Answer:
Approximately [tex]1.2 \times 10^{4}\; {\rm J}[/tex] (assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].)
Explanation:
The strength of the gravitational field near the surface of the earth is approximately constant: [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].
The change in the gravitational potential energy ([tex]{\rm GPE}[/tex]) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass [tex]m[/tex] increased by [tex]\Delta h[/tex], the [tex]{\rm GPE}[/tex] of that object would have increased by [tex]m\, g\, \Delta h[/tex].
In this question, the height of this object increased by [tex]\Delta h = 6.0\; {\rm m}[/tex]. The mass of this object is [tex]m = 200\; {\rm kg}[/tex]. Thus, the [tex]{\rm GPE}[/tex] of this object would have increased by:
[tex]\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}[/tex].
(Note that [tex]1\; {\rm N \cdot m} = 1\; {\rm J}[/tex].)
