Question

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

Answer 1

50+20=70

70÷5=14

Step-by-step explanation:

50×20=1000

1000÷5=200

Answer 2

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

Step-by-step 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} * 10\; {\rm N \cdot kg^(-1)} * 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}`.