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Work as $\int F\,dx$, the area under the force–position graph; spring work $\tfrac{1}{2}kx^2$; the work–energy theorem; and power $P = \vec{F} \cdot \vec{v}$.
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By the end of this lesson you will be able to compute the work of variable forces by integration, use spring energy and the work–energy theorem, and relate power to force and velocity.
From Physics 1 you know work as force times displacement, $W = Fd\cos\theta$, for a constant force, kinetic energy $\tfrac{1}{2}mv^2$, and the work–energy theorem. The last lesson showed that integrating Newton's law for a force of position gives exactly that theorem. From calculus you know definite integrals as areas. This lesson extends work to forces that change along the path, which covers springs, gravity at a distance, and most real pushes and pulls.
| Term | What it means |
|---|---|
| Work | $W = \int\vec{F} \cdot d\vec{r}$, energy transferred by a force acting along a displacement. |
| Hooke's law | $F = -kx$, the restoring force of a spring stretched or compressed by $x$. |
| Spring constant | $k$, in N/m, the stiffness of a spring. |
| Kinetic energy | $K = \tfrac{1}{2}mv^2$, the energy of motion. |
| Work–energy theorem | $W_{\text{net}} = \Delta K$. |
| Power | $P = dW/dt = \vec{F} \cdot \vec{v}$, the rate at which work is done, in watts. |
| Line integral | The sum of $\vec{F} \cdot d\vec{r}$ along a curved path. |
For a constant force along a straight displacement, $W = F\Delta x$. When the force changes with position, split the path into small steps $dx$; on each the force is nearly constant and does work $F\,dx$. Adding them gives
$$W = \int_{x_1}^{x_2}F(x)\,dx,$$
the area under the force–position graph. In more than one dimension only the component of force along the motion counts, $W = \int\vec{F} \cdot d\vec{r}$.
A spring obeys Hooke's law, $F = -kx$. The work you do stretching it from $x_1$ to $x_2$ is $\int kx\,dx = \tfrac{1}{2}k(x_2^2 - x_1^2)$, and the energy stored at stretch $x$ is $\tfrac{1}{2}kx^2$. The work–energy theorem, $W_{\text{net}} = \Delta K$, holds for any forces, constant or not, because it came from integrating Newton's second law. Power, the rate of doing work, is $P = dW/dt = \vec{F} \cdot \vec{v}$.
Another way: picture
Plot the force you need to hold a spring against the stretch. It is a straight line from the origin, rising as the spring stretches. The work done is the triangle under that line. Stretch twice as far and the triangle is twice as wide and twice as tall: four times the area, four times the work.
Another way: steps
A constant force times a displacement is the area of a rectangle on the force–position graph. When the force changes, the graph is a curve, and the region under it can be cut into thin strips of width $dx$ and height $F(x)$. Each strip is a nearly rectangular piece of work, and the limit of their sum is the integral.
Areas below the axis count as negative work: the force opposes the displacement there and takes energy away. On a graph made of straight segments, the area splits into triangles and trapezoids; on a curved one, integrate the formula. Either way the work over a stretch is not the force at the end times the distance, which is the most tempting shortcut and almost always wrong.
Hold a spring stretched by $x$ and it pulls back with $F_s = -kx$. To stretch it slowly you must pull with $+kx$, doing work $\int_0^xkx'\,dx' = \tfrac{1}{2}kx^2$, which is stored as elastic potential energy. When the spring relaxes, it does that work back on whatever it pushes.
The spring's own work between two stretches is $W_s = \tfrac{1}{2}k(x_1^2 - x_2^2)$: positive when the spring relaxes toward its natural length, negative when it is stretched further. The sign follows from the force pointing back toward $x = 0$. Because the energy goes as $x^2$, the second centimeter of stretch costs three times the first, and the tenth centimeter nineteen times.
Integrating $m\,dv/dt = F_{\text{net}}$ with $a = v\,dv/dx$ gives $\int F_{\text{net}}\,dx = \tfrac{1}{2}mv_2^2 - \tfrac{1}{2}mv_1^2$: the net work is the change in kinetic energy. Nothing in the derivation assumed the force was constant, so the theorem works for springs, drag, variable pushes and all of them together.
Its power is that it skips time entirely. To find a speed after a displacement, add up the works of all the forces, set the total equal to $\Delta K$, and solve. Solving the equation of motion in time for a spring pushing a block across a rough floor would take far longer; the energy route takes a line. When a question asks how fast at some position rather than at some time, the work–energy theorem is almost always the right tool.
Power is the rate at which work is done: $P = dW/dt$. Since $dW = \vec{F} \cdot d\vec{r}$, dividing by $dt$ gives $P = \vec{F} \cdot \vec{v}$. A car's engine delivering $60$ kW while the car moves at $30$ m/s must supply a forward force of $2000$ N. At the same power and half the speed, it can push twice as hard, which is why low gears climb hills.
