Elastic Potential Energy of a Spring Between 2 Blocks

AI Thread Summary
The discussion revolves around calculating the elastic potential energy stored in a spring between two blocks of different masses. The user successfully calculated the final velocity of block A using conservation of momentum, arriving at 3.6 m/s. However, they struggled with finding the spring constant (k) and the compression distance (x) needed to determine the potential energy stored in the spring. After exploring various equations, they discovered that the kinetic energy of the blocks after the spring decompresses could be used to find the potential energy, leading to a total kinetic energy of 8.64 J. This realization clarified the relationship between kinetic energy and the potential energy stored in the spring.
mac31905
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Homework Statement


Here's the question:
Block A in the figure below has mass 1.00 kg, and block B has mass 3.00 kg. The blocks are forced together, compressing a spring S between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 1.20 m/s.

Part A: It asked for the final velocity of block A, and I used the mavai + mbvbi = mavaf + mbvbf equation to find A's final velocity of 3.6 m/s.
mavai = 0, mbvbi = 0, ma = 1kg, mb = 3 kg, vbf = -1.20m/s
0 = (1 kg)(Vaf)+(3kg)(-1.20m/s)
0 = 1kg(Vaf) -3.6kgm/s
3.6kgm/s = 1kg(Vaf)
Vaf = 3.6 m/s

Part B: The second part of the question is asking for how much potential energy was stored in the compressed spring... and I have no idea how to find it! I've tried different equations but I have too many unknown variables and don't know where to start.

Homework Equations


For Part B:
PE(elastic) = 1/2kx^2 (this seems like the one to use, but how do I find k and x?)
KE = 1/2mv^2 (I don't think I have to find KE to find PE... though I do know m and v.)
GPE = mgh (and nope, this is gravitational potential energy. And plus, I don't have h.)

The Attempt at a Solution


So... if I do indeed use the PE(elastic equation), where in the world do I find k? I know it is the force (spring) constant, but I am unsure as to how to get it from the information given. Also, x is the compressed distance from it's uncompressed state.
*To find the force constant, I'd use this equation: F(spring) = -kx, but I don't have F, nor do I know x.

*And the question says nothing about how long the spring is when it's compressed/not compressed!

It just seems like I am not given enough information, even though that must be false since it's a homework question...

I've been stuck on this for a long time, and I'm so tempted to click "Show answer" on Mastering Physics but that won't help me to understand it, so hopefully one of you guys can help point me in the right direction :smile: I just need to figure out k and x and I'm good! I'm not asking for the answer, just an equation or something to get me started. Thanks! My test is tomorrow and I'd really appreciate some help!
 
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mac31905 said:
For Part B:
PE(elastic) = 1/2kx^2 (this seems like the one to use, but how do I find k and x?)

Yes, it is hard to use this equation if you know neither k nor x.

KE = 1/2mv^2 (I don't think I have to find KE to find PE... though I do know m and v.)

Let's add a new question saying find the KE of the two blocks.

Now (after the spring drops), you have some kinetic energy. What is the source of this kinetic energy?
 
naresh said:
Yes, it is hard to use this equation if you know neither k nor x.



Let's add a new question saying find the KE of the two blocks.

Now (after the spring drops), you have some kinetic energy. What is the source of this kinetic energy?

Hmm... it seems to me that the blocks would be moving after the spring drops. Therefore, the blocks would have some kinetic energy, right? Maybe the source of the KE would be the spring decompressing?
 
Let me see:

If I use the KE = 1/2mv^2 formula to find the KE of the blocks (if I'm going in the correct direction), then it'd be KE(block a) = 1/2(1kg)(3.6m/s^2); KE(block b) = 1/2(3kg)(-1.20m/s^2)

KEblocka = 6.48
KEblockb = 2.16

So maybe total KE = 6.48+2.16 = 8.64? I have no idea...
 
HOORAY! I tried my value of 8.64 and it worked! All that stress for a simple equation. Thanks Naresh! I can move on to my other problems now :)
 
I'm glad it worked, but do you understand why it worked? Why should the sum of the kinetic energies give you the potential energy?
 
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