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Homework Help: Simple hooke's law question

  1. Oct 30, 2005 #1
    Hi again, all. I thought of another question that I need help with. Again I have the answer but I'm not sure how they got to it. The question is:

    A 1.6kg block is supported on a horizontal surface. The block is attached to a sping with k = 1000 N/m. The block compresses the spring x = 0.02 m and is then released from rest. The surface provides 4.00 N constant frictional force. Determine the speed of the block when located at spring's equilibrium point.

    Now, I know to use W = (1/2)mv^2 - (1/2)mv^2 to find the velocity but I'm having trouble understanding how they got the total work. What they did was find the work done by the spring on the block and then find the work done by the friction and subtract the two to get the net work done.

    But I tried to do it by calculating the spring force (which I got 20N for), subtract the friction so net force is 16 N and then multiply that by the distance 0.02 to get the net work done. But I get the wrong answer and I don't understand why. Maybe I am not calculating the spring force right? I'm using F = -kx.

    Any help as to what I'm doing wrong would be greatly appreciated!
  2. jcsd
  3. Oct 30, 2005 #2
    maybe you need to take in consideration acceleration due to gravity?
    the mass 1.6 would be acted upon by gravity when it is let go right? so there you have downwards force
    Last edited: Oct 30, 2005
  4. Oct 30, 2005 #3


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    Force isn't constant. You'll need calculus to solve it that way.

    Potential energy of a spring:
    [tex]E = \frac{1}{2}kx^2[/tex]

    It's directed downwards. But then there's the support force of the surface upwards, which is equal but opposite to the force caused by gravity. Thus we only need to consider horizontal motion.
    Last edited: Oct 30, 2005
  5. Oct 30, 2005 #4
    So if the force isn't constant I have to find the work done by each force then add or subtract them to get the net work? Otherwise if the force is constant I can use the method I tried using in which I just find the net force and then multiply by the distance to find the work right?
  6. Oct 30, 2005 #5


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    As Päällikkö was trying to say (I think) you dont have to workout the individual forces, just the energies. In the initial position there is only potential energy from the spring, at the equilibrium there is only kinetic energy so you can write that:
    [tex] \frac{1}{2}mv_{f}^2 = \frac{1}{2}kx_{i}^2[/tex]
  7. Oct 30, 2005 #6


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    I don't think so Daniel, you're applying the Conservation of energy without taking care of friction.

    [tex] \Delta E = W_{friction} [/tex]
  8. Oct 30, 2005 #7


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    Sorry, I didn't notice the friction.
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