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Determine the speed of the block EPE

  1. Nov 28, 2006 #1
    A 2.5 kg block at rest on a tabletop is attached to a horizontal spring having constant 19.6 N/m. The spring is initially unstretched. A constant 20 N horizontal force is applied to the object, causing the spring to stretch. Determine the speed of the block after it has moved 0.900 m from equilibrium if the surface between the block and tabletop is frictionless.

    I used PE=1/2kx^2 and got 7.938

    Then, I used E=1/2mv^2 and plugged 7.938 in for E and solved for v. I found v to be 2.52.

    I tried the problem another way and got the same answer, I don't understand why it's wrong.
     
  2. jcsd
  3. Nov 28, 2006 #2
    Because you are completely ignoring the 20N force that is applied to the object.
     
  4. Nov 28, 2006 #3
    How would I work that into the solution?
     
  5. Nov 28, 2006 #4
    In line 2 of your solution you state that the potential energy of the spring is exactly equal the total energy of the system. Is that correct? Is there any other forces acting on the block other than the spring? How about that constant horizontal force being applied to the block? Is this force in the same direction as the potential force?

    Also note that it says constant force is being applied to the block. Which tells us what about acceleration? This implies net force = _______ .
     
    Last edited: Nov 28, 2006
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