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Potential energy and work by spring.

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data

    A 15.0 kg stone slides down a snow-covered hill (the figure ), leaving point A with a speed of 10.0 m/s. There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100m and then runs into a very long, light spring with force constant 2.50 N/m. The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively.

    Here is the picture: http://session.masteringphysics.com/problemAsset/1260235/2/YF-07-34.jpg

    A) What is the speed of the stone when it reaches point B?

    B) How far will the stone compress the spring?



    2. Relevant equations

    PE = mgh
    KE = .5mv^2

    spring = .5kx^2


    3. The attempt at a solution

    A) I did this:

    PE = (15.0)(9.80)(20.0) = 2943 J
    KE= .5(15.0)(10.0)^2 = 750 J

    2943+750 = 3693 J

    3693 = .5(15.0)(v)^2-.5(15.0)(10.0)^2
    v=19.80 m/s

    B) work done by spring = .5(2.50)(x)^2 = 1.25x^2
    work done by friction = (0.2)(15.0)(9.8)x = 29.4x

    Using quadratic:

    1.25x^2 + 29.43x - 3693 = 0

    x= 43.84 or -67.38
    x= only 43.84 meters.

    Do these answers seem right? thanks!
     
    Last edited: Oct 30, 2011
  2. jcsd
  3. Oct 30, 2011 #2

    ehild

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    You miss a square it the KE term: It has to be .5(15.0)(10.0)2

    I do not understand what you mean with the second term when you calculate the velocity at B.


    ehild
     
  4. Oct 30, 2011 #3
    yes, thanks. It should be .5(15.0)(10.0)^2 = 750 J.

    Wouldn't the total energy = KE + PE, so 750 Joules+ 2943 joules be the total work at point B? Then set 3693 joules equal to .5mv^2-.5mvo^2, which game me 19.80 m/s.
     
    Last edited: Oct 30, 2011
  5. Oct 30, 2011 #4

    ehild

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    So the total energy at B is 3693 J, what is the velocity there?

    ehild
     
  6. Oct 30, 2011 #5
    Oh ok. I think that I only need to set the joules equal to the final velocity, since the initial is 0.

    In that case, velocity would equal 22.2 m/s.
     
  7. Oct 30, 2011 #6

    ehild

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    Correct!
    Now, question B. Before the stone reaches the spring, it slides 100 m on the rough ground. So it has lost a lot of energy when reaching the spring.

    ehild
     
  8. Oct 30, 2011 #7
    thanks! I will try figure part B out now.

    ok, I added work done by friction. I received (15.0)(9.8)(0.2)(100)cos180 = -2940 joules.

    -2940 + 3693 joules = 653 joules.

    Would the new quadratic equation be 1.25d^2 + 29.4d -653 = 0?
     
    Last edited: Oct 30, 2011
  9. Oct 30, 2011 #8

    ehild

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    it looks all right, but the red 6 has to be 7.


    ehild
     
  10. Oct 30, 2011 #9
    ah ok. good catch. Thanks a lot for the help!
     
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