Question

mr. franklin uses a machine that has an efficiency of 80%. the machine applies a force of 800 n through a distance of 20 m. mr. franklin applies a force of 50 n on the machine. through what distance does mr. franklin apply the force?
4 m
16 m
260 m
400 m

Explanation:

Step1: Recall the efficiency formula

Efficiency ($\eta$) = $\frac{W_{output}}{W_{input}}$, and work ($W$) = $F\times d$ (force times distance). So $\eta=\frac{F_{output}\times d_{output}}{F_{input}\times d_{input}}$.

Step2: Rearrange the formula to solve for $d_{input}$

We know $\eta = 0.8$, $F_{output}=800$ N, $d_{output}=20$ m, $F_{input}=50$ N. Rearranging gives $d_{input}=\frac{F_{output}\times d_{output}}{\eta\times F_{input}}$.

Step3: Substitute the values

$d_{input}=\frac{800\times20}{0.8\times50}=\frac{16000}{40}=400$ m.

Answer:

400 m