Question

1. a motor uses 1,000 j of energy to lift a 20 kg object to a height of 3 m. what is the efficiency of the motor?

Explanation:

Step1: Calculate useful work done

The formula for gravitational - potential energy (useful work $W_{useful}$) is $W = mgh$, where $m = 20\ kg$, $g= 9.8\ m/s^{2}$, and $h = 3\ m$. So $W_{useful}=mgh=20\times9.8\times3= 588\ J$.

Step2: Calculate the efficiency

The formula for efficiency $\eta$ of a machine is $\eta=\frac{W_{useful}}{W_{input}}\times100\%$. Here, $W_{input}=1000\ J$. So $\eta=\frac{588}{1000}\times100\% = 58.8\%$.

Answer:

$58.8\%$