1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetics: Two Blocks and a Spring

  1. Feb 11, 2015 #1
    1. The problem statement, all variables and given/known data
    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].

    2. Relevant equations


    3. 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.
     
    Last edited: Feb 11, 2015
  2. jcsd
  3. Feb 11, 2015 #2

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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)?
     
  4. Feb 11, 2015 #3
    A quick question, would it be allowed to treat both blocks as one mass?
     
  5. Feb 11, 2015 #4

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Certainly.
     
  6. Feb 11, 2015 #5
    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])...
     
  7. Feb 11, 2015 #6

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    ... and, I will insert, "from the spring," right at this point ....
    Unless I've been too clever seeing through all the distractions built into the problem statement.
     
  8. Feb 11, 2015 #7
    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..
     
  9. Feb 11, 2015 #8
    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].
     
  10. Feb 11, 2015 #9

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
  11. Feb 11, 2015 #10
    Awesome, thanks for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Kinetics: Two Blocks and a Spring
Loading...