Work and kinetic energy

Solving the second law requires knowing the force at every instant of time, and most interesting forces are known instead as functions of position, which is the gap this lesson closes.

What is carried in is $\sum \vec{F} = m \vec{a}$ and the identity from the first lesson that $a = v d v / d x$ . Nothing else is assumed; energy is built here from scratch.

Integrating over distance

The first lesson noticed a trick and promised to come back to it. Write acceleration using the chain rule with position as the intermediate variable, $a = d v / d t = (d v / d x) (d x / d t) = v d v / d x$ , so that $a d x = v d v$ . Multiply the second law by $d x$ and integrate from one position to another:

\int_{x_{1}}^{x_{2}} F d x = \int_{v_{1}}^{v_{2}} m v d v = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}

Time has vanished. What is left on the left is a new quantity, the integral of force over distance, called the work. What is left on the right is another new quantity, $\frac{1}{2} m v^{2}$ , evaluated at the two ends, called the kinetic energy. The equation says that the work done equals the change in kinetic energy, and it is not an extra physical law: it is the second law integrated once, so anything it predicts could in principle be got the long way round.

The reason to bother is that the long way round is often impossible. Ask how fast a roller coaster reaches the bottom of a drop and the answer follows in one line; ask for the same thing by integrating the second law and you need the shape of the track, the time of every instant, and a numerical solver.

Work, and what the angle does

For a constant force $\vec{F}$ moving its point of application through a displacement $\vec{s}$ , the work is the product of the displacement with the component of the force along it:

W = F s \cos φ

where $φ$ is the angle between the force and the displacement. This combination is common enough to have a name, the scalar product of the two vectors, written $\vec{F} \cdot \vec{s}$ , and in components it is $F_{x} s_{x} + F_{y} s_{y}$ . It turns two vectors into a single number, and that number has no direction: work is a scalar, and a body has no notion of which way its kinetic energy points. The unit is the joule, one newton metre, and 1 J = 1 kg m² s⁻².

The cosine carries three cases worth separating. When the force has a component along the motion, $φ < 90^{\circ}$ and the work is positive: energy goes in. When it opposes the motion, as friction always does, the work is negative: energy comes out. And when the force is perpendicular to the motion, the work is exactly zero.

That last case is not a curiosity, it is most of mechanics. The normal force on a block sliding along a floor does no work, because the floor pushes up while the block moves sideways. The tension in a string does no work on a conical pendulum bob. The gravitational force on a satellite in a circular orbit does no work, which is why its speed is constant. Static friction on the driving wheel of a car does no work either, since the contact patch is instantaneously at rest, and the energy that accelerates the car comes from the fuel rather than from the road.

Example. A 20 kg crate is dragged 12 m along a floor by a rope pulling at 50 N, 30° above the horizontal, against $μ_{k} = 0.15$ . Find the work done by each force and the final speed, starting from rest.

The rope does $W = (50) (12) \cos 30^{\circ} = 519.6$ J. The normal force must be found before friction can be: vertically, $N + 50 \sin 30^{\circ} = m g$ , so $N = 196.2 - 25.0 = 171.2$ N, less than the weight because the rope lifts as well as pulls. Friction is then $0.15 \times 171.2 = 25.68$ N opposing the motion, doing $- 25.68 \times 12 = - 308.2$ J. Gravity and the normal force do no work, being perpendicular to the displacement. The net work is $519.6 - 308.2 = 211.5$ J, so $\frac{1}{2} (20) v^{2} = 211.5$ and $v = 4.60$ m s⁻¹.

Now you. The same crate is dragged 6.0 m by an 80 N rope at 25° above the horizontal, with the same $μ_{k} = 0.15$ . How much work does the rope do, and what is the work done against friction?

Answer

The rope does $(80) (6.0) \cos 25^{\circ} = 435$ J. The normal force is $196.2 - 80 \sin 25^{\circ} = 196.2 - 33.8 = 162.4$ N, so friction is 24.4 N and does $- 146$ J.

The work energy theorem

Collecting the result and stating it as a theorem: for any body, the total work done by all forces acting on it equals the change in its kinetic energy.

W_{net} = Δ K = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}

Kinetic energy is never negative and does not depend on the direction of travel, which is both the strength and the weakness of the method. It is a strength because a curved, complicated path can be handled without tracking the direction at every point. It is a weakness because one scalar equation cannot determine a two dimensional velocity: the theorem gives the speed at the bottom of a slide and says nothing about which way the rider is facing.

The dependence on $v^{2}$ is where the practical bite is. A 1500 kg car at 28 m s⁻¹ carries $\frac{1}{2} (1500) (784) = 588$ kJ. Stopping it in 50 m requires the brakes to do $- 588$ kJ of work, an average force of 11.8 kN, which is exactly what a coefficient of friction of 0.80 supplies, since $0.80 \times 1500 \times 9.81 = 11.8$ kN. Double the speed and the energy quadruples, so the same brakes need four times the distance. Every argument about speed limits and stopping distances is this quadratic, and it is worth noting that the energy has to go somewhere: 588 kJ deposited into brake discs is enough to raise 2 kg of steel by roughly 600 K, which is why brakes fade on long descents.

