Work & the Work-Energy Theorem

The question forces alone make painful

A block is launched up a rough ramp at $8.0\ \mathrm{m/s}$ and you want to know how far it gets before it stops. You can do this with $F = ma$: find the acceleration, then use a kinematic equation. Now curve the ramp so the slope changes as the block climbs. The acceleration is no longer constant, the kinematic equations die, and the honest route is a differential equation.

But something has not changed. The block started with a certain amount of motion and ended with none, and gravity and friction were what drained it. If you can keep an account of how much each force takes out of that stock of motion, you get the answer without ever knowing the acceleration at any instant. That account is called work, the stock of motion is called kinetic energy, and the rule connecting them is the last general-purpose tool you need before conservation of energy becomes available in the next lesson.

Orbits in the previous lesson were a case where speed and radius traded off in a way $F = ma$ described but never made obvious. Work is the language in which that trade is bookkeeping.

Only the part of the force along the motion counts

Push a crate $4.0\ \mathrm{m}$ across the floor with a steady $25\ \mathrm{N}$ in the direction it moves, and you have done $25 \times 4.0 = 100\ \mathrm{J}$ of work. One joule is one newton-metre. That is the whole definition when the force and the motion point the same way.

Now tilt your push. A child drags a sled with a rope at $30^\circ$ above the horizontal. The rope pulls partly forward and partly upward, and the sled only goes forward. The upward part of the pull is not lifting the sled anywhere: it just presses less weight onto the ground. Only the forward part of the tension travels with the sled, so only that part does work. Split the $40\ \mathrm{N}$ tension into a forward piece $40\cos 30^\circ$ and an upward piece $40\sin 30^\circ$, and keep the forward one:

$W = Fd\cos\theta$

where $F$ is the size of the force, $d$ the distance moved, and $\theta$ the angle between them. For the sled, $W = 40 \times 6.0 \times \cos 30^\circ$. In vector language this is the dot product, $W = \vec F \cdot \vec d$, which is the machine that picks out "the part of one vector along another".

Three consequences fall straight out of the cosine. At $\theta = 0$ you get $W = Fd$, the most work a force of that size can do. At $\theta = 90^\circ$, $\cos\theta = 0$ and the work is exactly zero: a normal force on a sliding block, the tension in a string swinging a ball in a circle, the track pressing sideways on a bead. These forces steer without feeding. And past $90^\circ$ the cosine goes negative. Friction on a sliding block points backwards, so $\theta = 180^\circ$, $\cos\theta = -1$, and the work is $-f d$. Negative work is a real answer, not a mistake: it means the force removed motion instead of supplying it.

There is a bonus hiding in $W = \vec F \cdot \vec d$. If the force is constant, only the straight-line displacement from start to finish matters, not the path. A bead is dragged along a quarter circle of radius $1.5\ \mathrm{m}$ from $(1.5, 0)$ to $(0, 1.5)$ while a constant $\vec F = (10\ \mathrm{N})\hat x$ acts. The path is curved, but the net displacement is $\Delta x = 0 - 1.5 = -1.5\ \mathrm{m}$ and $\Delta y = +1.5\ \mathrm{m}$. Only $\Delta x$ pairs with a force along $\hat x$, so $W = (10)(-1.5) = -15\ \mathrm{J}$. You never had to follow the arc.

When the force changes with position

$W = Fd\cos\theta$ assumes $F$ is the same the whole way. Springs, gravity far from a planet, and a friction coefficient that worsens along a track all break that. The repair is the standard one: chop the motion into steps so short that the force is effectively constant across each, multiply force by step, and add.

$W = \int_{A}^{B} \vec F \cdot d\vec r$

If you have not met an integral, read the symbol as "the area under the curve of $F$ against $x$". That is not a metaphor: it is what the sum of $(\text{height}) \times (\text{width})$ strips converges to. Area under an $F$ versus $x$ graph is work, and area below the axis counts as negative.

Two shapes cover most problems. A force growing linearly, $F = 6x$, from $x = 0$ to $x = 3.0\ \mathrm{m}$ traces a triangle of base $3.0$ and height $6 \times 3.0 = 18\ \mathrm{N}$, so the work is $\tfrac12 (3.0)(18) = 27\ \mathrm{J}$. A force dying away linearly, $F = 12 - 2x$ from $x=0$ to $x = 5.0\ \mathrm{m}$, runs from $12\ \mathrm{N}$ down to $2\ \mathrm{N}$: a trapezoid, whose area is the average height times the width, $\tfrac{12+2}{2} \times 5.0 = 35\ \mathrm{J}$.

