Suppose that yo do 1000 joules of work the you operate a can opener....?

Question

however, the can opener does only 500 joules of work in opening the can.

What is the efficiency of the can opener?

Please explain to me.

Answer 1

50%

.....................(Energy actually used to do the "job")
Efficiency = ------------------------- ---------------------------
.....................(Energy you "gave" to the can opener)

Efficiency = 500 / 1000 = 0.5 = 50%

Answer 2

A Joule = Newton-meter a measure of energy or work.

If you expend 1000 J to accomplish only 500 J of useful work, then you've lost 500 J to heat etc.

Effeiciency is by definition e = output/input; where the output is useful work = 500 J and the input is the total energy expended = 1000 J to get that useful work. Therefore your efficiency is .5 or 50%.

Answer 3

This question doesn't quite make sense to me.

You say you do 1000 j of work to operate the can opener.
The can requires 500 j of work to open it.

Then you are doing 1500 j of work to operate the can opener and open the can.

You need to know what the watt-hours are to operate the opener alone and watt-hours to operate it under load while opening the can. That way you will include any hysterisis loses.