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Conservation of energy/inclined plane w/ spring

  1. Oct 13, 2011 #1
    Problem: A 2 kg mass is held at the top of a ramp 6m above a spring which has a spring constant of 40 N/m (the 6 m distance is measured along the ramp surface). The ramp is at 30 degrees relative to the horizon. Find the speed of the mass as it just strikes the spring 6 meters below the point it was released from. Find how much the spring is compressed. Ignore friction.

    Formulae: PEspring=1/2kx2, KE=1/2mv2, GPE=mgh (we use 10 m/s2 for acceleration due to gravity), and conservation of energy

    The professor said this problem was tricky, but the solution I found seems like it may be too easy. I calculated the gravitational potential energy to be 60J at the top of the incline. At the end of the ramp, where theoretically all of the GPE would be kinetic energy now because μ=0, I calculated the velocity to be 7.75 m/s. Would this be the correct way to go, given that the incline is at an angle?
  2. jcsd
  3. Oct 13, 2011 #2
    The dimensions of the incline are 6m on the hypotenuse, 3m on vertical leg, 5.2m on horizontal leg by trigonometry.
  4. Oct 13, 2011 #3
    And also, k=40N/m.
  5. Oct 13, 2011 #4
    The solution will become fairly easy if you apply the work energy theorem.

    Initial configuration - ke is 0

    Final configuration (when spring is completely compressed) - ke is again 0

    So between these to configurations, the work done by all the forces should cancel out to be zero.

    W(gravity) {for 6m + compression} + W(spring){For Cmpression ONLY} = 0
  6. Oct 13, 2011 #5
    Thanks for the help! I looked in the textbook (there was an example problem that was exactly the same but done backwards and with different values) and they all lined up with your explanation of using the work-energy theorem, and it basically seemed to be the way I worked it at first. I guess the tricky part that he was talking about was figuring out that you need to find the vertical leg of the inclined plane in order to use that value to plug in to find the potential energy.

    Thanks again!
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