For curved force laws you need one calculus fact, the power rule for areas: the area under $x^n$ from $0$ to $b$ is $b^{n+1}/(n+1)$. So $F = 3x^2$ from $0$ to $2.0\ \mathrm{m}$ does $3 \times 2.0^3/3 = 8.0\ \mathrm{J}$. The same rule with $n = -2$ handles inverse-square forces, where the antiderivative of $1/x^2$ is $-1/x$. Take a force of magnitude $50/x^2$ pulling a probe back toward the origin, so the force component along $+x$ is $-50/x^2$. Moving out from $x = 1.0\ \mathrm{m}$ to $x = 5.0\ \mathrm{m}$:

$W = \int_{1}^{5} \left(-\frac{50}{x^2}\right)dx = \left[\frac{50}{x}\right]_{1}^{5} = \frac{50}{5} - \frac{50}{1} = 10 - 50 = -40\ \mathrm{J}$

Negative, as it must be: the probe moved away from something pulling it back.

Net work is the change in kinetic energy

Here is why any of this is worth tracking. Start from $F_{\text{net}} = ma$ for motion along a line and compute the work done by the net force:

$W_{\text{net}} = \int F_{\text{net}}\,dx = \int m\frac{dv}{dt}\,dx$

Now use a trick worth keeping. Since $dx/dt = v$, you can write $dx = v\,dt$, and the $dt$ cancels:

$\int m\frac{dv}{dt}v\,dt = \int_{v_i}^{v_f} mv\,dv$

The variable is now $v$ instead of $x$, and that last area (a triangle under the line $mv$) is $\tfrac12 mv_f^2 - \tfrac12 mv_i^2$. Define the kinetic energy $K = \tfrac12 mv^2$ and you have the work-energy theorem:

$W_{\text{net}} = \Delta K = \tfrac12 mv_f^2 - \tfrac12 mv_i^2$

Read it in words: add up the work done by every force acting on an object, and the total equals the change in its $\tfrac12 mv^2$. Speed up and the net work was positive. Slow down and it was negative. Move at constant speed, even around a corner, and the net work was zero.

Notice what is absent. There is no time in that equation, and no acceleration. That is the whole reason to use it. A $1200\ \mathrm{kg}$ car at $20\ \mathrm{m/s}$ comes to a stop; the brakes did $0 - \tfrac12(1200)(20)^2 = -240{,}000\ \mathrm{J}$ of work, and you do not need to know whether it took two seconds or ten.

Three worked examples

One: a sled with two forces acting. A $25\ \mathrm{kg}$ sled starts from rest on level ground. A rope pulls with $90\ \mathrm{N}$ at $25^\circ$ above the horizontal for $12\ \mathrm{m}$, and the coefficient of kinetic friction is $0.15$. Find the final speed.

On blank paper, write down every force first: tension, gravity, normal, friction. Gravity and the normal force are both perpendicular to the motion, so they do no work and you can drop them from the energy account, but you still need the normal force's size, because friction depends on it.

Vertically the sled does not accelerate, so $N = mg - T\sin 25^\circ = (25)(9.8) - 90(0.4226) = 245 - 38.0 = 207.0\ \mathrm{N}$. That upward tug of the rope matters: friction is $f = \mu_k N = 0.15 \times 207.0 = 31.05\ \mathrm{N}$.

Work by the rope: $W_T = (90)(12)\cos 25^\circ = (90)(12)(0.9063) = 978.8\ \mathrm{J}$. Work by friction: $W_f = -(31.05)(12) = -372.6\ \mathrm{J}$. Net work $= 978.8 - 372.6 = 606.2\ \mathrm{J}$.

The sled started at rest, so $\tfrac12 mv_f^2 = 606.2$, giving $v_f^2 = 2(606.2)/25 = 48.5$ and $v_f = 6.96\ \mathrm{m/s}$.

Two: a force that varies, ending in a speed. A $2.0\ \mathrm{kg}$ cart at rest is pushed along the $x$-axis by $F = 4x^2$ newtons, with $x$ in metres, from $x = 0$ to $x = 3.0\ \mathrm{m}$. Find the final speed.

You cannot use $Fd$, because $F$ runs from $0$ to $4(3.0)^2 = 36\ \mathrm{N}$. Take the area instead. With $n = 2$, the area under $4x^2$ from $0$ to $3.0$ is $4 \times 3.0^3/3 = 4 \times 27/3 = 36\ \mathrm{J}$.

Then $\tfrac12(2.0)v_f^2 = 36$, so $v_f^2 = 36$ and $v_f = 6.0\ \mathrm{m/s}$.

A close cousin: if friction itself varies with position, say $\mu = 0.050x$, the friction force is $\mu mg = 0.050x\,mg$, which grows linearly, and its work over a distance $d$ is the triangle $-\tfrac12 (0.050 d\, mg)(d)$. Setting that equal to $-\tfrac12 mv_i^2$ makes the mass cancel and leaves $0.050\,g\,d^2 = v_i^2$. That is a one-line answer to a problem that would be a nightmare with $F = ma$.