Variable forces, and the spring

The theorem was derived as an integral, so a force that varies with position costs nothing extra. The standard case is a spring. Robert Hooke published the relation in 1678 as the anagram ceiiinosssttuv, unscrambled as ut tensio sic vis, as the extension so the force:

F = - k x

with $x$ the displacement from the natural length and $k$ the stiffness in newtons per metre. The minus sign says the force opposes the displacement, which is what makes a spring restore rather than run away. To stretch a spring you must pull with $+ k x$ , so the work you do is

W = \int_{0}^{x} k x^{'} d x^{'} = \frac{1}{2} k x^{2}

That is the area of a triangle under the straight line $F = k x$ , which is a useful check: for a linear force law the work is the average force times the distance, and the average of a force rising from 0 to $k x$ is $\frac{1}{2} k x$ .

Hooke's law is a first order approximation and fails for large extensions, where a real spring stiffens and then yields permanently. The reason it works at all for small ones is the same reason it will reappear in the lesson on oscillations: any smooth restoring force looks linear close enough to equilibrium.

Example. A spring of stiffness 800 N m⁻¹ is compressed 0.15 m and released against a 0.50 kg block on a frictionless surface. How fast does the block leave the spring?

The spring does $\frac{1}{2} k x^{2} = \frac{1}{2} (800) (0.0225) = 9.0$ J of work on the block as it returns to its natural length. All of it becomes kinetic energy, so $\frac{1}{2} (0.50) v^{2} = 9.0$ and $v = \sqrt{36} = 6.0$ m s⁻¹.

Now you. A spring of stiffness 250 N m⁻¹ is stretched 0.20 m. How much work was done stretching it, and what speed would that give a 0.40 kg block?

Answer

$W = \frac{1}{2} (250) (0.04) = 5.0$ J. Then $\frac{1}{2} (0.40) v^{2} = 5.0$ gives $v^{2} = 25$ and $v = 5.0$ m s⁻¹.

Power

Work says how much energy moved; power says how fast. It is the rate of doing work,

P = \frac{d W}{d t} = \vec{F} \cdot \vec{v}

measured in watts, one joule per second. The second form follows because $d W = \vec{F} \cdot d \vec{s}$ and $d \vec{s} / d t = \vec{v}$ , and it is usually the more useful one, since it needs no integration.

James Watt, needing to sell engines to people who owned horses, defined the horsepower in the 1780s by measuring how much a mill horse could lift, and set it at 550 foot-pounds per second, which is 746 W. The figure is generous: a horse sustains rather less, and a fit human cyclist sustains about 200 to 300 W for an hour, with sprint peaks above 1000 W.

The $\vec{F} \cdot \vec{v}$ form explains the shape of a car's performance. A 100 kW engine at 30 m s⁻¹ can deliver at most $100000 / 30 = 3.3$ kN of force, and at 60 m s⁻¹ only 1.7 kN, which is why acceleration falls away at speed even with the throttle wide open. Top speed arrives when all the power is being spent against drag: with $C_{d} A = 0.66$ m² and air at 1.2 kg m⁻³, the drag at 30 m s⁻¹ is $\frac{1}{2} (1.2) (0.66) (900) = 356$ N, costing 10.7 kW. Since drag force goes as $v^{2}$ , drag power goes as $v^{3}$ , and doubling the top speed of a car requires eight times the power.

Example. A cyclist and machine of total mass 85 kg climb a 6 per cent gradient at a steady 5.0 m s⁻¹. What power is needed against gravity alone?

A 6 per cent gradient rises 6 m in 100 m along the road, so $\sin θ = 0.0599$ once the angle is worked out from $\tan θ = 0.06$ . At steady speed the rider's force along the road matches the gravitational component $m g \sin θ = (85) (9.81) (0.0599) = 49.9$ N, and the power is $49.9 \times 5.0 = 250$ W. That is a hard but sustainable effort for a trained rider, and it explains why hills are where cycling races are decided: on the flat the same 250 W buys a much higher speed, because it is fighting drag rather than gravity.

Now you. A lift of total mass 1200 kg rises at a constant 2.0 m s⁻¹. What power must the motor supply, ignoring any counterweight and friction?

Answer

At constant speed the cable tension equals the weight, $1200 \times 9.81 = 11772$ N, so $P = F v = 11772 \times 2.0 = 23.5$ kW. A real lift uses a counterweight of roughly the car plus half the load, which cuts this by most of its value.

What the theorem cannot do

The work energy theorem holds for every force, including friction, and that generality hides a distinction which the next lesson makes central.

Slide a block round a closed loop on a rough table and return it to its start. Gravity did no net work, since the height did not change. The normal force did none, being perpendicular throughout. Friction, though, did negative work on every centimetre of the journey, because it always opposes the motion, and over the whole loop it took out an amount proportional to the path length. Take a longer route and it takes out more.

So some forces have the property that the work they do between two points is the same for every route, and others do not. That distinction is not a technicality: it is the difference between a force for which a potential energy can be defined and one for which it cannot, and therefore between energy that can be recovered and energy that is gone. Where the energy goes when friction takes it is a question that mechanics cannot answer at all and thermodynamics was invented to answer.

The next lesson develops the recoverable case and gets a conserved total energy out of it, which turns out to be the single most useful equation in the subject.