Wattage an of engine sufficient to move a car

[razidan]

I am tutoring a student, and this question was given to her. I believe it is not clearly stated and wanted your opinion. I hope this is the best forum for this.

1. Homework Statement
What is the minimum wattage required to accelerate a car of mass m (don't remember the value) from 0 to 28 m/s in 6 seconds, on an incline of 10 degrees, with a constant drag force of 150N.

Homework Equations

The Attempt at a Solution

The solution, as was given by the teacher was: $P=F\cdot\frac{\Delta x}{\Delta t}$ where F is the net force (pointing downhill), $|F|=F_{drag} +mgsin(\alpha)$, and the average speed can be determined from kinematics equations.

I believe this is wrong. the question states that car is accelerated to 28 m/s. meaning, at the instant of t=6 sec, it has a velocity of 28 m/s, and thus some engine power went to kinetic energy. then the minimum wattage needed would have to have an addition of $\frac{\frac{mv^2}{2}}{6 sec}$ to the power that was calculated above.

Am i correct?

Thanks,
R

 

[jack action]

the question states that car is accelerated to 28 m/s. meaning, at the instant of t=6 sec, it has a velocity of 28 m/s

How do you know it is at t = 6 s? No acceleration is stated. And I don't think it is relevant to the problem.

I agree with your power and force equations, but don't agree with the use of the average velocity to find the minimum power required. The minimum power required will be the power required at maximum velocity where the acceleration will be zero. If the available power is lower than that value, then the car will never reach the maximum speed (i.e. the acceleration will be zero at a lower speed).

 

[haruspex]

It says nothing about constant acceleration. If it is constant then the power demand increases, so then it is unclear whether we are being asked for peak power or average power.
if the acceleration varies then we need to know the height gain.

 

[razidan]

How do you know it is at t = 6 s? No acceleration is stated. And I don't think it is relevant to the problem.

I agree with your power and force equations, but don't agree with the use of the average velocity to find the minimum power required. The minimum power required will be the power required at maximum velocity where the acceleration will be zero. If the available power is lower than that value, then the car will never reach the maximum speed (i.e. the acceleration will be zero at a lower speed).

Sorry, i edited the question. the acceleration is from 0 to 28 m/s in 6 seconds.

 

[jack action]

Minimum power (if it is what you mean by "wattage") can be calculated at the maximum velocity with no acceleration (i.e. your equation of force). Whatever happened before that, at a lower velocity, can be anything. The power can be equal or lower at anytime during the acceleration process.

It can also be higher if there are large accelerations during the journey. If you add the $ma$ component to your force equation, it is possible that the minimum power required at some velocity exceeds the minimum power required at maximum velocity. But you need to know the acceleration vs velocity relationship to determine power vs velocity over the entire journey.

Is the acceleration constant or is it varying with velocity?

 

[BvU]

Is the acceleration constant or is it varying with velocity?

Whatever is cheaper in power, as per the problem statement ...

 

[BvU]

And I fully agree with @razidan . Average acceleration $a$ is 28 /6 m/s², gravity adds $g\sin\alpha$.
Force $F = m(a + g\sin\alpha) + 150$ N.
SUVAT gives us distance d and $F\cdot d / t$ the average power. Less means it takes longer to reach 28 m/s. My own car bites the dust, but not by a great deal.

 

[razidan]

How do you know it is at t = 6 s? No acceleration is stated. And I don't think it is relevant to the problem.

I agree with your power and force equations, but don't agree with the use of the average velocity to find the minimum power required. The minimum power required will be the power required at maximum velocity where the acceleration will be zero. If the available power is lower than that value, then the car will never reach the maximum speed (i.e. the acceleration will be zero at a lower speed).

It says nothing about constant acceleration. If it is constant then the power demand increases, so then it is unclear whether we are being asked for peak power or average power.
if the acceleration varies then we need to know the height gain.

Given that this is high school level, let's assume a constant acceleration.

Is the power just the force times the average speed?
The engine had to provide enough power against the forces as well as providing kinetic energy.

 

[CWatters]

No not average power.

power = force * velocity

If we assume constant acceleration then it must still have enough power to accelerate at that rate when at max velocity.

