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Engine pulling Train

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Engine: 100 tonnes, with 1,000N frictional force
    Train: 400 tonnes, with 20,000N frictional force

    Speed = 80km/hour
    Engine is exerting a power of 4000kW

    Find the tension in the couplings between train and engine

    3. The attempt at a solution

    I'm not sure where to start, I thought of finding coefficient of friction but it didn't play out.
     
  2. jcsd
  3. Nov 22, 2009 #2

    mgb_phys

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    Draw a diagram with the forces for a train moving at constant speed
     
  4. Nov 22, 2009 #3
    Engine:
    -981,000N vertically both up and down
    -Not sure what force is to right (forward). Calculate from power?
    -T+1,000N to left

    Train:
    -3,924,000N vertically both up and down
    -T to right
    -20,000N to left

    wait, if the train isn't accelerating, does that mean T = 20,000N?

    if it is, why would they provide all the other data?
     
    Last edited: Nov 22, 2009
  5. Nov 22, 2009 #4

    mgb_phys

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    Exactly

    It could be there is another part to the question you haven't been given, or it could just be to confuse you!
     
    Last edited: Nov 22, 2009
  6. Nov 22, 2009 #5
    That's weird, the question is worth 11 marks!

    Okay, here's the exact wording:

    An engine of mass 100 tonne pulls a train of mass 400 tonne along a horizontal track. There is a horizontal frictional force of size 1 kN acting on the engine and a horizontal frictional force of size 20 kN acting on the train. Find the tension in the couplings between the engine and the train at the instant when the speed of the train is 80 km/h and the engine is exerting a power of 4000 kW.

    That doesn't indicate it's accelerating when it's at 80km/h, does it?
     
  7. Nov 22, 2009 #6

    mgb_phys

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    Correct it's not accelating.
    What i meant was that in the advanced class there might be a part 2 to the question that says, eg. now the train is moving up a 10% slope, find the maximum speed for the given engine power.
     
  8. Nov 22, 2009 #7
    No, there's no follow up question.

    Are you sure it's not accelerating? Can I work out the driving force of the Engine from the power it exerts?

    Maybe the engine is accelerating and I have to work out what the tension is as ir reaches 80km/h, as it does specify the instant[/i] it is travelling at 80km/h
     
  9. Nov 22, 2009 #8

    mgb_phys

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    Yes the point was that quite often a textbook or a professor will reuse the easy part of a more advanced question in an easier class.
    So it's not unusual for their to be extra information given that you don't need.
    It's good practice in spotting which bits of data are relavent for the problem you are studying
     
  10. Nov 22, 2009 #9
    so, is it possible to calculate acceleration from the power the engine is exerting?
     
  11. Nov 22, 2009 #10

    mgb_phys

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    Yes,
    The energy lost to friction is friction_force*distance
    So with the speed you can work out the rate of energy lost to friction = power.
    Then with the mass you can work out the change in kinetic energy for the power thats left
     
  12. Nov 22, 2009 #11
    80km/h = 22.222222...m/s

    so at this speed, the engine is losing (22.222222.... x 1,000) joules per second

    =-22,222 J/s
    =-22,222 W

    So that leaves 3,977,778 W

    The kinetic energy of the engine at this speed is 24,691,358 J [ii]

    so, do I add to [ii], calculate the new speed, and take difference between the speeds to be the acceleration?
     
  13. Nov 22, 2009 #12

    mgb_phys

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    21,000N of friction
    At 80km/h it's doing 22.2m/s or 1m in 0.045s

    So to move 1m with 21,000N of friction takes 21KJ of energy, in 0.045s = 21,000/0.045 = 470KW
    That means the engine at full power has 4000-470KW spare = 3530KW

    At 80km/h it has a kinetic energy of = 1/2 mv^2 = 0.5 * 500,000kg * 22.2^2 = 123MJ
    In one second it can supply 3530KJ extra energy so KE goes to 123,000+3530 KJ
    an then working back with 1/2 mv^2 you can get the new v^2
     
  14. Nov 23, 2009 #13
    power = force * velocity
     
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