Question

A box weighing 100 newtons is pushed up an inclined plane that is 5 meters long. It takes a force of 75 newtons to push it to the top, which has a height of 3 meters. Work Output? Work Input? Efficiency?

Answer 1

The work output can be given as,

`W_O=Fh`

Substitute the known values,

`\begin{gathered} W_O=(75\text{ N)(3 m)(}\frac{1\text{ J}}{1\text{ Nm}}) \\ =225\text{ J} \end{gathered}`

Thus, the work output of the procedure is 225 J.

The work input of the procedure is,

`W_i=wd`

Substitute the known values,

`\begin{gathered} W_i=(100\text{ N)(5 m)(}\frac{1\text{ J}}{1\text{ Nm}}) \\ =500\text{ J} \end{gathered}`

Thus, the work input of the procedure is 225 J.

The efficiency of the work done is given as,

`e=(W_O)/(W_i)`

Substitute the known values,

`\begin{gathered} e=\frac{225\text{ J}}{500\text{ J}} \\ =0.45 \end{gathered}`

Thus, the efficiency of procedure is 0.45 or 45%.