How Much Power is Needed for a Car to Climb an 8% Grade?

[falcon0311]

1700kg automobile moving at a constant speed of 15 m/s, the motor supplies 16 kW of power to overcome friction, wind resistance, etc. What power must be supplied if the car is to move up an 8.0% grade (8.0 m vertically for every 100 m horizontally).

If power = force x velocity, can I somehow use gravity to help find power?

I'm mainly looking for a jumpstart as I seem to be having a lot of problems with energy problems. Any help is much appreciated. Thank you.

 

[cookiemonster]

What's the force that's resisting the car's motion?

cookiemonster

 

[gnome]

Power is also energy (or work) per unit time.

 

[falcon0311]

The force acting against it moving up the incline would be gravity (I'm going to assume the friction's the same as before). I've got the angle of the incline as 4.57 degrees or something like that. So the force of gravity acts against it as it moves up this small incline, the engine has to produce a little more power, right?

 

[cookiemonster]

Correct.

cookiemonster

 

[gnome]

Think about what happens to the car's energy as it climbs at constant speed.

 

[falcon0311]

Think about what happens to the car's energy as it climbs at constant speed.

It loses kinetic energy and gains potential energy? I'm not sure what set of tracks you're trying to get me on.

For some reason I keep coming up with 35.9 kW. Does this seem right to anyone?

I take mg*sin(theta) and add that to the original force I got from taking 16000 W / 15 m/s = 1067 N + 1327 N = 2394 N * 15 m/s = 35.9 kW.

 

[gnome]

At constant speed kinetic energy is constant. Only potential energy changes.

I get the same result this way:

Looking at the ramp (or whatever) as a right triangle you have the horizontal leg 100 m, the vertical leg 8 m, so the sloped surface is √(10064) & sin$$\theta$$ = 8/√(10064).

The vertical component of velocity v_y=15sin$$\theta$$=15*8/√(10064) = 1.196m/s

Potential energy u = mgh
Rate of increase of potential energy:
u/t = mgh/t = mgv_y = 1700*9.8*1.196=19.9kJ/s=19.9kW
this is the amount of power needed to increase height at that speed at that angle.
Add to the power needed to overcome friction, etc. which you already had:
16 + 19.9 =35.9kW

 

[falcon0311]

Thanks for the help!