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anonymous
 4 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 ?
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Power = Force x velocity Use this and you can do both on a level road and an inclined road

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0got it? If you have any doubt, please specify.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01 second am doing that now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay got the first part , what about the 2nd ? how to do that ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Resolve 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.......

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1349437062303:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0got it? Is it clear now?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0kind of , what is gradient 1 in 10 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0tan(theta) = 1/10 i guess..gradient is slope and slope=tan(theta)
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