Finding Frictional Force and Power Using Work and Energy

[RushBJJ]

Homework Statement

To measure the combined force of friction (rolling friction plus air drag) on a moving car, an automotive engineering team you are on turns off the engine and allows the car to coast down hills of known steepness. The team collects the following data: (1) On a 2.70° hill, the car can coast at a steady 20 m/s. (2) On a 5.50° hill, the steady coasting speed is 30 m/s. The total mass of the car is 1150 kg.
(a) What is the magnitude of the combined force of friction at 20 m/s (F20) and at 30 m/s (F30)?

(b) How much power must the engine deliver to drive the car on a level road at steady speeds of 20 m/s (P20) and 30 m/s (P30)?

(c) The maximum power the engine can deliver is 46 kW. What is the angle of the steepest incline up which the car can maintain a steady 20 m/s?

(d) Assume that the engine delivers the same total useful work from each liter of gas, no matter what the speed. At 20 m/s on a level road, the car gets 14.2 km/L. How many kilometers per liter does it get if it goes 30 m/s instead?

Homework Equations

W(ext) = Delta E(mech) + Delta E(therm)
Delta E(mech) = Delta K + Dela U
Delta E(therm) = F(friction) * s

The Attempt at a Solution

Parts c and d are the the ones I can't figure out.

For part a, I used W(ext) = Delta E(mech) + Delta E(therm).

Since there are no external forces, Delta E(therm) = - Delta E(mech).

Delta E(mech) = Delta K + Delta U, and since the car is coasting at a constant speed, Delta K is 0.

I set the gravitational potential energy to be 0 at the top of the hill, so Delta U = U(final), which is -mgssin(theta) if s is the length of the hypotenuse of the incline.

So then Delta E(therm) = mgssin(theta), and since Delta E(therm) = F(friction) * s, F(friction) = mgsin(theta).

Plugging the numbers, I got that the frictional force at 20 m/s was 531N and at 30 m/s was 1080N, which was correct.

For b, I just used P = F * v to find the power at 20 m/s and 30 m/s.

For c, I plugged the equation for F into the Power equation, which gave me P = mgsin(theta) * v.

Then I solved for theta, and got theta = arcsin(Power / (mgv)).

I tried plugging in 46000 W for power along with the other given information to get theta = 11.8 degrees, and I also tried plugging in my answer for b at 20 m/s in, neither of which were right.

For d, I'm not really sure where to start, or what equations/relationships to use.

Maybe I could use the ratio between the power needed at 20 m/s and 30 m/s somehow?

Any help would be appreciated.

 

[TSny]

Welcome to PF!

For (c), you assumed P = mgsin(theta) * v. Is this the correct expression for the power of the engine? What force should you use here?

For (d), suppose you let L be the number of liters of gas used when the car engine does an amount of work W. How are L and W related? That is, are they directly proportional? inversely proportional? something else?

 

[RushBJJ]

Oh, I see, I plugged in the equation for the frictional force into the power equation for (c), but I'm not sure what the expression for the force of the engine would be.

It would have to involve the angle in some way, right? I'm just not sure how . . .
The force supplied by the engine works in the direction opposite the frictional force, so maybe, using Newton's 2nd law, F[Engine] = F[Friction] - ma? But then what would a be?

For (d), I suppose the amount of work would increase at the same rate as the number of liters used does, so I can set up a ratio between the work and the number of liters?
But I don't have the work for each, I have the power. But since power is change in work over change in time, can I use the power at 20 m/s and 30 m/s in the ratio instead?

I'm still rather befuddled . . .

 

[TSny]

The force supplied by the engine works in the direction opposite the frictional force, so maybe, using Newton's 2nd law, F[Engine] = F[Friction] - ma? But then what would a be?

Yes, use the 2nd law for the forces that act parallel to the incline. Be sure to include all forces that act parallel to the incline. You can assume the car moves at constant speed.

By the way, this would also be a good approach for part (a), although the method that you used is OK.

For (d), I suppose the amount of work would increase at the same rate as the number of liters used does, so I can set up a ratio between the work and the number of liters?

Yes. Is this ratio the same for 20 m/s as for 30 m/s?

But I don't have the work for each, I have the power. But since power is change in work over change in time, can I use the power at 20 m/s and 30 m/s in the ratio instead?

In the ratio of the work to the number of liters, how can you express the work in terms of distance traveled, d? You don't need to use power in this part.

 

[RushBJJ]

Ah, I understand now.

Thanks for all the help!