Work

Hold a $20\ \text{kg}$ suitcase motionless for a minute and your arms ache, yet the suitcase has gained no energy at all. Carry it up a flight of stairs and it has. In physics, work is the energy a force transfers into or out of a system by acting while the system moves, and it is the link between forces (Unit 2) and energy (Topic 3.1).

Only the part of the force along the motion does work

When a force $\vec{F}$ acts while its point of application moves through a small displacement $d\vec{r}$, the work it does is the dot product $dW = \vec{F}\cdot d\vec{r}$. For any two vectors, the dot product is a scalar, $\vec{A}\cdot\vec{B} = AB\cos\theta$, where $\theta$ is the angle between them when they are drawn tail to tail. Adding up the small pieces along the whole path gives the definition.

The $\cos\theta$ is doing real work of its own. Pull a wagon with $50\ \text{N}$ along a handle tilted $60^\circ$ above the ground for $10\ \text{m}$: only $50\cos 60^\circ = 25\ \text{N}$ points along the path, so you do $W = (50)(10)\cos 60^\circ = 250\ \text{J}$. The upward $43\ \text{N}$ component lightens the wagon's load on the ground but transfers no energy to it, because the wagon doesn't move up.

The sign tells you which way the energy flows. A force with a component along the motion (positive work) puts energy into the object; kinetic friction on a sliding box points against the motion and does negative work, taking energy out. A force can do zero work three ways: the object doesn't move, the force is perpendicular to the motion, or the force acts at a point that doesn't move. That last case is why holding the suitcase still does no work on it, however tired you get.

Perpendicular forces steer without changing speed

A force perpendicular to the velocity can change the direction of motion, but not the speed, because it never has a component along the motion. The normal force on a car rounding a banked curve, the tension in a string swinging a ball in a horizontal circle, and gravity on a satellite in a circular orbit all do zero work. Each one bends the path while leaving $\frac{1}{2}mv^2$ unchanged, which is exactly what uniform circular motion (Topic 2.10) looks like.

The normal force from a fixed surface does zero work on an object sliding along it for the same reason, even on a curved ramp: at every instant it is perpendicular to the surface, and the object moves along the surface.

Variable forces: integrate, or find the area

When the force changes along the path, $Fd\cos\theta$ no longer applies, and you go back to the integral. In one dimension, along the $x$-axis, $W = \int_{x_1}^{x_2} F_x\,dx$, and that integral is the signed area under the force-position graph, with $F_x$ on the vertical axis and $x$ on the horizontal. Area above the axis is positive work; area below it is negative.

A spring is the standard case. A spring of constant $k$ stretched by $x$ pulls back with $F_x = -kx$ (Topic 2.8). As a block on its end moves from $x_1 = 0.10\ \text{m}$ to $x_2 = 0.20\ \text{m}$ with $k = 200\ \text{N/m}$, the spring does

It is negative because the spring pulls toward $x = 0$ while the block moves away. On the graph of $F_x = -kx$, it's the area of a trapezoid below the axis. Using $F d$ with the force at either end would give $-2.0\ \text{J}$ or $-4.0\ \text{J}$, both wrong; the integral averages the changing force correctly.

The work-energy theorem

Add up the work done by every force on an object and you get its change in kinetic energy. The derivation is the one in Topic 3.1: Newton's second law gives $F_{net}\,dx = mv\,dv$, and integrating from $v_i$ to $v_f$ gives $\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.

The theorem replaces "find the acceleration, then use kinematics" with one equation, and it keeps working when the force varies, where kinematics equations don't apply.

When is something "an object"? Work is computed with the displacement of the point where the force is applied. If that point moves exactly as the system's center of mass does, the force can only speed up or slow down the whole system, and the system can be modeled as an object: all the work shows up as kinetic energy. If the point of application moves differently from the center of mass, the force can change the system's shape or internal energy instead. Two cases:

• Squeeze a spring between your hands. The ends move inward and your forces do positive work, but the spring's center of mass stays put and it gains no kinetic energy. The work went into the spring's configuration, stored as potential energy (Topic 3.3).
• A skater pushes off a wall. The wall's force on her hands acts at a point that never moves, so the wall does zero work, yet she leaves with kinetic energy. That energy came from inside her (chemical energy in her muscles), so she is not a single object for this energy analysis.

Friction takes energy out over the whole path. Kinetic friction of magnitude $f_k$ always points against the sliding, so the energy it dissipates is $f_k$ times the length of the path traveled, not the displacement. That energy becomes thermal energy in the surfaces (and some sound). AP Physics C asks you to track the mechanical energy lost; you only need to know where it goes, not to analyze the heating.

Conservative and nonconservative forces

Lift a $2.0\ \text{kg}$ book $1.5\ \text{m}$ straight up, or carry it up a winding staircase to the same height: gravity does $-mg\Delta y = -(2.0)(9.8)(1.5) = -29.4\ \text{J}$ either way, because only the vertical part of each step contributes. Bring the book back down to where it started and gravity does $+29.4\ \text{J}$, for a total of zero around the closed loop.

That defines a conservative force: the work it does depends only on the starting and ending positions, not on the path, so it is zero for any round trip. Gravity and ideal spring forces are conservative. Because the work depends only on position, you can store it as a function of position, a potential energy. Potential energies exist only for conservative forces, and Topic 3.3 builds them.

A nonconservative force does work that depends on the path. Drag a box across a rough floor and back, and friction does negative work on both legs; the round trip total is not zero, and a longer route costs more. Friction and air resistance are the nonconservative forces you will meet most often. No potential energy can be defined for them, because there is no single number to assign to a position.

Worked examples

Worked exampleReading work from a force graph

A $3.0\ \text{kg}$ cart moves along a straight, horizontal, frictionless track. The graph shows the net horizontal force on it as a function of position. At $x = 0$ the cart moves in the $+x$ direction at $2.0\ \text{m/s}$.

(a) Find the net work done on the cart from $x = 0$ to $x = 5\ \text{m}$, and the cart's speed at $x = 5\ \text{m}$.

(b) Find the cart's speed at $x = 6\ \text{m}$.

(c) At what position is the cart's speed greatest? Justify your answer.

Show solution

(a) Work is the area under the curve, split into simple shapes: the triangle from 0 to 2 m, $\frac{1}{2}(2)(8) = 8\ \text{J}$; the rectangle from 2 to 4 m, $(2)(8) = 16\ \text{J}$; the triangle from 4 to 5 m, $\frac{1}{2}(1)(8) = 4\ \text{J}$. The net work is $28\ \text{J}$. The cart started with $K_i = \frac{1}{2}(3.0)(2.0)^2 = 6.0\ \text{J}$, so by the work-energy theorem $K = 34\ \text{J}$ and

$$v = \sqrt{\frac{2(34)}{3.0}} = 4.8\ \text{m/s}.$$

(b) From 5 to 6 m the area is a triangle below the axis, $-\frac{1}{2}(1)(8) = -4\ \text{J}$. Now $K = 30\ \text{J}$ and $v = \sqrt{\frac{2(30)}{3.0}} = 4.5\ \text{m/s}$.

(c) At $x = 5\ \text{m}$. Up to there the force points in the direction of motion, so it does positive work and the kinetic energy keeps rising. Beyond $5\ \text{m}$ it points backward and does negative work, so the kinetic energy falls. The maximum is where the force changes sign, not where the force is largest.

Where students lose points