Question

A person in a wheelchair exerts a force of 25 newtons to go up a ramp that is 10 meters long. The weight of the person and wheelchair is 60 newtons and the height of the ramp is 3 meters. Work Output? Work Input? Efficiency?

Answer 1

The work output of the procedure is given as,

`W_O=wh`

Substitute the known values,

`\begin{gathered} W_O=(60\text{ N)(3 m)(}\frac{1\text{ J}}{1\text{ Nm}}) \\ =180\text{ J} \end{gathered}`

Thus, the work output is 180 J.

The work input of the procedure is,

`W_i=Fd`

Substitute the known values,

`\begin{gathered} W_i=(25\text{ N)(10 m)(}\frac{1\text{ J}}{1\text{ Nm}}) \\ =250\text{ J} \end{gathered}`

Thus, the work input is 250 J.

The efficiency of the person is given as,

`e=(W_O)/(W_i)`

Substitute the known values,

`\begin{gathered} e=\frac{180\text{ J}}{250\text{ J}} \\ =0.72 \end{gathered}`

Thus, the efficiency of person is 0.72 or 72%.