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Conservation of Energy 2

  1. Apr 4, 2007 #1
    2. Relevant equations

    Total Energy = Kinetic Energy + Potential Energy
    T.E = 1/2mv^2 + m(g)(h)

    3. The attempt at a solution

    Distance Child travels
    Sin 30 = 4/ x
    x = 8m

    T.E at top
    = 1/2mv^2 + m(9.8)(h)
    = 40m
     
    Last edited: Apr 4, 2007
  2. jcsd
  3. Apr 4, 2007 #2
    Conservation of energy only works for non-conservative forces! You could do E = Work_net = Work_conservative + Work_non-conservative, but going back to Newton's 2nd law should give an easier time.
     
  4. Apr 4, 2007 #3
    well you're on the right track, and assume from your last calculation that you're rounding g to 10m/s^2.

    TE at bottom=pe+ke where pe=0

    however, there is energy lost in going from top to bottom in form of friction.

    so TE at top=TE at bottom plus frictional energy. You are given a magnitude for friction and have computed the distance right, can you finish from here?

    edit: this was more or less simultaneous post, I think its actually easier using energy eqn, but solveable from either approach.
     
  5. Apr 4, 2007 #4
    40m = 1/2mv^2 + 1/4m
    v = 12.6 - m

    is this correct?

    And I dont know how to calculate the mass.
     
    Last edited: Apr 4, 2007
  6. Apr 4, 2007 #5
    closer. this is how I approached the problem, and always BTW much better to complete the algebra before posting numbers--both for purposes here and on an exam as a simple number mistake made early will cost dearly;

    mgh=1/2mv^2+Ff*distance where Ff=frictional force

    we know from problem, that frictional force = 1/4mg

    so mgh-1/4*(mg)*8m= 1/2mv^2
    (8m from your calculations involving length of slide)
     
  7. Apr 4, 2007 #6
    This is what I got:
    40m-1/4*(m)*8m= 1/2mv^2
    40m - 2m = 1/2mv^2
    76m / m = v^2
    v = 8.7 m/s

    Question stated that friction is 1/4m not 1/4mg. Or is it suppose to be 1/4mg?
     
  8. Apr 4, 2007 #7

    hage567

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    The question said the frictional force was one quarter of the child's weight. Weight is m*g, expressed in newtons. So it should be (1/4)mg.
     
  9. Apr 4, 2007 #8
    This is what I got now:
    40m-1/4*(m)*8m= 1/2mv^2
    40m - 20m = 1/2mv^2
    20m / m = v^2
    v = 4.47 m/s
     
  10. Apr 4, 2007 #9

    hage567

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    Where did the 1/2 from the kinetic energy go? You dropped it in the second last line.
     
  11. Apr 4, 2007 #10
    O ya thanks I forget to multiply by 2.

    This is what I got now:
    40m-1/4*(m*g)*8m= 1/2mv^2
    40m - 20m = 1/2mv^2
    40m / m = v^2
    v = 6.3 m/s
     
  12. Apr 4, 2007 #11

    hage567

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    I agree with that answer.
     
  13. Apr 4, 2007 #12
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