Why is work done by force $+mgh$ in the situation of throwing something up?

Question

If there is a particle at point A(at rest) and a force moves it to point B(Above point A vertically)(final velocity = 0 at this point), the work done by gravity is $-mgh$. This I understand as the force of $mg$ is downwards while displacement is upwards so negative work is done. But the work done by the force that brings it above to point B is $+mgh$. I dont think its necessary. That will mean that the force making it move up is $+mg$ but it can also be something else. This is the part that I do not get. Maybe its very simple and I am not getting the concept. Can somebody explain why is that both intuitively and by theory? refer to https://www.youtube.com/watch?v=9gUdDM6LZGo at 5:40

Answer 1

You are right, the work applied is not necessarily equal to $mgh$. To see this, let's look at energy conservation:

$$K_0+U_0+W_{\text{n.c.}}=K_f+U_f$$

where $K_0$ and $K_f$ are the initial and final kinetic energy, respectively, $U_0$ and $U_f$ are the initial and final potential energy, respectively, and $W_{\text{n.c.}}$ is the work done by any non-conservative forces.

Assuming the only potential energy in the system is gravitational potential energy $U=mgh$ where $h=0$ is at the starting position, and saying $W_{\text{n.c.}}=W_\text{you}$, then if $W_{\text{n.c.}}=mgh$ we get

$$K_0+0+mgh=K_f+mgh$$ $$K_0=K_f$$

In other words, the object needs to be moving at the same speed at the start and at the end of the path (although it can change in the middle). Otherwise, $W_{\text{n.c.}}\neq mgh$

As an aside, note that if we did have $W_{\text{n.c.}}= mgh$, that wouldn't necessarily mean the applied force is equal to $mg$, although certainly in the case where the forces are equal we would have $W_{\text{n.c.}}=mgh$

Answer 2

System: particle.

Two forces acting on the particle, gravitational attraction due to Earth, magnitude $mg$, downwards and external force exerted by you, magnitude $mg$, upwards.
If the particles rises a vertical distance $h$ then the work done by the gravitational force is $-mgh$ and that by you is $+mgh$.
Thus the net work done on the particle is zero and hence its change in kinetic energy is zero.

If the particles is thrown upwards with a velocity $v$ and rises a maximum height $h$, ie has zero kinetic energy at that position, then the work done on the particle by the gravitational field of the Earth is $-mgh$ and the change in kinetic energy is $0-mv^2/2=-mv^2/2.

Answer 3

But the work done by the force that brings it above to point B is $+mgh$. I dont think its necessary. That will mean that the force making it move up is $+mg$ but it can also be something else.

If the particle starts and ends at rest, then the work done by the external agent responsible for raising the object must be $mgh$ since the net work done must be zero. The negative work done by gravity (a conservative force) is $-mgh$ and depends only on the displacement, and not how the particle got there or whether or not it is at rest. But the upward force applied by the external agent causing the particle to rise need not be $mg$.

If the force is constantly acting up (say a person is lifting the particle) with the the particle initially at rest at point A and ending at rest at point B which, based on what you said, is a height $h$ above A, then the average upward lifting force is $mg$.

I say average because the lifting force must be initially greater than $mg$ in order to accelerate the particle from rest, and then become less than $mg$ before reaching $h$ such that the particle decelerates and comes to rest at $h$. That requires the average force to be $mg$ for a change in kinetic energy of zero.

On the other hand, if you are throwing the particle up and releasing it before reaching $h$, it means that the work done by the thrower is only done while the force is applied, and equals $F_{ave}d$, where $F_{ave}$ is the average upward throwing force and $d$ is the vertical displacement of the particle while the force is applied. This work must equal the increase in potential energy, $mgh$, as the particle has no kinetic energy at $h$ and there is no change in kinetic energy.

So for the thrown particle

$$F_{ave}d=mgh$$

$$F_{ave}=\frac{h}{d}mg$$

Where $d\lt h$.

Therefore, $F_{ave}\gt mg$.

Note that the net work done (by the person throwing the object and gravity) on the particle is still zero since the change in kinetic energy is zero. But average force must be greater than $mg$ only because the force is not applied over the entire vertical displacement.

Hope this helps.

Answer 4

I don't quite understand the question. You have a force $\vec{F}$ acting on the object to take it from A to B. This force is an external force, which goes against the force of gravity, and also this force is not conservative, so it will not have an associated potential. However you can calculate the gravitational potential in each case since the gravitational force is conservative. You can calculate it as $\int \vec{F} d\vec{r}$. In the case of taking the ball from A to B, you will have negative $\vec{F}$ and positive $d\vec{r}$, in total, negative sign. When the ball falls from B to A due to gravity, you will have negative $\vec{F}$ and negative $d\vec{r}$, in total, positive. So total work therefore 0.