Question

How long will it take a 50w motor to lift a 20 kg object 5 m?

Answer 1

Answer:

50+20=70

70÷5=14

Explanation:

50×20=1000

1000÷5=200

Answer 2

Answer:

Approximately $20\; {\rm s}$, assuming no energy loss and that $g = 10\; {\rm N \cdot kg^{-1}}$.

Explanation:

If the gravitational field strength is constantly $g$, lifting an object of mass $m$  by a height of $\Delta h$ would increase the gravitational potential energy ($\text{GPE}$) of that object by $m\, g\, \Delta h$.

In this question, $m = 20\; {\rm kg}$ and $\Delta h = 5\; {\rm m}$. Assume that $g = 10\; {\rm N \cdot kg^{-1}}$. Hence, the $\text{GPE}$ that this object would eventual gain would be:

$\begin{aligned}m\, g\, \Delta h &= 20\; {\rm kg} \times 10\; {\rm N \cdot kg^{-1}} \times 5\; {\rm m} \\ &= 1000\; {\rm N \cdot m} \\ &= 1000\; {\rm J}\end{aligned}$.

Thus, this motor would need to do (at least) $1000\; {\rm J}$ of work to lift this object up by $5\; {\rm m}$. The power rating of this motor, $P = 50\; {\rm W} = 50\; {\rm J \cdot s^{-1}}$, means that (assuming no energy loss) this motor could do $50\; {\rm J}$ of work every second.

The time it would take for this motor to do $1000\; {\rm J}$ of work under these assumptions would be:

$\begin{aligned}t &= \frac{\text{work}}{\text{power}} \\ &= \frac{1000\; {\rm J}}{50\; {\rm J \cdot s^{-1}}} \\ &= 20\; {\rm s}\end{aligned}$.