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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Power = Force x velocity
Use this and you can do both on a level road and an inclined road
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okay got the first part , what about the 2nd ? how to do that ?
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.......
got it? Is it clear now?
kind of , what is gradient 1 in 10 ?
tan(theta) = 1/10 i guess..gradient is slope and slope=tan(theta)