I know that gravity does a work of an amount equal to $MGH$ because a more formal definition of work $W$ done by a net force $F$ is $$\int_a^bF(x)dx$$ where $F(x)$ is the net force acting on a specific distance $x$. In the case of the work done by gravity on an object and neglecting friction, buoyant force, etc. the net force is equal to $MG$ (mass x acceleration). Hence $$\int_a^bF(x)dx = -\int_h^0F(h)dh = -MG\int_h^0dh = MGH$$ by letting $H = -(0-h)$ and $F(h)$ is the net force at a specific height $h$.
Suppose I do it the other way around, lifting the object from the ground ($h=0$) upwards and suppose I have to do it in two case scenarios:
A. Apply initial force $F_o \gt MG$ at $h=0$ for it to gain an initial velocity and then at $h>0$, $F(h) = MG$ such that it would only possess a constant velocity.
B. Apply a constant force $F>MG$ and $F-MG=F_n\lt MG$ from the ground upward. In this case, the object is gaining more speed and attains maximum velocity at $h$.
I was told that the work required would still be the same for these two cases, $MGH$, but I don't know if this is valid. I tried doing the same process as above but I get work close to $0$ for case A since the net force is $0$ for $h>0$, and for case B, $$\int_a^bF(x)dx = \int_0^hF(h)dh = F_n\int_0^hdh\lt MGH $$. Is this correct?
Thank you for the reply. Also, I'm sorry for the late response. In the case of the object falling due to the pull of gravity, does it mean that the net work done is $0$ if suppose at $h=H$ and $h=0$, the velocities are both zero using the work-energy theorem? Also, I think it would be more beneficial to "make up" sub-cases for this scenario to further clarify not only my understanding but also for others who don't have the detailed knowledge regarding this topic. Let us assume that at $h=H$, the velocity of the object is 0, and then after gaining kinetic energy, the ball has attained positive velocity at the immediate point after dropping, call this $h = H_i\lt H$. At the point immediately prior to impact $h=H_f\gt 0$, the object should have attained maximum velocity so that at $h=0$, the object is already at rest. Also let the work done be a function of any pair of these four points: $$W=W(P_1,P_2)$$
Of these 6 possible pairs, the following 4 are considered to be the most rewarding: $(H,0)$,$(H,H_f)$,$(H_i,H_f)$,$(H_i,0)$. The work done across these pairs are: $$W(H,0)=0$$ since the initial and final velocities are the same. $$W(H,H_f)=MGH$$ This should be true since the object has lost all of its gravitational potential energy at $h=H_f$. $$0\lt W(H_i,H_f)<MGH$$ In this case $W(H_i,H_f)$ is close to $MGH$. Finally, $$W(H_i,0)<0$$ because the velocity at $H_i$ is greater than the velocity at $h=0$. Also $W(H_i,0)$ is close to 0.
Plotting the Force-Distance graph (I can't show you sorry, you'll have to imagine it) and comparing the above results with the "more formal" definition of work previously considered $W=\int_a^b F(h)dh$, it appears that it is impossible to achieve the result of $W(H,0)=0$ and $W(H_i,0)\lt 0$ because no matter how I set up the integral for the above sub-cases, the result is always $MGH$, that is, the area under the curve of $F(h)$ is always equal to $MGH$ given that the net force on any point between the boundaries, $F(h)=MG$ and in the boundaries, $F(H)=MG$ and at $h=0$,$F(0)$ is greater in magnitude than $MG$ but oppositely directed to stop the object. Does this mean that $\int_a^b F(h)dh$ fails to describe "fully" the work done across two boundaries?