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 the output work (\(W_{out}\)) to the input work (\(W_{in}\)) multiplied by 100% to get a percentage. The work done is calculated as force (\(F\)) multiplied by distance (\(d\)), so \(W = F\times d\). Thus, \(\eta=\frac{W_{out}}{W_{in}}\times100\%=\frac{F_{out}\times d_{out}}{F_{in}\times d_{in}}\times100\%\)

Step2: Identify the given values

We are given: \(F_{in} = 100N\), \(d_{in}=8.5m\), \(F_{out}=121N\), \(d_{out}=3.6m\)

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

\(W_{out}=F_{out}\times d_{out}=121N\times3.6m = 435.6J\) (Joules, since \(1N\times1m = 1J\))

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

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

Step5: Calculate the efficiency

\(\eta=\frac{W_{out}}{W_{in}}\times100\%=\frac{435.6J}{850J}\times100\%\)
First, calculate \(\frac{435.6}{850}=0.51247\) (approx)
Then, multiply by 100%: \(0.51247\times100\% = 51.247\%\approx51.25\%\)

Answer:

The machine's efficiency is approximately \(51.25\%\) (or more precisely, using the exact values: \(\frac{121\times3.6}{100\times8.5}\times100\%=\frac{435.6}{850}\times100\% = 51.247\%\approx51.25\%\))