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Work problem

  1. Jan 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A car (mass 900 kg) runs 12 km along a level road at a constant speed of 80 km / h. The friction force is 500N. Calculate the work done by the engine

    2. Relevant equations
    W= F x displacement



    3. The attempt at a solution
    W= (900x 9.81) N x 12000m
    W=10594800 J
    But this is wrong .answer should be 6 x10^6 J
     
  2. jcsd
  3. Jan 18, 2016 #2

    SammyS

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    (900 kg)(9.81 m/s2) is the forge due to gravity. It acts in a direction perpendicular to the displacement of the car, so that force does zero work.

    What force must the engine overcome to maintain the car's sped?
     
  4. Jan 18, 2016 #3
    oh friction force
    so
    Work = Friction force x (12000m)
    W= 500 N x 12000m
    W=6 x10^6 Joule
     
  5. Jan 18, 2016 #4

    SammyS

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    Correct !
     
  6. Jan 18, 2016 #5
    Thank you very much . Where on forum should i ask if i want explanation about vectors components?
     
  7. Jan 18, 2016 #6

    Ray Vickson

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    First, look in your textbook. If you cannot find a satisfactory answer there, go next to the internet; Google "vectors" or "vector components" or "vectors in physics", or similar entries. If you still have not found a satisfactory answer, THEN come to this Forum and ask a specific question (not a general question like 'can someone tell me about vector components?')
     
  8. Jan 18, 2016 #7

    SammyS

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    If it's a from a homework problem or closely related to one, then this is the correct forum.

    If it's more general, then either this forum or the General Physics section of the Physics section.
     
  9. Jan 18, 2016 #8

    haruspex

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    Calling the resistance to a car's normal motion simply 'friction' is one of my pet hates. It's so confusing for students. What the question means is some combination internal friction (bearings etc.), rolling resistance and drag. The friction between the tyres and the road is precisely what allows the car to get anywhere. In the context of the problem, it would be acting in the forward direction.
     
  10. Jan 19, 2016 #9

    CWatters

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    I agree, calling it friction is misleading but just for completeness... If we ignore internal/bearing friction then the forward pointing friction between tyre and road is equal to the rearward forces such as rolling resistance and drag.
     
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