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ashna Group Title

An engine pulls a 1500kg car on a level road at a constant speed of 5m/s against a frictional force of 500N Calculate the power expended by the engine . What extra power has the engine to expend in order to maintain the same speed of the car up an inclined plane having a gradient of 1 in 10 ?

  • one year ago
  • one year ago

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  1. AbhimanyuPudi Group Title
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    Power = Force x velocity Use this and you can do both on a level road and an inclined road

    • one year ago
  2. AbhimanyuPudi Group Title
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    got it? If you have any doubt, please specify.

    • one year ago
  3. ashna Group Title
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    1 second am doing that now

    • one year ago
  4. ashna Group Title
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    okay got the first part , what about the 2nd ? how to do that ?

    • one year ago
  5. AbhimanyuPudi Group Title
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    Resolve the weight mg of the car into components; one through the inclined plane and the other in the plane of the Normal Contact force....you will get N=mg x cos(theta) and the force acting opposite to the motion of thecar will be frictional force + (mg x sin(theta) ) So, now you have an extra force of mgxsin(theta) opposing the car's motion other than the frictional force...Extra power = Extra Force x velocity.......

    • one year ago
  6. AbhimanyuPudi Group Title
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    |dw:1349437062303:dw|

    • one year ago
  7. AbhimanyuPudi Group Title
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    got it? Is it clear now?

    • one year ago
  8. ashna Group Title
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    kind of , what is gradient 1 in 10 ?

    • one year ago
  9. AbhimanyuPudi Group Title
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    tan(theta) = 1/10 i guess..gradient is slope and slope=tan(theta)

    • one year ago
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