Using power to find velocity (a car meets a hill)

[bopll]

Homework Statement

A car encounters an inclined plane

Given- Weight of car in N (6500), velocity on the flat surface (22.5 m/s), power of the engine (78000), incline of the hill (8.1 degrees)

Want to find- velocity on the hill (power and restistive forces remain constant)

Homework Equations

P = Fvcos(theta)

The Attempt at a Solution

I found F by plugging in the power and velocity. I then subtracted the gravitational force due to the hill from this number to get the resultant force. plugged this into P = Fvcos(theta) and got a number bigger than the original...

urgent help would be greatly appreciated since HW is due in 8 minutes, but I'm more worried about the concept for the test tomorrow. thanks.

 

[Astronuc]

The engine power is constant so the car is traveling at constant speed so that resistance is constant.

When going up hill - the car starts increasing its gravitational potential energy - and if the car's power output is constant, then the car's kinetic energy must be decreasing.

 

[bopll]

okay, so if i have to use kinetic energy, does that need i need to find the mass of the car (1/2mv^2)? that doesn't seem right...

 

[ShawnD]

Given- Weight of car in N (6500), velocity on the flat surface (22.5 m/s), power of the engine (78000), incline of the hill (8.1 degrees)

Want to find- velocity on the hill (power and restistive forces remain constant)

edit: Ok I think I have it this time

Your flat ground situation is just P = FV. Move it around to get F = P/V. This is the force coming from the engine; it does not change. When you get on the hill, gravity applies a force against the motor as Wsin(theta). With this new net force, you find the new velocity.

Flat ground:
P = FV (start with this)
F = P/V (solve for force)

Hill:
P = (F + gravity)V

P is the same, F you find out, gravity is Wsin(theta), V is your answer. They are ADDED together because F and gravity represent DRAG as opposed to the force you are applying.

 

[bopll]

edit: Ok I think I have it this time

Your flat ground situation is just P = FV. Move it around to get F = P/V. This is the force coming from the engine; it does not change. When you get on the hill, gravity applies a force against the motor as Wsin(theta). With this new net force, you find the new velocity.

Flat ground:
P = FV (start with this)
F = P/V (solve for force)

Hill:
P = (F + gravity)V

P is the same, F you find out, gravity is Wsin(theta), V is your answer. They are ADDED together because F and gravity represent DRAG as opposed to the force you are applying.

i tried this also, maybe i made a calculation error...