Answer:
Approximately [tex]20\; {\rm s}[/tex], assuming no energy loss and that [tex]g = 10\; {\rm N \cdot kg^{-1}}[/tex].
Explanation:
If the gravitational field strength is constantly [tex]g[/tex], lifting an object of mass [tex]m[/tex] by a height of [tex]\Delta h[/tex] would increase the gravitational potential energy ([tex]\text{GPE}[/tex]) of that object by [tex]m\, g\, \Delta h[/tex].
In this question, [tex]m = 20\; {\rm kg}[/tex] and [tex]\Delta h = 5\; {\rm m}[/tex]. Assume that [tex]g = 10\; {\rm N \cdot kg^{-1}}[/tex]. Hence, the [tex]\text{GPE}[/tex] that this object would eventual gain would be:
[tex]\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}[/tex].
Thus, this motor would need to do (at least) [tex]1000\; {\rm J}[/tex] of work to lift this object up by [tex]5\; {\rm m}[/tex]. The power rating of this motor, [tex]P = 50\; {\rm W} = 50\; {\rm J \cdot s^{-1}}[/tex], means that (assuming no energy loss) this motor could do [tex]50\; {\rm J}[/tex] of work every second.
The time it would take for this motor to do [tex]1000\; {\rm J}[/tex] of work under these assumptions would be:
[tex]\begin{aligned}t &= \frac{\text{work}}{\text{power}} \\ &= \frac{1000\; {\rm J}}{50\; {\rm J \cdot s^{-1}}} \\ &= 20\; {\rm s}\end{aligned}[/tex].
