## ashna 3 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 ?

1. AbhimanyuPudi

Power = Force x velocity Use this and you can do both on a level road and an inclined road

2. AbhimanyuPudi

got it? If you have any doubt, please specify.

3. ashna

1 second am doing that now

4. ashna

okay got the first part , what about the 2nd ? how to do that ?

5. AbhimanyuPudi

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

|dw:1349437062303:dw|

7. AbhimanyuPudi

got it? Is it clear now?

8. ashna

kind of , what is gradient 1 in 10 ?

9. AbhimanyuPudi

tan(theta) = 1/10 i guess..gradient is slope and slope=tan(theta)