Question

efficiency of machines quick check
ronald pushes a box up an inclined plane with 30% efficiency. if 254 joules of energy went into moving the box, how much energy did ronald put into the inclined p
154.2 joules
177.8 joules
76.2 joules
76.4 joules

Explanation:

Step1: Recall Efficiency Formula

Efficiency ($\eta$) is given by $\eta = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\%$. Here, useful energy (moving the box) is 254 J, efficiency is 30% (0.3), and we need to find total energy input (energy Ronald put in).

Step2: Rearrange Formula

From $\eta = \frac{E_{useful}}{E_{input}} \times 100\%$, we get $E_{input} = \frac{E_{useful}}{\eta/100\%}$. Substitute $E_{useful} = 254$ J and $\eta = 30\%$ (0.3): $E_{input} = \frac{254}{0.3} \approx 846.67$ J? Wait, no, wait—wait, the problem says "30% efficiency", useful energy is 254 J? Wait, no, maybe I misread. Wait, the question is: Ronald pushes a box up an inclined plane with 30% efficiency. If 254 joules of energy went into moving the box (useful energy), how much energy did Ronald put into the inclined plane (total input energy)? Wait, no, efficiency is (useful energy)/(input energy) *100. So input energy = useful energy / (efficiency/100). So 254 / 0.3 ≈ 846.67, but the options—wait, the options are 154 J, 180 J, 847 J, 76 J? Wait, maybe the useful energy is the output, and input is what we need. Wait, let's recalculate: $\eta = 30\% = 0.3$, $E_{useful} = 254$ J. So $0.3 = \frac{254}{E_{input}} \implies E_{input} = \frac{254}{0.3} \approx 846.67 \approx 847$ J. So the correct option is the one with 847 J (assuming the option is 847 J, like "847 joules").

Answer:

The correct option (assuming the option is labeled with 847 joules, e.g., if the option is "847 joules", then that's the answer. Based on the calculation, the energy Ronald put in is approximately 847 joules.