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

step 1: calculate the total work required to move up the box using the formula U = mgh

step 2: calculate: work required/work performed, then convert to a percentage to find the efficiency of the mover

Answer 2

As Vocaloid said, the formula is basically the following.

\[W = F \times D\]

Where,
W = Work
F = Force
D = Distance

The force and the distance must be parallel. Work is equal to force done over some distance. Since the only forces active is gravity, we will calculate the force of gravity, over the vertical height.

\[W = mg \times h\]
Where,
mg = the force of gravity (mass times the acceleration of gravity)
h = height

For a nice pretty number, lets say g = 10. You may get different numbers if you use a more approximate number, such as 9.8.

\[W = 10(25) \times 4\]
\[W = 250 \times 4\]
\[W = 1000 J\]

So the work required to move this box is 1000 Joules.
If the worked used 1250, we can gain a percentage to see how efficient this mover is, as Vocaloid said.

\[\frac{ Work Required }{ Work Done }\]
\[\frac{ 1000 }{ 1250 }\]
\[0.8 = 80%\]

Answer 3

80%

Answer 4

@aketron2