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Spring constants

  1. Oct 7, 2003 #1
    A block of mass 12.0 kg slides from rest down a frictionless 37.0 degree incline and is stopped by a strong spring with k=3.50e4 N/m. The block slides 3.00 m from the point of release to the point where is comes to a rest against the spring. When the block comes to rest, how far has the spring been compressed?

    I really wanted to use 1/2kx^2 = W to get this, but I know I didn't have all of the values. I thought of using KE = 1/2 mv^2, but I don't know how to get velocity without a time value. So I ended up using an equation for the incline, g sin theta = acceleration, got 5.90 m/s^2 for that, plugged mass and acceleration into f=ma to get a force of 70.8 N, plugged values into W=F S Cos theta to get work as 170 J, and finally plugged values into 1/2kx^2 = W to get x as .0984, but this is wrong. For every other equation I try to use, I'm missing 2 values and don't know how to get either of them. Where did I go wrong?
  2. jcsd
  3. Oct 7, 2003 #2
    You're overcomplicating.
    Use conservation of energy.
    When an object of mass m goes down a vertical distance (height) h, then it gains the energy E = mgh (g is gravitational acceleration).
    This energy goes into the spring, which will thus compress by a length x, where E = kx. OK?
  4. Oct 7, 2003 #3
    But it's not exactly a vertical distance since it's on an incline, is it?
  5. Oct 7, 2003 #4
    Come on. You know the distance, and you know the angle...
  6. Oct 7, 2003 #5
    So I took the sin of 37 times 3 to get the height. I got 1.805. Then I multiplied that times 12.0 and 9.8 to get E. I got 212.3. Then I divided that by 3.50e4. Got .00607. And it isn't right. I've been doing physics since 8 AM so I'm really tired, sorry. What am I missing?
  7. Oct 8, 2003 #6


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    Yes, that is the correct height (in meters).

    Yes, that is the change in potential energy (in Joules).

    Why? Did you forget the formula 1/2kx^2 = W that you gave before?

    x2= 2W/k or, using the figures you got
    x2= 2(212.3)/(35000). Solve for x.
  8. Oct 8, 2003 #7
    thank you!!!
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