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 force and \(d\) is distance. So, \(\eta=\frac{W_{out}}{W_{in}}\times100\%=\frac{F_{out}\times d_{out}}{F_{in}\times d_{in}}\times100\%\)

Step2: Identify the given values

\(F_{out} = 121\,\text{N}\), \(d_{out} = 3.6\,\text{m}\), \(F_{in} = 100\,\text{N}\), \(d_{in} = 8.5\,\text{m}\)

Step3: Calculate output work (\(W_{out}\))

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

Step4: Calculate input work (\(W_{in}\))

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

Step5: Calculate efficiency

\(\eta=\frac{W_{out}}{W_{in}}\times100\%=\frac{435.6}{850}\times100\%\)
First, calculate \(\frac{435.6}{850}=0.51247\) (approx)
Then, multiply by 100%: \(0.51247\times100\% = 51.25\%\) (approx)

Answer:

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