Question

1. a machine produces 121n of force when 100n is input. this mechanical device moves 3.6m of each 8.5m of input force. what is the machine’s efficiency?

Explanation:

Step1: Recall the formula for machine efficiency

The efficiency (\(\eta\)) of a machine is given by the ratio of output work (\(W_{out}\)) to input work (\(W_{in}\)) multiplied by 100% to get a percentage. The work done is calculated as \(W = F \times d\), where \(F\) is the force and \(d\) is the distance moved. So, \(\eta=\frac{W_{out}}{W_{in}}\times100\%=\frac{F_{out}\times d_{out}}{F_{in}\times d_{in}}\times100\%\).

Given: \(F_{out} = 121\space N\), \(d_{out}=3.6\space m\), \(F_{in}=100\space N\), \(d_{in}=8.5\space m\) (assuming the "8.5m of input force" is a typo and should be "8.5m of input distance" as force distance is needed for work).

Step2: Calculate output work

\(W_{out}=F_{out}\times d_{out}=121\space N\times3.6\space m = 435.6\space J\)

Step3: Calculate input work

\(W_{in}=F_{in}\times d_{in}=100\space N\times8.5\space m = 850\space J\)

Step4: Calculate efficiency

\(\eta=\frac{W_{out}}{W_{in}}\times100\%=\frac{435.6}{850}\times100\%\)
\(\eta = 0.51247\times100\% \approx 51.25\%\)

Answer:

The machine's efficiency is approximately \(51.25\%\)