Work

The work done by a force is how much energy that force feeds into a moving object. For a force along the motion it is just force times distance; in general it is $W = Fd\cos\theta$, where $\theta$ is the angle between the force and the displacement. Force is in newtons, distance in meters, and the work comes out in joules (J).

The $\cos\theta$ is the whole story of this topic. It keeps only the part of the force that points along the motion. A force straight along the path ($\theta = 0$) does the full $Fd$; a force at a right angle ($\theta = 90^\circ$) does nothing; a force that points partly backward does negative work.

Three habits cost points: dropping the $\cos\theta$ and writing $W = Fd$ for an angled force (§2), using one force times the distance when the force changes as the object moves (§3), and assuming work is always positive (§4).

Throughout this lesson, $g = 10$ m/s$^2$ where it is needed.

The most common slip is to write $W = Fd$ for every force, even one that pushes partly sideways. Only the component of the force along the motion does work, so the right formula is $W = Fd\cos\theta$.

Picture pulling a wagon with a rope held at an angle above the ground. The pull has a forward part, $F\cos\theta$, and an upward part, $F\sin\theta$. The wagon moves forward, so only the forward part does work: $W = (F\cos\theta)d = Fd\cos\theta$. The upward part does nothing, because the wagon does not move up.

Two angles are worth knowing cold. At $\theta = 0$ the force is along the motion and $W = Fd$, the most work possible. At $\theta = 90^\circ$ the force is perpendicular and $\cos 90^\circ = 0$, so $W = 0$: carrying a tray level across a room, or the normal force on a sliding box, does no work.

The fix is to always ask what angle the force makes with the motion, then multiply by $\cos\theta$. Writing $W = Fd$ is the special case $\theta = 0$, not the general rule.

The formula $W = Fd\cos\theta$ assumes the force is constant. When the force changes as the object moves, you can no longer use a single force times the distance. Instead the work is the area under the force-versus-position graph.

For a constant force that area is just a rectangle, height $F$ and width $d$, which gives back $W = Fd$. For a force that varies, the graph is no longer flat, and you read off the area under whatever shape it makes. In calculus form $W = \int F\,dx$, but for the graphs in this course the area is a triangle or a rectangle you can find directly.

The spring is the standard case. A spring obeys $F = kx$: the force grows in a straight line from zero as you stretch it. The area under that line from $0$ to $x$ is a triangle, $\tfrac{1}{2}(\text{base})(\text{height}) = \tfrac{1}{2}(x)(kx) = \tfrac{1}{2}kx^2$. So the work to stretch a spring a distance $x$ is $W = \tfrac{1}{2}kx^2$.

The trap is to grab the final force, $kx$, and multiply by the distance, getting $kx^2$. That is the rectangle, and it is twice too big, because the force started at zero. A varying force needs the area, not the last value times the distance.

Work carries a sign, and the sign comes from the $\cos\theta$. When the force has a component along the motion, $\cos\theta > 0$ and the work is positive: the force adds energy. When the force opposes the motion, the angle is more than $90^\circ$, $\cos\theta < 0$, and the work is negative: the force removes energy.

Friction is the everyday example. Kinetic friction always points opposite the motion, so it does negative work, $W = -fd$, draining energy as heat. Slide a box $3$ m against $6$ N of friction and friction does $-18$ J on it.

Gravity changes sign with the direction of travel. Lifting an object, gravity points down while the motion is up, so gravity does negative work. Lowering it, gravity and the motion line up, so gravity does positive work.

The trap is to treat work as always positive, adding up sizes and ignoring direction. Always check whether each force helps the motion (positive), is perpendicular to it (zero), or opposes it (negative).

A work question nearly always turns on the angle a force makes with the motion, whether that force holds steady, or the sign the work carries.

An angled force. Find the angle between the force and the motion, then multiply by $\cos\theta$: $W = Fd\cos\theta$. A force along the motion gives $Fd$; a perpendicular force gives zero.

A varying force. Do not use one force times the distance. Take the area under the force-versus-position graph. For a spring, $F = kx$, that area is the triangle $\tfrac{1}{2}kx^2$.

The sign. Decide whether each force helps the motion, is perpendicular to it, or opposes it. A force that opposes the motion, like friction, does negative work.

Setup. (a) A crate is pulled $4$ m along the floor by a $50$ N rope held $37^\circ$ above the horizontal ($\cos 37^\circ = 0.8$); find the work done by the pull. (b) A spring with $k = 200$ N/m is stretched $0.5$ m; find the work done on it. (c) While the crate moves the $4$ m, friction of $10$ N opposes it; find the work done by friction.

(a) An angled force. Only the part along the motion does work, so multiply by $\cos\theta$: $W = Fd\cos\theta = (50)(4)(0.8) = 160$ J. Using $W = Fd = 200$ J would wrongly keep the whole force, including the upward part, which does no work.

(b) A varying force. The spring force grows from $0$ to $kx = 100$ N along a straight line, so the work is the area under it, a triangle: $W = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.5)^2 = 25$ J. Multiplying the final $100$ N by $0.5$ m gives $50$ J, twice too much, because the force started at zero.

(c) A force that opposes the motion. Friction points opposite the motion, so its work is negative: $W = -fd = -(10)(4) = -40$ J. It drains $40$ J from the crate.

The traps to avoid. Do not drop the $\cos\theta$ on the angled pull (it is $160$ J, not $200$ J), do not use one force times the distance for the spring (it is the area, $25$ J, not $50$ J), and do not make friction positive (it is $-40$ J).