A-level Physics/Forces and Motion/Force, work and power

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

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

where $$W$$ is work and $$F$$ is force on the object and $$x$$ is displacement. Essentially this integral is the component of the force in question 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

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

where $$W$$ is work and $$F$$ is force on the object and $$x$$ is displacement. Take note of the dot product.

We say that W is the "work done by the force, F." Notice that $$F$$ need not be the total force on an object, just the force we are looking at. It makes sense to ask what is the work done by a given force on an object. Notice also that the work done by the sum of two forces acting on an object is the sum of the work done by the forces acting individually on the object. This gives rise to the interpretation that work is that it is the energy transferred to the body by a force that acts on it. (Of course negative work is energy transferred from the body). This is the whole point of even considering work.

For, say we had a total force $$\vec{F}$$ acting on an object. Then the work is

$$ \int \vec{F} \cdot d\vec{r} = \int m \frac{d \vec{v}}{dt} \cdot d\vec{r} = m \int \frac{d\vec{v}}{dt} \cdot \vec{v} dt = m \int \vec{v} \cdot d\vec{v} = \Delta \left(\frac{1}{2}mv^2\right) = \Delta KE $$

This simply uses Newton's second law in the first step and a 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. For example, if you hold an apple, then move the apple down a little bit then stop, what is happening? Surely the potential energy of the apple has changed, so someone is doing work even though there is no change in kinetic energy—how can that be? We must consider all the forces. Gravity did work on the apple, but the apple did work on you (you did negative work on the apple) -- you have absorbed the energy! So there really is no paradox after all.

In a very special case, it happens that 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 basically states that the total energy of the system is constant. This is in fact why such a force is called conservative—it conserves mechanical energy!

Dissipative forces, such as friction (it always 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. But for all of our purposes, there are no non-conservative forces!

However, just to quantify everything, we have the work done by a non conservative force is the change in the total energy of the body.

Power
Power is the rate of doing work. Thus we have

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

So,

$$P = \frac{d}{dt}(\vec{F} \cdot \vec{x})$$,

and for forces that do not vary over time becomes

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

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

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

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 $$\rho$$ 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 acceleration—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

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

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

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

Pressure
Pressure is the force per unit area.

$$ P = \frac{F}{A}$$

Torque
$$ \vec{\tau} = \vec{r} \; \mathbf{x} \; \vec{F} \ $$

$$ \tau = rF \sin \theta \ $$

Torque is the "rotational force" applied as part of circular motion, such as the force making the wheels of a car turn. In the SI unit system, torque is measured in Newton-metres.