Question

A mover uses 1250 J of energy to move a 25 kg box up an inclined plane to a height of 4 m. What was the mover’s efficiency while moving the box?

Answer 1

Moved to Physics.

Answer 2

** mover’s efficiency ** = meaningless metric. we can re-ask the question to see why.

so we instead query the efficiency of the lifting process. and then the question compares the energy \(E\) given to the box, where \(E = \text{1250 J}\), compared to the actual increase in energy of the box.

**assuming** that the only increase is the increase is in it's gravitational potential energy, \(\Delta U\), so it ends up at rest/ has no added kinetic energy ....on that basis, we can say that its increase in PE is: \(\Delta U = m g \Delta h\)

we also define efficiency as \(\eta = \dfrac{\text{actual increase in energy of the box}}{\text{energy given to the box}}\)

And so:

\(\eta = \dfrac{25 \cdot g \cdot 4}{1250} = \text{depends on what you count "g" as}\)

That's the efficiency of the process of moving the box. Nothing to do with the mover.