The efficiency of a vehicle engine

AI Thread Summary
The discussion revolves around calculating the useful energy supplied by a vehicle engine with a power output of 6.2 KW, fueled by a substance releasing 45 MJ/kg. Participants highlight the need to determine the fuel consumption rate to find the energy consumed over time, specifically at a speed of 30 m/s and a fuel efficiency of 18 km/kg. There is confusion regarding the definition of "useful energy," particularly since the scenario involves no change in height, suggesting that energy calculations may be straightforward. Additionally, questions arise about the clarity of the problem statement, especially concerning the time interval needed for calculations. Overall, the conversation emphasizes the importance of precise information for accurate energy and efficiency calculations.
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Homework Statement



A vehicle engine has a power output of 6.2 KW and uses fuel which releases 45 MJ per kilogram when burned. At a speed of 30ms ^-1 on a level road, the fuel usage of the vehicle is 18km per kilogram. Calculate: the useful energy supplied by the engine in this time.

Homework Equations



efficiency= (useful energy) / (energy supplied)

The Attempt at a Solution



Well, first i know that the time taken by the vehicle to travel 18 km at 30ms^-1 is 600s using s=d/t. i also know that (useful energy) is equal to (mg x change in height). the problem is that I am not given the value of mg(N) therefore i have to work it out. so i know F=mxa but again, we're not given the value of m and a. i might have confused my self.
 
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The efficiency is also the rate of work output / rate of energy input. That is easier to work with here. You are given the rate of work output. All you need to do is figure out how much energy it consumes per unit time (denominator).

To calculate the denominator, first work out the number of kilograms of fuel it consumes per second. Then multiply it by the amount of energy per kilogram.

AM
 
heavenjai2007 said:
1.Calculate: the useful energy supplied by the engine in this time.

i also know that (useful energy) is equal to (mg x change in height).


That's an interesting definition of "useful energy". The change in height is zero isn't it? In that case you should be able to work out the useful energy.
 
heavenjai2007 said:

Homework Statement



A vehicle engine has a power output of 6.2 KW and uses fuel which releases 45 MJ per kilogram when burned. At a speed of 30ms ^-1 on a level road, the fuel usage of the vehicle is 18km per kilogram. Calculate: the useful energy supplied by the engine in this time.
Welcome to Physics Forums.

What is the problem actually asking for? The useful energy supplied (as asked in the above post), or the efficiency of the engine (as mentioned in the thread title)?

And if the question is to find the useful energy supplied "in this time", that is troublesome because there is no time interval specified. Nor is it possible to calculate a time interval; one would need more information such as the distance traveled or the available mass of fuel.
 
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