What is the efficiency of a lever if it uses 40J of input work to complete 80J of work?

To determine the efficiency of a lever, one can use the formula for mechanical efficiency, which is the ratio of the output work to the input work, expressed as a percentage. In this case, the lever takes 40 joules of input work and produces 80 joules of output work.

Efficiency is calculated as:

Efficiency = (Output Work / Input Work) × 100%

Plugging in the values from the problem:

Efficiency = (80 J / 40 J) × 100% = 2 × 100% = 200%

This indicates that the lever is producing more output work than it is inputting, which is theoretically impossible because it violates the conservation of energy principle. It shows that the given work values might have been interpreted incorrectly.

Typically, the efficiency of a machine cannot exceed 100%, meaning it cannot output more energy than it consumes. Therefore, an efficiency of 200% would indicate a misunderstanding of how work and efficiency are calculated.

Normally for typical situations: if input is less than output proportionately, you'd categorize the lever as having optimal performance or approaching ideal efficiency. In reality, the efficiency percentage can fall between 0% - 100%, based on real-world conditions experiencing friction and other losses