# Physics with Calculus/Mechanics/Work and Energy

## Work

Work is a special name given to the (scalar) quantity

${\displaystyle W=\int {\vec {F}}\cdot d{\vec {x}}.}$

where ${\displaystyle W}$ is work, ${\displaystyle F}$ is force on the object and ${\displaystyle x}$ is displacement. Since the the dot product is a projection, the work is the component of the force in the direction of the displacement times the displacement. If the force is constant and the object travels in a straight line, this reduces to

${\displaystyle W={\vec {F}}\cdot {\vec {x}}=Fx\cos \theta \ }$

where ${\displaystyle W}$ is work and ${\displaystyle F}$ is force on the object and ${\displaystyle x}$ is displacement.

We say that W is the "work done by the force, F." Let us derive a useful relationship between work and kinetic energy.

Say we have a total force ${\displaystyle {\vec {F}}}$ acting on an object. Then the work is

${\displaystyle \int {\vec {F}}\cdot d{\vec {r}}=\int m{\frac {d{\vec {v}}}{dt}}\cdot d{\vec {r}}=m\int {\frac {d{\vec {v}}}{dt}}\cdot {\vec {v}}dt=m\int {\vec {v}}\cdot d{\vec {v}}=\Delta (mv^{2}/2)=\Delta K}$

I simply used Newton's second law in the first step and a nice substitution in the integral. This states that the work done by the total force on an object is the change in kinetic energy of the object. In fact, this is sufficient to be the motivation for defining K to be ${\displaystyle mv^{2}/2}$.

For example, if you hold an apple, then move the apple up a little bit then stop, what is happening? The potential energy of the apple changes, so someone is doing work even though there is no change in kinetic energy -- and by our theorem, this means that there was no work done -- how can that be? Gravity did (negative) work on the apple, but you also did (positive) work on the apple. That is, the total work done is zero, although the work you did is non zero. Always remember that it is the total force that changes kinetic energy.

Another useful property of work is that it is linear in the force. That is, the work done by the sum of two forces is the sum of the work done by each force. Because of this, you can interpret the work as how much kinetic energy each force is giving to the object. In the apple example, at first the force from your hand is greater than the force of gravity, so the kinetic energy increases and the apple accelerates up. Then as you slow down, gravity does more work so the total work is negative and the apple decreases its kinetic energy and comes to a stop.

In a very special case, the the quantity of work does not depend on how you move a particle around, but only on the beginning and ending points. Such a field is called "conservative." It means that we can introduce a potential. Gravity is such a conservative force, amazingly, which is why we can talk about the "potential energy" of an object. It is just shorthand for saying the work it takes to move the object from somewhere (the reference point) to wherever we are talking about. Consequently, the change in kinetic energy equals the negative change in potential energy, which means that the total energy of the system is constant. This is in fact why such a force is called conservative -- it conserves mechanical energy! The converse, however, is not true. That is, if a system conserves mechanical energy, it is not necessarily a conservative force field.

Dissipative forces, such as friction (it eats up energy) are sometimes called non-conservative forces. This is somewhat of a mistake because on the molecular level, the forces really are conservative. However, it is often nicer to just say that energy is not conserved in a given scenario, even though we know full well that it is disappearing into the motion of atoms, or heat. You will hear many people say that energy is not conserved in a given situation, but of course it is; energy is always conserved.

It turns out that a force is conservative if and only if the force is "irrotational," or "curl-less" which has to do with vector calculus. I means that if you put a paddle wheel in, it won't spontaneously start to turn. It is an interesting fact that there are no non-conservative forces!

To quantify everything, we have the work done by a nonconservative force is the change in the total energy of the body. By total energy, we mean potential energy plus kinetic energy, since the total force can be split into conservative (which appears in the potential energy term) and nonconservative (the new term).

## Power

Power is the rate of doing work. To come up with a useful expression for this, consider a short amount of time ${\displaystyle \Delta t}$. How much work is done in that time? Well, by the definition of power, it is very nearly ${\displaystyle P\Delta t}$. By the definition of work, this is ${\displaystyle {\vec {F}}\cdot {\vec {\Delta x}}}$ where ${\displaystyle \Delta x}$ is the displacement which occurs in ${\displaystyle \Delta t}$. We have

${\displaystyle P\Delta t={\vec {F}}\cdot {\vec {\Delta x}}}$

So,

${\displaystyle P={\vec {F}}\cdot {\frac {\vec {\Delta x}}{\Delta t}}}$,

which in the limit as ${\displaystyle \Delta t}$ goes to zero (which is when our "equations" become exact),

${\displaystyle P={\vec {F}}\cdot {\frac {d{\vec {x}}}{dt}}={\vec {F}}\cdot {\vec {v}}}$.

Equivalently, we could have gotten the expression by simply differentiating our expression for work. No matter the derivation, because it simply does not matter that much; we have a useful expression for power.

This means that if the force is acting perpendicular to the velocity, the speed does not change, because the work is zero so the change in kinetic energy is zero. But wait, how can that be, since a force necessarily accelerates something? It is accelerating it, it is changing the direction of travel -- acceleration means the derivative of the vector velocity, not the magnitude of velocity. In fact, this tells us that the component of force in the same direction as velocity is responsible for (and only for) changes in the magnitude of the velocity, and the component of force perpendicular to the velocity is responsible for (and only for) changes in the direction of the velocity. Just to quantify this a little bit, it can be shown that

${\displaystyle {\vec {a}}=||v||{\vec {T}}+{\frac {||v||^{2}}{\rho }}{\vec {N}}}$

where a is acceleration, v is the velocity, T is the unit tangent vector (tangent to the path of the particle and consequently parallel to the velocity vector), N is the unit normal vector (perpendicular to the tangent vector and in the direction of the derivative of the tangent vector, which you can picture by drawing two pretty close tangent vectors on a curve), and ${\displaystyle \rho }$ is the radius of curvature, which is essentially the radius of the circle which closest fits the path at the point (the radius of curvature of a circle is the radius of the circle, and the radius of curvature of a straight line is infinity). All this business is not really necessary for understanding physics, but if you understand it it will help you understand what is going on. Notice that the second term is the centripetal acceration -- this is in fact where we get the formula for it.

Finally, just writing out the definition of power to look pretty, if the work is done at a changing rate, then

${\displaystyle P={\frac {dW}{dt}}}$

If the work is done at a constant rate, then this becomes

${\displaystyle P={\frac {W}{\Delta t}}={\frac {\Delta E}{\Delta t}}}$.