 

[razidan]

No,

power = force * velocity

If we assume constant acceleration then it must still have enough power to accelerate at that rate when at max velocity.

here is how i think about this, and please tell me why it's wrong:
the initial energy is zero. the energy the car has after it has reached the max velocity at some point x along the incline is $E_{tot}=mgh+F_{drag}x+\frac{mv^2}{2} = mgxsin{\alpha}+F_{drag}x+\frac{mv^2}{2}$.
This energy was provided by the engine in the duration of 6 seconds. therefore, the average power provided by the engine is $P=\frac{\Delta E}{\Delta t}=\frac{E_{tot}}{6}$.
The difference between the two approaches (F*v vs the change in energy) is the addition of kinetic energy.

 

[jbriggs444]

There is a tricky (cheating?) technique for solving this.

Set your engine to compressing a spring of arbitrarily high strength and outfit the car with tires of arbitrarily strong grip. Set the brakes. After six seconds, release the brakes and quickly accelerate to 28 m/s.

The impulsive acceleration is insensitive to both slope and drag. The required engine power is easily calculated as the required kinetic energy divided by the allowed time.

 

[jbriggs444]

SUVAT gives us distance

For variable acceleration, it does not.

 

[haruspex]

Whatever is cheaper in power, as per the problem statement ...

Good point. If we take that literally there is enough information to solve for constant power. However, it leads to the equation $v+\frac{Gt}{m}=\frac PG\ln(1-\frac{vG}P)$, where G is the total downslope force from gravity and drag. So no closed form solution for P.

 

[jack action]

here is how i think about this, and please tell me why it's wrong:
the initial energy is zero. the energy the car has after it has reached the max velocity at some point x along the incline is $E_{tot}=mgh+F_{drag}x+\frac{mv^2}{2} = mgxsin{\alpha}+F_{drag}x+\frac{mv^2}{2}$.
This energy was provided by the engine in the duration of 6 seconds. therefore, the average power provided by the engine is $P=\frac{\Delta E}{\Delta t}=\frac{E_{tot}}{6}$.
The difference between the two approaches (F*v vs the change in energy) is the addition of kinetic energy.

This will give you the average power, but not the maximum power. It is the equivalent of doing $P_{avg}=F_{avg}v_{avg}$ ¹ where $F_{avg}$ is calculated based on $a_{avg}$.

If the acceleration is constant, then $P_{max}=F_{avg}v_{max}$, since the acceleration is the same at any speed. One can easily see that the maximum power is twice the average power, thus the minimum required.

¹ $P_{avg}= \frac{\Delta E}{\Delta t}= \frac{F_{avg}\Delta x}{\Delta t}=F_{avg}v_{avg}$

 

[haruspex]

here is how i think about this, and please tell me why it's wrong:
the initial energy is zero. the energy the car has after it has reached the max velocity at some point x along the incline is $E_{tot}=mgh+F_{drag}x+\frac{mv^2}{2} = mgxsin{\alpha}+F_{drag}x+\frac{mv^2}{2}$.
This energy was provided by the engine in the duration of 6 seconds. therefore, the average power provided by the engine is $P=\frac{\Delta E}{\Delta t}=\frac{E_{tot}}{6}$.
The difference between the two approaches (F*v vs the change in energy) is the addition of kinetic energy.

If the question intends minimum average power then it is poorly worded. "Minimum wattage" implies to me that we have an electric motor with a quoted power output. That will be its peak, not its average.
If it intends constant acceleration then there was no need to specify minimum power. As has been noted, you would simply find the acceleration, then the power required to maintain that acceleration having reached the given speed.

 

[BvU]

There is a tricky (cheating?) technique for solving this.
The impulsive acceleration is insensitive to both slope and drag. The required engine power is easily calculated as the required kinetic energy divided by the allowed time.

This may be slightly deadly for any passengers, but the scenario is well within the bounds of the problem statement: "to accelerate a car". It sure means teacher is dead wrong and razidan comes out shining !

Teacher can be convinced by asking for the wattage for a horizontal road and no friction. Zero ?

I notice engineering toolbox evaluates the same average expression, no matter how big $\Delta v$ or $\Delta t$ (the word average appears only once, in small print and much earlier on the page).

Alll in all an unfortunate idea for a simple exercise in a simple mechanics course.

And I wholeheartedly agree with @jack action #14 and @haruspex #15.

Also have to confess my car has only a fraction of the power needed for this (#7)