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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?

UPDATE:

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?

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  • $\begingroup$ Are you familiar with the work-energy theorem? $\endgroup$
    – Bob D
    Aug 29 '19 at 17:49
  • $\begingroup$ I'm not sure I understand the difference between A and B. For scenario A at what point is the force $F_o$ reduced to $MG$? $\endgroup$
    – Bob D
    Aug 29 '19 at 20:09
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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).

I’m assuming you are describing the situation involving a mass that falls freely from height $H$ to the ground in the absence of air friction, and that the mass is initially at rest at $h=H$. If that’s the case, then gravity does positive work equal to $MGH$. The work is positive because the direction of the gravitational force is the same as the direction of motion of the mass. Since no other forces act on the mass, this is the net work done on the mass. According to the work-energy theorem, the net work done on an object equals its change in kinetic energy. Therefore the net work done by gravity will equal the kinetic energy of the mass just prior to impact with the ground, or

$$MGH=\frac{MV^2}{2}$$.

Where $V$ is the velocity upon impact. In essence, the gravitational potential energy of the earth/mass system is converted into kinetic energy.

Suppose I do it the other way around, lifting the object from the ground ($h=0$) upwards

Before addressing your scenarios A and B, which both appear to involve the mass having a velocity when it reaches $H$, consider the following scenario where the mass has no velocity (is at rest) upon reaching $H$:

Let the mass begin at rest ($h=0$). Then you, an external (to the mass/earth system) agent raise the mass and bring it to rest a height $H$. To do this you had to initially exert a force slightly greater than gravity to get it moving and give it some kinetic energy. Once you get it moving you reduce your force to equal the gravitational force so that now the net force is zero and the mass continues to move at constant velocity.

In order to bring the mass to rest at $H$, prior to reaching $H$ you need to reduce your force to less than $MG$ to decelerate it. In doing so the mass loses the initial kinetic energy you gave it, so that the overall change in kinetic energy is zero. Per the work-energy theorem the net work done on the mass is zero. The positive work you did of $+MGH$ equals the negative work done by gravity of $-MGH$, so the net work is zero. The work you did was positive because your force was in the same direction as the motion of the mass. The work gravity did is negative because its force is in the opposite direction of the motion of the mass. In essence, gravity takes the energy you gave the mass and stores it as gravitational potential energy in the earth-mass system.

Scenarios A and B:

The details of both these scenarios and how they differ from each other is unclear to me. I assume the mass starts at rest at $h=0$ for both scenarios. Per the work-energy theorem, the net work done on the mass for either scenario going from $h=0$ to $h=H$ will equal the change in kinetic energy of each mass. That will depend only on the velocity of each mass when it reaches $H$. The details of what goes on between $h=0$ and $h=H$ are irrelevant. Or

$$W_{A}=\frac{MV_{A}^2}{2}$$

$$W_{B}=\frac{MV_{B}^2}{2}$$

Where

$W_A$ and $W_B$ is the net work done on the mass in scenario A and B, respectively

$V_A$ and $V_B$ is the velocity of the mass at $H$ in scenario A and B, respectively

As always, the total energy of a mass is the sum of its kinetic and potential energy. Therefore

$$E_{A(tot)}=\frac{MV_{A}^2}{2}+MGH$$

$$E_{B(tot)}=\frac{MV_{B}^2}{2}+MGH$$

The kinetic energy of each is due to the net work done on the mass in each scenario. The potential energy is due to the negative work done by gravity taking the energy the external agent did raising the mass and storing it as gravitational potential energy in the earth-mass system.

Hope this helps.

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