Derive an expression for potential energy near the surface of the earth

Explanation

The gravitatinal potenetial energy (U) at the same height h is equal to the amount of work requried to take the object from ground to the height h with constant veocity. Let us consider a body of mass m being moved from ground to the height h aginst the gravitational force as shown.
The gravitational force `vecF_(2)` , acting on the body is, `vecF_(g)=-mghatj` (as the force is in y direction unit vector `hatj` is used). Here negative sign implies that the force is acting vertically downwards were the body without acceleration (or with constant velocity), an external applied weapol in magnitude but opposite to that of gravitational force `vecF_(g)`, has to be applied on the body i,.e `vecF_(a)=-vecF_(g)` . This implies that F, = + mg). The positive sign implies that sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains unchanged and thus its kinetic energy also remains constant
The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from the ground to that height it.
`U=intvecF_(a).dvecr=underset(0)overset(h)int|vecF_(a)||dvecr|costheta` Since the displacement and the applied force are in the same upward direction, the angle between them, ` theta= 0^(@)" Hence " cos0^(@) = 1 and |vecF_(a)|= mg and |dvecr| = dr.`
`U=mgunderset(0)overset(h)intdr`
`U=mg[r]_(0)^(h)=mgh`