Calculating the Efficiency of a Car Engine: A Scientific Approach

[Izekid]

Here I have a hard one... A car with the mass 1100kg accelerates from 0km/h to 50km/h in 9,0s. In this time intervall the engine exhausts an power of 57kW. How many procent of the power the engine exhauts is used to increase the moveable energy...

For this I take :

Wres = Fres = Wk

Wk= 1/2 M * v^2

First I take Km/h and transform it to m/s 50km/h = 13,8 m/s

then

550kg *13,8^2 = 104742 J Totalty Cars Moveable power

Then to get Fres I use

S=1/2(v-v0)t

S=13,8/2 * 9 = 62,1m

To get Fres I use Wres = F*S

62,1 * 104742 = 6504478,2J = 6504kJ

57kJ/6504kJ=0,0008 = 0,8 procent

Which Is TOTALY WRONG please help me what shall I do!? It shall be around 21%

 

[Doc Al]

You started out OK, but then went off track. How much power actually goes to increasing the speed of the car? Hint: What's the increase in KE? How much time does that take?

 

[Izekid]

The KE is 104742J = 104kJ then shall I take 57/104 which is... 0,54 which is 54% which is also totally wrong ...

 

[Doc Al]

Of course it's wrong. You need to compare power (total) with power (used to accelerate car), not with the KE of the car.

You found the KE; now find the power. (That's why I asked you "How much time does that take?")

 

[Izekid]

Thx man forgot that effect is P = W/T! You should have sade Effect and it should have popped up directly ... well anyway you helped me thanks man
So The answer is 104742/9 = 11638W = 11,6 kW

11,6kW/57 = 0,209= 0,21 :)