# What is the average mechanical power output of the engine?

## Homework Statement

A car with mass 1000 kg accelerates from 0 m/s to 40 m/s in 10.0 s. Ignoring air resistance. The engine has a 22% efficiency, which means that 22% of the thermal energy released by the burning gasoline is converted into mechanical energy.

a) What is the average mechanical power output of the engine?

b) What volume of gasoline is consumed? Assume the burning of 1 L of gasoline releases 46 MJ of thermal energy.

## Homework Equations

$$\Delta$$x = [(v + v0) * t] / 2

Poweravg = work/time = energy transformed / time

Efficiency = Wout/Ein = Pout/Pin

## The Attempt at a Solution

a) $$\Delta$$x = [(v + v0) * t] / 2

=> $$\Delta$$x = 200 m

Work = m g x = 1000*9.81*200 = 1962 kW.

The engine has a 22% efficiency, which means that 22% of the thermal energy released by the burning gasoline is converted into mechanical energy.

Wout = 0.22 * 1962kW = 431.64 kW.

1hp = 746 W => 431.64 kW = 579 hp = WRONG answer

b) I have no ideas to start. What is MJ?

Last edited:

Galileo
Homework Helper

## Homework Equations

$$\Delta$$x = [(v + v0) * t] / 2

Poweravg = work/time = energy transformed / time

Efficiency = Wout/Ein = Pout/Pin

## The Attempt at a Solution

a) $$\Delta$$x = [(v + v0) * t] / 2

=> $$\Delta$$x = 200 m

Work = m g x = 1000*9.81*200 = 1962 kW.

mgx is the work done done on an object by the gravitational force as it falls a vertical distance x. The gravitational force is irrelevant in this problem; the force is supplied by the engine. Also, work has units of energy (Joules) not power (Watts).
A way to find the average power is, indeed, to find the work done by the engine: Work=Fx. You already found x, assuming the accerelation is constant. Since you're looking for average power, that's ok. Then you need to know the average force, which you can find from the average acceleration.
An easier way to do it is to look at the initial and final energy (kinetic only in this case). Then you know the total mechanical energy suppilied in those 10 s, from which you can get the rest.
The engine has a 22% efficiency, which means that 22% of the thermal energy released by the burning gasoline is converted into mechanical energy.

Wout = 0.22 * 1962kW = 431.64 kW.
In (a) you calculated the mechanical energy. The thermal energy is much larger, since the mechanical energy is 22% of this. So you shouldn't take 22% of the answer in a).

b) I have no ideas to start. What is MJ?
MJ is a megajoule: $10^6$ J.

mgx is the work done done on an object by the gravitational force as it falls a vertical distance x. The gravitational force is irrelevant in this problem; the force is supplied by the engine. Also, work has units of energy (Joules) not power (Watts).
A way to find the average power is, indeed, to find the work done by the engine: Work=Fx. You already found x, assuming the accerelation is constant. Since you're looking for average power, that's ok. Then you need to know the average force, which you can find from the average acceleration.
An easier way to do it is to look at the initial and final energy (kinetic only in this case). Then you know the total mechanical energy suppilied in those 10 s, from which you can get the rest.

In (a) you calculated the mechanical energy. The thermal energy is much larger, since the mechanical energy is 22% of this. So you shouldn't take 22% of the answer in a).

MJ is a megajoule: $10^6$ J.

a) Ok. So total mechanical energy suppilied in those 10 s is

W = Kfinal - Kinitial = 1/2 mv2 - 1/2 m(v(0))2 = 1/2*1000*40^2 = 8*105 J.

Then the average mechanical power output of the engine is:

Power average = 8*105 / 10 = 8*10^4 W = 107 hp final answer for (a).

b) The engine has a 22% efficiency, which means that 22% of the thermal energy released by the burning gasoline is converted into mechanical energy.

Thermal Energy = 8*10^5 - (8*10^5 * 0.22) = 624000 J

Assume the burning of 1 L of gasoline releases 46 MJ of thermal energy, so:

624000/(46*10^6) = 0.014 L = answer.

Char. Limit
Gold Member

I'm just not sure if you converted mechanical energy to thermal energy, considering that thermal energy should be over three times as big as mechanical energy, and yours is lower.

Edit: Yeah...

$$E_m=.22E_t$$

Where E_m is mechanical energy and E_t is thermal energy.

I'm just not sure if you converted mechanical energy to thermal energy, considering that thermal energy should be over three times as big as mechanical energy, and yours is lower.

I'm really confused on (b). So

The engine has a 22% efficiency, which means that 22% of the thermal energy released by the burning gasoline is converted into mechanical energy means:

Thermal Energy = 8*10^5 + (8*10^5 * 0.22) = 976000 J

Assume the burning of 1 L of gasoline releases 46 MJ of thermal energy, so:

976000/(46*10^6) = 0.21 L = answer.

How is it???

I'm just not sure if you converted mechanical energy to thermal energy, considering that thermal energy should be over three times as big as mechanical energy, and yours is lower.

Edit: Yeah...

$$E_m=.22E_t$$

Where E_m is mechanical energy and E_t is thermal energy.

=> Et = 8*10^5 / 0.22 = 3636363.636 J

Assume the burning of 1 L of gasoline releases 46 MJ of thermal energy, so:

3636363.636 / (46*10^6) = 0.079 L.

Is this correct?

Finally I got (b) correct. Thanks all ;)