Integrating power over time gives back the work: $W = \int P\,dt$. A motor whose power rises linearly from zero to $P_0$ over a time $T$ does $\tfrac{1}{2}P_0T$ of work. Power is measured in watts, and the kilowatt-hour on an electric bill is $3.6$ MJ of work, delivered at one kilowatt for an hour.
In two or three dimensions, $W = \int\vec{F} \cdot d\vec{r} = \int(F_x\,dx + F_y\,dy + F_z\,dz)$: only the component of force along the path does work. A normal force on a curved track does none, because it is always perpendicular to the motion; neither does the tension in a pendulum's string.
For gravity near the ground, $\vec{F} = -mg\hat{y}$, and the work over any path is $-mg\Delta y$: it depends only on the change in height, not on the route. Forces with that property are conservative, and they have potential energies, which the next lesson builds. Friction is different: sliding a box around a longer route takes more work against friction, because friction always opposes the motion.
Checking an answer. A work in joules must be N times m. A force that points along the motion throughout must do positive work. For a linear force, the work must equal the average of the end forces times the distance. And for a spring, doubling the stretch must quadruple the stored energy.
In the laboratory, force and position are often measured together: a force sensor on a cart and a motion sensor tracking it. Plotting force against position and finding the area under the curve gives the work, whatever the shape. Software adds the areas of the thin trapezoids between data points.
Comparing that work with the measured change in kinetic energy tests the work–energy theorem directly. In a good experiment on a low-friction track, the two agree within a few percent; the difference measures the work done by friction and air, which the sensors do not see. The same method is used in engineering to test springs, shock absorbers and car crumple zones, whose force–displacement curves are far from straight lines.
Near the Earth's surface gravity is nearly constant, and its work is $-mg\Delta y$. Far from the surface the force falls off as $F = GMm/r^2$, pointing toward the Earth's center, and the work must be integrated. Moving outward from $r_1$ to $r_2$, gravity does $W = -\int_{r_1}^{r_2}\frac{GMm}{r^2}\,dr = GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$, which is negative because the motion is against the pull.
Lifting a $1000$ kg satellite from the surface to $400$ km altitude against this force takes $3.7 \times 10^9$ J; the constant-gravity estimate $mgh$ gives $3.9 \times 10^9$ J, about six percent too much, because gravity weakens with height. For a trip to the Moon the difference is huge: the constant-$g$ estimate would be off by a factor of sixty. The same integral, taken out to infinity, gives the energy needed to escape the Earth entirely, which the gravitation lesson turns into the escape speed.
Every October, on Bridge Day, the New River Gorge Bridge in West Virginia is closed to traffic and opened to BASE jumpers and bungee jumpers. A bungee cord hangs slack for its first stretch of fall and then behaves roughly like a spring. At the bottom the jumper is momentarily at rest, so the work done by gravity over the whole fall equals the energy stored in the cord: $mg(L + x) = \tfrac{1}{2}kx^2$.
For a $70$ kg jumper on a cord with $20$ m of slack and $k = 200$ N/m, the cord stretches about $15.6$ m, for a total fall near $36$ m. The quadratic makes the design sensitive: a jumper twice as heavy does not simply double the stretch. Operators choose cords for each jumper's weight and leave a generous margin above the river, because the real cord is stiffer at first and softer later than a perfect spring.
When a car hits a barrier, its front end crushes, and the force on the passenger compartment depends on how far the crush has gone. The work done by that force over the crush distance must absorb the car's kinetic energy: at $15$ m/s, a $1500$ kg car carries $169$ kJ. Crushing over $0.6$ m needs an average force of about $280$ kN.
Engineers at the Insurance Institute for Highway Safety's test center in Virginia measure force against crush distance in crash tests and integrate to find the work. A good design keeps the force nearly constant, a rectangle rather than a spike, because for a given area a flat curve has the lowest peak. The whole field of crashworthiness is, at root, shaping the area under a force–displacement curve.
Because work is force times distance, it seems that twice the stretch should take twice the work. But a spring's force grows as it stretches, so the second stretch is done against a larger force. The work is $\tfrac{1}{2}kx^2$: twice the stretch takes four times the work, and the extra stretch alone takes three times the first.
A related error is to compute the work of any variable force as the final force times the distance. Only a constant force allows that; otherwise integrate, or use the average force, which for a linear force is the mean of the two end values.
A force $F = 6x^2$ (N, m) acts from $x = 0$ to $x = 2.0$ m. Write the work.
$W = \displaystyle\int_0^{2.0}6x^2\,dx$
The force changes along the path.
Find the antiderivative.
$W = 2x^3\Big|_0^{2.0}$
$\int x^2\,dx = x^3/3$.
Evaluate at the limits.
$W = 2 \times 8.0 - 0 = 16\ \text{J}$
Upper limit minus lower.
Compare with the force at the end times the distance.
$F(2.0) \times 2.0 = 24 \times 2.0 = 48\ \text{J}$
Three times too large.
Find the average force.
$\bar{F} = \dfrac{16}{2.0} = 8.0\ \text{N}$
Work over distance.
A spring with $k = 400$ N/m is compressed $0.10$ m against a $0.20$ kg ball. Find the stored energy.
