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Finding speed of a block on a ramp/pulley

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data


    A 20kg block (m1) slides on a 40 degree frictionless ramp. This block is connected to a 30kg block (m2) by a string that passes over a frictionless pulley. The 30kg block is 20cm above the floor and connected to a spring of negligible mass with spring constant 250N/m. The spring is initially unstretched and connected to the floor. The 20kg block is pulled a distance of 20cm down the ramp (so that the 30kg block is 40cm above the floor) and is released from rest. Calculate the speed of each block when the 30kg block is again 20cm above the floor (spring unstretched).



    2. Relevant equations

    Usi + Ugi +KEi = Usf + Ugf + KEf

    3. The attempt at a solution

    I assumed that you had to use conservation of energy to find v:
    Usi + Ugi +KEi = Usf + Ugf + KEf,
    where Us is Spring potential, Ug is gravitational potential and KE is kinetic energy.

    KE is initially zero so we have
    0.5k(xi)^2 + mghi = 0.5k(xf)^2 + mghf + 0.5mv^2

    xi is 0.2m, xf is 0.4m for block 2.
    I worked everything out and got 10v^2 = -64...which I'm pretty sure is incorrect. Can anyone guide me to the right direction??

    Thanks!!
     
  2. jcsd
  3. Dec 3, 2009 #2

    kuruman

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    First off, why are you so sure that your answer is incorrect? Secondly, assuming that you are wrong, we need to see the details of your work and your detailed calculations. Otherwise, it will not be easy to pinpoint where you might have gone astray.
     
  4. Dec 3, 2009 #3
    Ok, this is my work:

    0.5k(xf)^2 + m2ghf + 0.5m1v^2 = 0.5k(xi)^2 + m2ghi
    0.5k(0.4)[tex]^{2}[/tex] + m2g(0.4) + 0.5m1v[tex]^{2}[/tex] = 0.5k(0.2)[tex]^{2}[/tex] + m2g(0.2) + 0

    0.5(250)(0.4)[tex]^{2}[/tex] + (30)(9.8)(0.4) + 0.5(20)v[tex]^{2}[/tex] = 5 + 59
    20 + 117.6 + 10v[tex]^{2}[/tex] = 64
    10v[tex]^{2}[/tex] = -73.6

    :/
     
  5. Dec 3, 2009 #4

    kuruman

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    You forgot two things. (1) Mass m1 is also moving, so it has final kinetic energy and (2) mass m1 has an initial and final potential energy as it rises above its initial position.
     
  6. Dec 3, 2009 #5
    OK. So to find PE of m1 before and after, I need the height of m1 before and after.
    I assumed that xi for m1 is 0.2m
    Using trigonometry:

    sin40 = [tex]\frac{hi}{0.2meters}[/tex]
    hi = 0.13m
    and hf = 0m
    Right?
     
  7. Dec 3, 2009 #6

    kuruman

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    Right.
     
  8. Dec 3, 2009 #7
    Does m2 have KE as well, or only m1?
     
  9. Dec 3, 2009 #8

    kuruman

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    Yes. Any mass that is moving has KE.
     
  10. Dec 3, 2009 #9
    So that means I need vf for m1 and m2, but that gives me two variables.
    How can I solve them individually?
    Do they have the same speed since they are connected?

    Thanks for the help by the way.
     
  11. Dec 4, 2009 #10

    kuruman

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    Yes, they have the same speed. What would happen if they did not have the same speed? They would have non-zero relative velocity which implies that the connecting shrink is either stretched or compressed. An implicit assumption is that the connecting string maintains the same length throughout the motion.
     
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