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Spring Energy-a block slides down a frictionless incline plane

  1. Jan 7, 2009 #1
    Spring Energy--a block slides down a frictionless incline plane

    A block of mass m = 13 kg is released from rest on a frictionless incline angled of angle θ = 30°. Below the block is a spring that can be compressed 2.0 cm by a force of 270 N. The block momentarily stops when it compresses the spring by 5.8 cm.

    What is the speed of the block when if first touches the spring?




    Ug = mgh, Us = 1/2kx2, K = 1/2mv2



    By K = 1/2kx^2, I know the spring constant. 270 N * 0.02m = 1/2k(0.02)2. So, k = 27,000. I solved for the h in mgh--> 1/2kx^2 = mgh, 1/2(27,000)(0.058)^2 = 13 * 9.81 * h. h = 0.356 m. The equation when it first touches the spring is 1/2mv^2 = mgh. To find this new h value, I used the 30-60-90 triangle. When h was 0.356 m, the length of the relative incline is 2 * 0.356, or 0.712. So, the length of the new incline is 0.712 - the length of compression. So, 0.712 m - 0.058 m = 0.654 m, which is the length of the incline. Using 30-60-90 triangles, the relative vertical height of the incline is 0.654/2 = 0.327 m.
    So, if mass cancels, 1/2(v^2) = 9.81 * 0.327. v = 2.53 m/s


    But my online homework thing says it's wrong??
     
  2. jcsd
  3. Jan 7, 2009 #2
    Re: Spring Energy--a block slides down a frictionless incline plane

    If I read the question correctly, the spring is oriented along the incline. If this is the case, think about what the final length of the spring will be when the system is at equilibrium.

    Also, I'm not sure that the equation Ug = mgh is necessarily relevant here, as the height from which the block is released is not part of the question. If you look at the kinetic energy of the block when it first contacts the spring and the potential energy of the spring when it is compressed to its maximum amount, you should have all the information needed to determine the velocity of the block.
     
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