Question

1. calculating efficiency the input work you do on a can opener is 12 j. the output work the can opener does is 6 j. what is the efficiency of the can opener? 5. calculating efficiency suppose the efficiency of a manual pencil sharpener is 58%. if the output work needed to sharpen a pencil is 4.8 j, how much input work must you do to sharpen the pencil?

Explanation:

Step1: Recall efficiency formula

Efficiency ($\eta$) is given by $\eta=\frac{W_{output}}{W_{input}}\times100\%$, where $W_{output}$ is output work and $W_{input}$ is input work.

Step2: Solve problem 4

Given $W_{input} = 12\ J$ and $W_{output}=6\ J$. Substitute into the formula: $\eta=\frac{6}{12}\times 100\%$.
$\eta = 50\%$

Step3: Solve problem 5

Given $\eta = 58\%=0.58$ and $W_{output}=4.8\ J$. Rearrange the formula $\eta=\frac{W_{output}}{W_{input}}$ to solve for $W_{input}$. So $W_{input}=\frac{W_{output}}{\eta}$.
$W_{input}=\frac{4.8}{0.58}\approx8.28\ J$

Answer:

1. The efficiency of the can - opener is $50\%$.
2. The input work needed to sharpen the pencil is approximately $8.28\ J$.