs in series or parallel in the text and we didn't discuss it in lecture. But the k value will be different than simply adding these two individual springs two:

http://scienceworld.wolfram.com/physics/SpringsTwoSpringsinSeries.html" [Broken]

Are we supposed to use the formula for springs in series or solve it based on what we are expected to know from lecture and the text?
He replied:
If it is done correctly either method will work. I think it is clearest by thinking of the two springs as two separate PE terms with the same deflection. Then you don't have to know or remember a special case of springs in parallel or series ( they are in parallel in this problem)
As far as part C, it wants to know the height above the uncompressed spring(s). I have been on this one for almost 10 hours now and ready to jump off a bridge.

I came up with:
Work of spring on block should equal PE2 of block... unless the hints are misleading again. Setup is
Uelas1 - Uelas2 = PE2
1/2kx12 - 1/2kx22 = mgh2
1/2(214N/m)(0.663)2 - 0 = (2.20kg)(9.81m/s2)(h2)
h = 2.17m

He said about that answer: "You've calculated a correct height based in your datum but the question asks about a different datum."

I don't know what to do. They didn't cover this in lecture and I can't find anything similar on the internet.
 
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PhanthomJay

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Re: Spring compression -- Conservation of Energy

I wrote to the prof:


He replied:


As far as part C, it wants to know the height above the uncompressed spring(s). I have been on this one for almost 10 hours now and ready to jump off a bridge.

I came up with:
Work of spring on block should equal PE2 of block... unless the hints are misleading again. Setup is
Uelas1 - Uelas2 = PE2
1/2kx12 - 1/2kx22 = mgh2
1/2(214N/m)(0.663)2 - 0 = (2.20kg)(9.81m/s2)(h2)
h = 2.17m

He said about that answer: "You've calculated a correct height based in your datum but the question asks about a different datum."

I don't know what to do. They didn't cover this in lecture and I can't find anything similar on the internet.
I think the prof may be right...yes, if 2 springs are in parallel, their deflection is the same and the stiffness of the springs are additive, so I guess I misread the picture, they are not on top of each other, but rather, one is behind the other....in this case, the springs are in parallel.

Now regardless of the correct k value, the conservation of energy equation does not change.
Don't jump! You took the initial Grav PE as 0 at the compressed level of the spring, which is good...so when you calculate h, it is measured from that elevation....it's height above the uncompressed length of the spring is less. Just subtract out that compressed length, whatever it may be. BTW, please write out the full conservation of energy equation when you do your work...it avoids errors...K_initial + U_elas_initial + U_grav_initial = K_final + U_elas_final + U_grav_final. Or if you prefer, Delta K + Delta U_elas + Delta U_grav = 0.
 
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