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ashna

  • 2 years ago

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 ?

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  1. AbhimanyuPudi
    • 2 years ago
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    Power = Force x velocity Use this and you can do both on a level road and an inclined road

  2. AbhimanyuPudi
    • 2 years ago
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    got it? If you have any doubt, please specify.

  3. ashna
    • 2 years ago
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    1 second am doing that now

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

  5. AbhimanyuPudi
    • 2 years ago
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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.......

  6. AbhimanyuPudi
    • 2 years ago
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    |dw:1349437062303:dw|

  7. AbhimanyuPudi
    • 2 years ago
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    got it? Is it clear now?

  8. ashna
    • 2 years ago
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    kind of , what is gradient 1 in 10 ?

  9. AbhimanyuPudi
    • 2 years ago
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    tan(theta) = 1/10 i guess..gradient is slope and slope=tan(theta)

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