Question

a team of students builds a lever as a science project. they expend 10,300 joules of energy to lift some bricks with the lever. if 6,283 joules of energy are applied to the bricks, what is the levers efficiency? (1 point)
○ 0.61%
○ 164%
○ 61%
○ 39%

Explanation:

Step1: Recall Efficiency Formula

Efficiency (\(\eta\)) of a machine is given by the formula \(\eta=\frac{\text{Useful Energy Output}}{\text{Total Energy Input}}\times100\%\). Here, useful energy applied to bricks is \(6283\) J (output), total energy expended is \(10300\) J (input).

Step2: Substitute Values

Substitute into the formula: \(\eta=\frac{6283}{10300}\times100\%\).

Step3: Calculate the Fraction

First, calculate \(\frac{6283}{10300}\approx0.61\).

Step4: Convert to Percentage

Multiply by \(100\%\): \(0.61\times100\% = 61\%\).

Answer:

61% (corresponding to the option "61%")