Three: reeling up a hanging rope. A uniform chain of mass $3.0\ \mathrm{kg}$ and length $2.0\ \mathrm{m}$ hangs from a hook. How much work does it take to pull all of it up onto the hook?

The trap is to reach for $mgh$ and not know what to use for $h$. Different pieces of the chain have to be lifted different distances: the top link barely moves, the bottom link travels the full $2.0\ \mathrm{m}$. So slice it.

Let $y$ be the depth of a slice below the hook. The chain has mass per unit length $\lambda = m/L = 3.0/2.0 = 1.5\ \mathrm{kg/m}$, so a slice of thickness $dy$ has mass $\lambda\,dy$ and must be raised by $y$. Its share of the work is $(\lambda\,dy)gy$. Adding over $y$ from $0$ to $L$ is the area under a line again:

$W = \lambda g \int_0^L y\,dy = \lambda g \frac{L^2}{2} = \frac{mgL}{2}$

Numerically, $W = (3.0)(9.8)(2.0)/2 = 29.4\ \mathrm{J}$.

Look at what $mgL/2$ says: it is the whole weight lifted through $L/2$, the distance the chain's centre of mass rises. That shortcut is general. For any rigid or floppy collection of pieces, the work gravity takes is $mg \times (\text{rise of the centre of mass})$.

Chains, ropes, and things that move different distances

The centre-of-mass shortcut earns its keep whenever only part of an object moves. Suppose a length $\ell$ of a chain lying on a table hangs over the edge, with mass per unit length $\lambda$. The part on the table is already at table height and does not need lifting. The hanging part has mass $\lambda \ell$ and its centre of mass sits $\ell/2$ below the edge, so pulling it all back up costs

$W = (\lambda \ell) g \frac{\ell}{2} = \frac{\lambda g \ell^2}{2}$

Run the same accounting backwards and you get the falling case. Release a chain that is partly over the edge and let it slide off a frictionless table. Gravity does positive work equal to the weight times the drop of the centre of mass between the two configurations, and by the work-energy theorem that equals $\tfrac12 mv^2$ for the whole chain, which all moves at the same speed because it is inextensible. Find both centre-of-mass positions, subtract, multiply by $mg$, and solve for $v$. The only real work is being careful about where each centre of mass is.

Power, and one place it saves you

Work per unit time is power, measured in watts. Differentiate $W = \vec F \cdot \vec d$ with respect to time and the displacement turns into velocity:

$P = \frac{dW}{dt} = \vec F \cdot \vec v = Fv\cos\theta$

The same "component along" idea, one derivative up. It is also the cleanest way to handle rope constraints. If a rope runs from a cart over a smooth pole, the speed at which the free end is reeled in equals the component of the cart's velocity along the rope, because that component is the rate at which the cart-to-pole distance shrinks.

So: a cart on a level track is pulled by a rope over a pole $4.0\ \mathrm{m}$ up, and the rope is hauled in at $3.0\ \mathrm{m/s}$. When the cart is $3.0\ \mathrm{m}$ from the base of the pole, the rope's length to the cart is $\sqrt{3.0^2 + 4.0^2} = 5.0\ \mathrm{m}$, and the rope makes an angle with the track whose cosine is $3.0/5.0 = 0.6$. The cart's speed $v$ satisfies $v\cos\theta = 3.0$, so $v = 3.0/0.6 = 5.0\ \mathrm{m/s}$. The cart moves faster than the rope is pulled, and it does so because only part of its velocity shortens the rope.

How to recognise a work-energy problem

Look for these signals in the problem statement.

Speeds and distances, but no times. "Launched at $8.0\ \mathrm{m/s}$, how far does it go?" or "how fast is it moving after $5\ \mathrm{m}$?" There is no clock in the question, so use the tool with no clock in it.

A force given as a function of position. $F = 6x$, $F = 200/x^2$, $\mu = 0.050x$. The moment a force depends on $x$ rather than $t$, integrating over $x$ is natural and integrating over $t$ is not.

A curved or awkward path with a question only about the endpoints. If you are asked for the speed at the bottom of a ramp of unspecified shape, nobody wants you tracking the direction of motion. Add up the work.

Something that moves in pieces, like a chain or a rope, where "the distance moved" is different for different parts. Slice it, or use the centre of mass.

Reach for $F = ma$ instead when the question asks when something happens, or asks for a force, or for an acceleration at a specific instant. Work-energy throws away the timeline in exchange for a scalar equation, which is a very good trade right up until the timeline is what you were asked about.

One limitation to carry forward. In every example here you had to compute the work by each force explicitly, and for gravity and springs that computation turned out to depend only on the start and end points. Forces with that property let you skip the integral entirely by storing the answer in advance, which is what potential energy is.