Is work done = change in KE, or change in mechanical energy?

Question

Starting from $F = ma$ we get:

$$ F=m\frac{dv}{dt} = m\frac{ds}{dt}\frac{dv}{ds} = mv\frac{dv}{ds} $$

which leads to the work done:

$$ W = \int Fds = \int mvdv = m\frac{v^2}{2} - m\frac{u^2}{2} \tag{1} $$

which is equal to the change in KE.

If a body is lifted against gravity by a variable force which is zero at time $t=0$, and also zero after having lifted the body a distance $h$, then is it correct that the work done against gravity is $mgh$?

Why does this not violate the equation (1) above? Why does the PE does not appear in equation (1)?

Answer 1

What you have found is the total work, let's call it $W_{total}$.

By lifting a book to a shelf, gravity does negative work on it while trying to hold it down, but you do positive work on it by lifting it. Since initial and final speed both are zero, there is no total work done as you have shown, $W_{total}=W_{gravity}+W_{you}=0$

That this is so is not violating the fact that $W_{gravity}$ itself is found by: $W_{gravity}=mgh$.

If you wanted to investigate gravity alone you should consider a situation where only gravity works. A free falling book e.g., where the initial and final velocities are not the same, and $W_{total}=W_{gravity}>0$.
And yes, it is still true that $mgh$ is the work done by gravity alone: Gravity is a constant force, so $W_{gravity}=\int F ds=F\int ds=F(s_2-s_1)=Fh=mgh$. The two terms $mgs_1$ and $mgs_2$ are then defined to be called potential energy, so $U_{gravity}=mgs$.

Answer 2

The equation written is valid for any force and at this point a mere definition of work. Potential energy comes into play when you consider a conservative force, that is by definition a force for which $\int Fds$ does not depend on the specific path taken.

This implies the existence of a function $U$ such that $F=-U'$ which is called a potential. Using this relation one is able to express equation (1) in a different way $$ W = \int F ds = -\int U' ds = U(r_0)-U(r) \equiv - \Delta U = \Delta E_{kin} $$

One can express the work $W$ as either the change in kinetic energy or the negative change in potential energy. More importantly it implies that $$ \Delta E_{kin} + \Delta U =0$$ Conservation of mechanical energy.

Answer 3

I have put the summary at the beginning.

Suppose a mass $m$ starting from rest at a height $h$ from the Earth's surface and the kinetic energy of the mass $m$ is to be found when the mass reaches the surface.
The gravitational field strength is $g$ and making the following assumptions:
- the height $h$ is much less than the radius of the Earth so the gravitational field strength can be assumed to be constant - the mass $m$ is much less than the mass of the Earth so the kinetic energy of the Earth can be assumed to be negligible.

For the Earth-mass system

$$KE_\text{surface} + PE_\text{surface} = KE_\text{height h} + PE_\text{height h}$$

$$\Rightarrow KE_\text{surface} =PE_\text{height h} - PE_\text{surface} = mgh $$

For the mass $m$ system

The work done by the external force (gravitational attraction of the Earth) is equal to the change in kinetic energy of the mass.
This is the work-energy theorem.

$$mg \times h = KE_\text{surface}$$

---

If you want to discuss potential energy then you must include at least two bodies.

In this case it will be mass $m$ and the Earth and these are now assumed to be the system.
If the mass $m$ is at the surface of the Earth of radius $R$ and mass $M$ and an external force increases the separation between the mass and the Earth by a distance $h$ with $h\ll R$ then the external force has done work equal to $\displaystyle \int^{R+h}_R \left ( \dfrac{GMm}{R^2}\right) dR \approx mgh$ in lifting the mass $m$ a distance $h$.
This assumes that $h \ll R$ and hence $g \approx \dfrac{GM}{R^2}$; the gravitational field strength is constant near the surface of the Earth.
Note that we should expect the work done by the external force to be a positive quantity as the external force direction and its displacement are in the same direction – upwards.
That work done by the external force increases the gravitational potential energy of mass $m$ (and the Earth) by $mgh$.
Normally one does not go through all this mathematics and just assumes that the gravitational field strength is constant near the surface of the Earth.

---

Consider a mass $m$ as the system.
On the Earth the mass $m$ has a force $mg$ downwards on it due to the gravitational attraction of the Earth.

Suppose that an external force of $mg$ upwards is exerted on the mass $m$.

The net force on the mass $m$ is now zero and if the mass moves up a distance $h$ the work done on the mass is zero and so the change in kinetic energy of the mass is zero.
This is the work-energy theorem in action.
You can consider the two forces individually.
The external force does positive work $mgh$ as the external force and its displacement are in the same direction and the gravitational attraction does negative work $-mgh$ as the gravitational force and its displacement are in opposite directions.

With the mass $m$ now at a height $h$ if the mass is released and allowed to drop to the surface of the Earth then the work done by the external force (gravitational attractive) is $+mgh$ (force and displacement both in the same direction) and this will result in the mass gaining kinetic energy by the work-energy theorem.

In terms of the mass-Earth system this last change can be described by stating that the mass-Earth system loses gravitational potential energy and the mass (and the Earth) gains kinetic energy but overall there is no change in the total energy of the system.
Here internal forces are doing work.
In this last statement there is a hint that usually the assumption $M \gg m$ is made and the kinetic energy gained by the Earth is not included as it is so small compared with the kinetic energy gained by the falling mass $m$.

Answer 4

In equation (1), you have no gravitational potential or any force field; you are just accelerating free body from speed $u$ to speed $v$. If we have a force like gravity here mentioned, then it is this force that can do work on a body. If you have some path, some kind of a hill, and on the top of that hill there is a body, you can now let the body roll down the hill. No matter what the path may be, work done by a gravitational force on this body is the same if the height is the same, id est, if the starting and end points are the same. So work done by a gravitational force depends only on the end points of the path not on the path itself. Force fields like this are called conservative fields. Now, when you move this same body up a hill, you are doing work also(you are applying a force on some path, and that is it). If you are moving the body at a constant speed then the force you are applying is the same as the gravitational pull. In a uniform gravitational field like near the earth, work done is just the force times the height ( the shortest path from point one to point two). So force is, as you know, $mg$, and the path is just $H$ (height). Of course, you have to define a point of zero potential energy and this is just something we choose freely but of course, in a useful way.Now, because we are talking about conservative force field, we now see that energy of a body in this force field is just a function of its position, its a scalar field actually. And why this does not change the equation you wrote? Firstly, if the body has velocity zero at any given point, of course it can not have kinetic energy. Secondly, of course you would have to modify equations to include the fact that there is some kind of gravitational force. But you should bear in mind that sum of kinetic and potential energy is constant for a given system. So if there is no force field there is no potential energy. Free body in space can not have potential energy. You can accelerate it by applying force but that is it. So it all depends how you are going to write equations. I can also write: $$mgH = W = \int mg.ds$$, integral going from one point to the other. If $mg$ is constant, and it is, you just get $mgH_1 - mgH_2 = U =mg\Delta{H}$...If you WANT you CAN write it as the equation (1), bearing in mind that $$H=\dfrac{v^2}{2g}$$. If you plug that in $mgH$, you get the kinetic energy of body falling from a height $H$.