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Energy and Power

  1. Jul 12, 2010 #1
    1. The problem statement, all variables and given/known data
    A train of mass 2.4x10^6kg enters a 5km stretch of track with a vertical rise of 40m at a s speed of 1.0m/s and leaves at 3.0m/s. Assuming that frictional drag is negligible, find:

    a) Each term of the equation Win = delta K + delta U + Wout


    2. Relevant equations
    Delta K = 1/2mvf^2 - 1/2mvi^2

    Delta U = mghf - mghi

    Wout = work removed from the system such as air drag.

    Win = not sure??


    3. The attempt at a solution
    I found Wout and that's equal to 0.

    I found Delta K. Not too sure if I used the right velocities. Final velocity, i used 3.0m's and inital velocity, i used 1.0m/s.

    I found Delta U, with the final height being 40m and the initial height being 0m.

    Am i doing these correctly? Can someone guide me please???
     
  2. jcsd
  3. Jul 12, 2010 #2

    Doc Al

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    Looks like you're doing fine.

    Since the energy increases, there must be some work input. That's what your equation will tell you.
     
  4. Jul 14, 2010 #3
    so is work input = delta k + delta U?
     
  5. Jul 14, 2010 #4

    Doc Al

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    Yes.
     
  6. Jul 14, 2010 #5
    part b of the question asks for the average mechanical power delivered by the engine for the climb if it is done at constant acceleration.

    I know power is work over time. We don't have time, but can we find that since they said it's done at constant acceleration? but don't know how to start...
     
  7. Jul 14, 2010 #6

    Doc Al

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    What's the average speed during the climb?
     
  8. Jul 14, 2010 #7
    speed = d/t
    where, speed is 1.0m/s and distance is 5m

    so t = d/speed?
     
  9. Jul 14, 2010 #8

    Doc Al

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    If the speed uniformly goes from 1 m/s to 3 m/s, what's the average speed?
     
  10. Jul 14, 2010 #9
    Average speed is equal to the total distance travelled divided by time.

    The average speed is 2km?
     
  11. Jul 15, 2010 #10

    Doc Al

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    Last edited by a moderator: Apr 25, 2017
  12. Jul 15, 2010 #11
    So to answer part b,

    Work = delta K + delta U
    = 1/2m(vf-v1)^2 + mg(h2-h1)

    doing the calculation for Work, I got 9x10^8J.

    Average speed is equal (vi + vf)/2,
    therefore, it is 2m/s.

    speed = distance/time
    time = distance/speed
    = 5/2
    = 2.5s

    so POWER = work / time

    = 9x10^8 / 2.5
    = 4x10^8Watts

    Is that correct?
     
  13. Jul 15, 2010 #12

    Doc Al

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    I get closer to 9.5 x 10^8 J.

    OK.

    Careful. The distance is 5 km, not 5 m.
     
  14. Jul 15, 2010 #13
    ok. That's a silly mistake. I should've noticed that. Thanks.

    ok...here's what I get:

    average speed = 2m/s
    distance = 5000m

    therefore, time = 2500s

    POWER = 9.5x10^8J / 2500s = 3.8 x 10^5W

    Thanks for your help. :smile:
     
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