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Spring Compression? (Mastering Physics Problem 10.69)

  1. Dec 2, 2006 #1
    Spring Compression??? (Mastering Physics Problem 10.69)

    1. A 21.0 kg box slides 4.0 m down the frictionless ramp shown in the figure, then collides with a spring whose spring constant is 130 N/m.

    2. 1. What is the maximum compression of the spring? (in meters)

    3. I have attempted to determine the max compression of the spring by finding the Potential energy of the block [Ug=mgh]. Then using the potenital energy of the block [Ug = Us] (potential energy of the srping) to find the distance the spring compresses [Us=(1/2)k(∆s)^2]. I messed up on my calculations and found 3.3 meters was close. I found that by calculating Ug=(21kg)(9.8m/s^2)(cos(30))...the height should be sin(30) so I don't know how I got close. Any suggestions?
  2. jcsd
  3. Dec 2, 2006 #2


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    You could also define the 'zero level' of potential energy at the point at which the spring is compressed by an amount ∆. Further on, use energy conservation from the initial point to that point and you should get a quadratic equation which you can solve for ∆. (The potential energy of the block is completely transformed into the potential energy of the compressed string.)
  4. Dec 3, 2006 #3
    so I got that the compression of the spring will be around 3.3 meters. So we add the ∆s from [Us=(1/2)k(∆s)^2] using Ug=(21kg)(9.8m/s^2)(cos(30)) for the Ug of the block and then the ∆s for the block sitting on the spring U=(21kg)(9.8)(cos(30)). My answer is still a little off (by like 0.1). Any idea's?
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