$U = \tfrac{1}{2} \times 400 \times 0.10^2 = 2.0\ \text{J}$
Work done compressing it.
Write the spring's work as it relaxes fully.
$W_s = \tfrac{1}{2}k(0.10^2 - 0^2) = 2.0\ \text{J}$
Positive: it pushes along the motion.
Apply the work–energy theorem on a frictionless track.
$\tfrac{1}{2}mv^2 = 2.0\ \text{J}$
From rest.
Solve for the speed.
$v = \sqrt{\dfrac{2 \times 2.0}{0.20}} = \sqrt{20} = 4.5\ \text{m/s}$
Take the root.
Find the speed when the compression has fallen to $0.05$ m.
$\tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 400 \times (0.01 - 0.0025) = 1.5 \Rightarrow v = 3.9\ \text{m/s}$
Three-quarters of the energy is released in the first half of the travel.
A $1500$ kg car climbs a road rising $30$ m. Find the work done by gravity.
$W_g = -mg\Delta y = -1500 \times 9.8 \times 30 = -4.41 \times 10^5\ \text{J}$
Independent of the road's shape.
Its speed stays $20$ m/s, and drag and rolling friction do $-1.2 \times 10^5$ J. Find the engine's work.
$W_{\text{engine}} = \Delta K - W_g - W_f = 0 + 4.41 \times 10^5 + 1.2 \times 10^5$
The work–energy theorem with no change in kinetic energy.
Evaluate the engine's work.
$W_{\text{engine}} = 5.61 \times 10^5\ \text{J}$
Positive: the engine supplies energy.
The climb takes $40$ s. Find the average power.
$\bar{P} = \dfrac{5.61 \times 10^5}{40} = 1.40 \times 10^4\ \text{W}$
Work over time.
Find the average forward force.
$F = \dfrac{P}{v} = \dfrac{1.40 \times 10^4}{20} = 701\ \text{N}$
$P = Fv$.
Convert the power to horsepower.
$\dfrac{1.40 \times 10^4}{746} = 18.8\ \text{hp}$
A modest share of a typical engine's rating.
Write the work.
$W = \tfrac{1}{2}k(x_2^2 - x_1^2)$
The integral of $kx$.
Substitute the values.
$W = \tfrac{1}{2} \times 300 \times (0.04 - 0.01)$
Square each stretch.
Evaluate the work.
Stretching a spring from its natural length by $x$ takes $8$ J of work. How much more work does it take to stretch it from $x$ to $2x$?
Complete the worked solution: a pull along a track weakens as the cart moves, $F = 38 - 3x$ (newtons, meters). Find the force at $x = 3$ m, the work it does from $x = 0$ to $x = 3$ m in joules, and its average value over that stretch in newtons.
Evaluate the force at the end of the stretch.
$F(L) =$ f
Substitute the position.
Integrate the force over the stretch.
$W = aL - \tfrac{1}{2}bL^2 =$ w
The area of a trapezoid under a falling line.
Divide the work by the distance.
$\bar{F} = \dfrac{W}{L} =$ m
For a linear force it is the mean of the end values.
Check the average against the end values.
$\text{average} = \text{midpoint of the starting and ending forces}$
True only because the force changes linearly with position.
Match each quantity to its expression.
| $\int F\,dx$ | $\tfrac{1}{2}kx^2$ | $\vec{F} \cdot \vec{v}$ | $W_{\text{net}} = \Delta K$ | |
|---|---|---|---|---|
| work of a variable force | ||||
| spring energy | ||||
| power | ||||
| work–energy theorem |
A force along the $x$ axis is $F = 9x^2$ (newtons, meters). Fill in the work it does from $x = 0$ to $x = 3$ m, the work from $x = 3$ m to $x = 6$ m, and its average value over the first stretch, in newtons.
| value | |
|---|---|
| work from 0 to the first mark (J) | |
| work over the next stretch (J) | |
| average force (N) |
A spring-like force $F = 4x$ (newtons, meters) acts on an object moved from $x = 0$ to $x$. Write the work it does, in joules, as a formula in $x$.
Answer:
A spring of stiffness $400$ N/m is compressed $8$ cm. It pushes a block as it relaxes to a compression of $5$ cm. How much work does the spring do on the block, in joules?
Answer: J from the spring
On Bridge Day at the New River Gorge in West Virginia, a $55$ kg jumper uses a cord with $18$ m of slack that then stretches like a spring of stiffness $120$ N/m. How far does the cord stretch at the lowest point, in meters? Use $g = 9.8$ m/s².
Answer: m of cord stretch
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A spring-like force $F = 8x$ (newtons, meters) acts on an object moved from $x = 0$ to $x$. Write the work it does, in joules, as a formula in $x$.
Answer:
You can compute the work of variable forces. Explain to someone why the second centimeter of a spring's stretch takes three times the work of the first.
19. Your turn: find the work to stretch a spring with $k = 300$ N/m from $0.10$ m to $0.20$ m., step 3
$W = 4.5\ \text{J}$
Three times the work of the first $0.10$ m.