# Car going up the slope, Power of Engine

1. Apr 1, 2012

### CroSinus

1. The problem statement, all variables and given/known data
Hello everybody,

can you help me with the following problem please. A car is going up the slope with a constant acceleration a (unknown). The friction force equals to Ffr = const. + 0,7v^2, where v is the velocity of the car. The angle of the slope is given and equals $\theta$. The mass of the car is given and equals m. What is the power of the car engine?

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Apr 1, 2012

### LawrenceC

You must first show your attempt before anyone on this forum can help. That said, this is a problem where keeping aware of units behooves you.

3. Apr 1, 2012

### CroSinus

Well, here's my attempt. The car engine has to overcome three forces. The first one is the force of the slope which equals G*sin theta. The second one is the friction force which is velocity dependent and equals Ffr = const. + 0,7v^2. The car engine has also to accelerate the vehicle. So the accelerating force equals to m*a. Where m is the car mass and a is the acceleration of the car.

In order to find the power of the engine I write: P = W/t. Power equals work divided by time. So I have three forces: G*sin theta + const. + 0,7*v^2 + m*a and I need the path of the car calculated as the length of the slope on which the car is going upwards.

Am I on the right path? Any hints please?

Thanks,
CroSinus

Last edited: Apr 1, 2012
4. Apr 2, 2012

### LawrenceC

Hint:

In SAE units, power is ft-lbs/sec which can also be written as lbs-(ft/sec).

5. Apr 3, 2012

### CroSinus

Thanks for trying to help me.

Well, power is defined as P = work done / (time taken to do work). P = dW/dt.
dW = F*dr
P = F dr/dt = F*v

Can you explain to me the meaning of the equation P = F*v. I think I can use this equation only when the Force is constant. But in my case F is changing with speed (F = cons. + 0,7v^2), or better to say with time. How can I apply the equation P = F*v then? Please help me!

Thank you very much,
CroSinus

6. Apr 3, 2012

### LawrenceC

Your power requirement will have several components. One is drag, another is the fact that it is accelerating, another is due to the vehicle increasing its potential energy.

Because you have drag as a function of velocity, the power requirement necessarily must be a function of velocity. It must increase as speed increases due to friction force increasing with velocity. A similar argument can be made for the rate of change of potential energy which is power.

"Can you explain to me the meaning of the equation P = F*v. I think I can use this equation only when the Force is constant."

If you stick your hand out of a car window when the car is moving 60 mph (88ft/sec), the power used is 88*F ft-lbs/sec. F would come from drag coefficient, air density, frontal area, and the velocity squared. So the force is speed dependent which makes the power requirement a function of the velocity cubed.

7. Apr 3, 2012

### CroSinus

Thank you very much for you answer. But how can I calculate power from the data given in my example. A) angle of the slope (theta) B) Ffr = 218 N + 0,7v^2 C) car mass = m

Thanks for help,
CroSinus

8. Apr 3, 2012

### LawrenceC

Because they do not provide you with actual numbers, your answer will have to be in the form of a function of V,m,theta, and a. Obviously, if theta or mass or acceleration increase, the power requirement will increase.

If theta is larger, the rate of change of potential energy increases.
If acceleration is greater, the power requirement to cause that increases due to Newton's second law.
If the vehicle has greater mass, the power requirement increases due to rate of change of potential energy as well as Newton's second law (more mass involved).
As V increases, the power requirement increases due to drag and rate of change of potential energy.

If they provided you with the actual numbers for theta, V, mass, and acc., you could come up with a number for the power needed at that instant in time.

9. Apr 3, 2012

### CroSinus

Great. Thank you very much again.