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Kinetics: Two Blocks and a Spring

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Homework Statement


Suppose a block of Uranium and a block of Radium, both with edge length [itex] \lambda [/itex], are pushed to the head of an un-stretched spring (spring constant [itex] k [/itex], un-stretched length [itex] \delta _0 [/itex]). The combination, pushed [itex] \delta [/itex] farther, compresses the spring. Upon release, the combination accelerates in the opposite direction. At what point do the blocks separate from the spring (not from each other)? What is their velocity at this instant? Assume the left block is Radium and the right block is Uranium. [itex] \lambda = 10cm [/itex], [itex] \rho _{radium} = 5 \frac{g}{cm^{3}} [/itex], [itex] \rho _{uranium} = 19.1 \frac{g}{cm^{3}} [/itex], [itex] k = 2410 \frac{dyne}{cm} [/itex], [itex] \delta _0 = 30 cm [/itex], [itex] \delta = 10 cm [/itex].

Homework Equations




The Attempt at a Solution


I have to say I am sort of stumped on this problem. I know that I can find the volume of each of the two blocks (assuming that they are cubes) by cubing the edge length, certainly I'll need that information later on in the problem. I was thinking that perhaps conservation of energy would be one way to approach the problem, but I believe that the instructor meant for us to use Newton's Law (this is for Mechanics class and we are doing the Force/Acceleration unit). I have my free body diagrams for the blocks drawn (and can post them if it would help). I was sort of hoping that someone could push me in the right direction and hopefully from there I'll be able to figure it out. Any help is appreciated, thanks.
 
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Answers and Replies

  • #2
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The detailed description of the "mass" is furnished as a distraction. Presumably you're operating on the traditional frictionless surface with a massless spring. When/where does the relaxing spring stop pushing the mass/block(s)?
 
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The detailed description of the "mass" is furnished as a distraction. Presumably you're operating on the traditional frictionless surface with a massless spring. When/where does the relaxing spring stop pushing the mass/block(s)?
A quick question, would it be allowed to treat both blocks as one mass?
 
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Certainly.
 
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Certainly.
Ok, so the blocks will separate once the spring returns to length [itex] \delta _0 [/itex] and then I could just find the velocity by using conservation of energy (solve for [itex] v [/itex])...
 
  • #6
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Ok, so the blocks will separate
... and, I will insert, "from the spring," right at this point ....
once the spring returns to length δ 0
Unless I've been too clever seeing through all the distractions built into the problem statement.
 
  • #7
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... and, I will insert, "from the spring," right at this point ....
Ironically, when this problem was first assigned that wasn't specified, so I sat down for like an hour trying to figure out some way that would allow me to figure out when the blocks would separate from each other (luckily the instructor finally specified what he meant) lol...Anyhow, thanks for the help, it doesn't seem nearly as hard as I originally thought it was..
 
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Ok here is what I did for my solution: I said [itex] x = 0 [/itex] when the spring is un-stretched and that x is positive to the right. Therefore, [itex] x_0 = -10 cm [/itex]. Using the formula for density and the fact that edge length cubed gives volume I found the total mass of both blocks (assumed to be one mass) is 29100 g. Then I applied conservation of energy. [itex] \frac{1}{2}mv_{0}^{2} + \frac{1}{2}kx_{0}^{2} = \frac{1}{2}mv^{2} + \frac{1}{2}kx^{2} [/itex]. I said that [itex] x = 0 [/itex] (the spring is unstretched at this point) and [itex] v_{0} = 0 [/itex]. After some manipulation I found: [itex] v = \sqrt{\frac{k}{m}x^{2}} [/itex]. Plugging in the known values I found [itex] v = 2.9 \frac{cm}{s} [/itex].
 
  • #9
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Looks like a good set-up; I haven't checked your mass calculation, but everything else works out, so I'll take your word for it, and I get the same number. Holler if it turns out I gave you a bum steer.
 
  • #10
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Looks like a good set-up; I haven't checked your mass calculation, but everything else works out, so I'll take your word for it, and I get the same number. Holler if it turns out I gave you a bum steer.
Awesome, thanks for the help
 

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