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 ?
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Power = Force x velocity Use this and you can do both on a level road and an inclined road
Not the answer you are looking for? Search for more explanations.
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)