Question

A 100-watt motor (60% efficient) is available to raise a load 5 meters into the air. If the task takes 65 seconds to complete, how heavy was the load in units of kilograms?

Answer 1

Efficiency

$n=\frac{P_{out}}{P_{in}}$

if

the power supplied to the motor

$P_{in}=100W$
efficiency

$n=0.60$

finally

$n=\frac{P_{out}}{P_{in}}\Rightarrow P_{out}=P_{in}\cdot n=100W\cdot 0.60=60W$

the potency is equal to

$P_{out}=\frac{E}{t}$

where:

energy is equal to the potential energy

$E=m\cdot g\cdot h$
time

$t=65s$
mass expression

$P_{out}=\frac{m\cdot g\cdot h}{t}\Rightarrow m=\frac{P_{out}\cdot t}{ g\cdot h}$

evaluating numerically

$m=\frac{60W\cdot 65s}{ 9.81\frac{m}{s^{2}}\cdot 5m}=79.5kg$