What is the efficiency of the engine?

[Jaybee85]

if a engine must do 500 joules of work to pull a 10 kg load up to a height of 4 meters, what is the efficiency of the engine?

 

[Delphi51]

Figure out the potential energy given to the load. Any additional work is lost.

 

[nlightner]

The efficiency of an engine is defined as the ratio of output work to input work. In this scenario, the output work is 500 joules and the input work is the energy consumed by the engine. We can calculate the input work by multiplying the force applied by the engine (which is equal to the weight of the load) by the distance it is lifted (4 meters). This can be expressed as 10 kg x 9.8 m/s^2 x 4 m = 392 joules.

Therefore, the efficiency of the engine can be calculated as: (output work/input work) x 100% = (500 joules/392 joules) x 100% = 127.55%. This means that the engine is approximately 127.55% efficient in converting the input energy into work. This may seem counterintuitive, but it is possible for an engine to have an efficiency greater than 100% due to factors such as friction and heat loss. In this case, it is likely that the engine is able to generate additional energy from the fuel it consumes, resulting in an efficiency greater than 100%. However, it is important to note that this calculation is based on ideal conditions and may not accurately reflect the actual efficiency of the engine in real-world scenarios.