# What is work done

1. Jul 24, 2014

### Greg Bernhardt

Definition/Summary

Work is a measure of change of energy.

The net work done equals the change in Kinetic Energy (this is the Work-Energy Theorem).

Work is the integral of the scalar product (dot-product) of two vectors: Force and Displacement. "Displacement" means the change in position of the point at which the force is applied.

So Work is a scalar (an ordinary number), with dimensions of mass times distance-squared over time-squared.

The SI unit is the amount of Work done by a Force of one newton acting over a displacement of one metre, and is called the joule (J), or newton-metre (N-m).

The SI unit of Power, which is the rate of Work done, is one joule per second, and is called the watt (W).

Equations

Work is the integral of the dot product of force and displacement.

$$W\,=\,\int_{\mathbf{a}}^{\mathbf{a}+\mathbf{d}} \mathbf{F} \cdot d\mathbf{r}$$

For a constant force, work is the dot product of the force with the total displacement.

$$W\,=\,\mathbf{F}\cdot\mathbf{d}\;.$$

The above work equals the magnitude of the force times the magnitude of the displacement times the cosine of the angle between the force and displacement:

$$W\,=\,Fd\cos(\theta)\;.$$

In a uniform gravitational field, Work done by gravity on a body moving along any path C starting at a height $h_1$ and ending at a height $h_2$ is:

$$W\,=\,\int_{(x,y,h_1)}^{(x',y',h_2)} (-mg\hat z)\cdot d\mathbf{r}\,=\,mg(h_1 - h_2)$$

In an inverse-square gravitational field, Work done by gravity on a body moving along any path C starting at a height $r_1$ and ending at a height $r_2$ is:

$$W\,=\,\int_{r_1}^{r_2} -GmM\frac{\mathbf{r}}{r^3}\cdot d\mathbf{r}\,=\,GmM\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$$

which, if $r_1$ is very close to $r_2$, is approximately the same as the previous formula, with $$g\,=\,\frac{GM}{r_1^2}$$

Extended explanation

Conservative Force:

If the work done after a total displacement of zero is zero (so the change in Kinetic Energy is zero), then the force is said to be conservative (for example, friction is not conservative, because a body moving in a full circle under friction loses Kinetic Energy, but gravity is conservative).

For a completely non-conservative force, work equals loss of mechanical energy (the energy lost generally becomes radiation or heat).

Potential energy:

For a conservative force, work done depends only on position and not on the path taken.

Potential energy is another name for work done by a conservative force.

Potential energy depends only on position, and is the work done relative to some arbitrarily-chosen position (the position of zero potential energy, chosen so as to make calculations easy).

For example, in a uniform gravitational field of strength g, when a mass m is moved by any path through a height h, the work done is mgh.

Gearing:

A machine has gearing G if the force out is G times the force in: $F_1\,=\,G F_0$

If no energy is lost, then the work out equals the work in.

Since work equals force times displacement (strictly, the inner product of force and displacement), that means that the displacement of the point of application of the force out is 1/G times the displacement of the point of application of the force in: $$d_1\,=\,\frac{d_0}{G}$$

Conversely, if a system has $d_1\,\neq\,d_0$, then $$F_1\,=\,\frac{d_0}{d_1} F_0$$

For example, the gearing of a lever is the ratio of the lengths of its two "lever arms".

So a lever, or a pulley system, in which the displacement out is less than the displacement in, can lift a heavy object with a force less than its weight.

Derivation of Work-Energy Theorem:

$$\Delta\,W\ =\ \int d\mathbf{x} \cdot \mathbf{F}_{\rm net}\ =\ \int (\mathbf{v}\,dt) \cdot \left( \frac{d(m\mathbf{v})}{dt} \right)\ =\ \int d \left( \frac{1}{2} m\mathbf{v}^2 \right) \ =\ \Delta\,KE$$

Relativistic version:

$$\Delta\,W\ =\ \int d\mathbf{x} \cdot \mathbf{F}_{\rm net}\ =\ \int (\mathbf{v}\,dt)\cdot\left(\frac{d(m\mathbf{v}/\sqrt{1 - \mathbf{v}^2/c^2})}{dt}\right)$$

$$=\ \int d\left(mc^2\sqrt{1 - \mathbf{v}^2/c^2}\right)\ \ +\ \int d\left(\frac{m\mathbf{v}^2}{\sqrt{1 - \mathbf{v}^2/c^2}}\right)\ =\ \int d\left(\frac{mc^2}{\sqrt{1 - \mathbf{v}^2/c^2}}\right)\ =\ \Delta\,E$$

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