Potential energy and conservation

Friction takes more energy out of a body the longer the route it travels, while gravity takes the same amount whatever route is taken between two heights, and that difference is what makes energy conservation possible.

Carried in from the previous lesson: work is the integral of force over distance, the net work on a body equals its change in kinetic energy $\frac{1}{2} m v^{2}$ , and stretching a spring by $x$ takes $\frac{1}{2} k x^{2}$ of work.

The test that divides forces in two

Take a body from point A to point B by two different routes and compute the work done by one particular force on each route. For some forces the two answers agree, always, for every pair of routes. Such a force is called conservative, and an equivalent statement is that the work it does round any closed loop is zero.

Gravity near the ground passes the test immediately. The force is $m g$ downward, so the work it does is $- m g$ times the height gained, whatever the horizontal wandering, because horizontal displacement is perpendicular to the force and contributes nothing. Walk to the top of a hill by the steep path or the long zigzag and gravity has taken the same $m g h$ from you. An ideal spring passes too, since its work depends only on the extensions at the two ends.

Friction fails, and fails structurally rather than by accident. Kinetic friction always opposes the motion, so its work is negative on every segment of every path and can never cancel on the way back. Round a closed loop it gives $- μ_{k} N L$ where $L$ is the total path length, which is not zero and grows without limit as the path is made longer. Air resistance fails for the same reason.

The distinction is the whole lesson. For a conservative force, the work done between two points is a property of the two points, so it can be tabulated in advance as a function of position and looked up rather than integrated. For a non-conservative force it cannot, because there is no such function to tabulate.

Potential energy

For a conservative force, define the potential energy $U$ so that the work the force does going from 1 to 2 equals the drop in $U$ :

W_{1 \to 2} = U_{1} - U_{2} = - Δ U

Equivalently, $U (x) = - \int F d x$ , and differentiating back,

F = - \frac{d U}{d x}

Force is minus the slope of the potential energy curve. That single relation replaces a great deal of reasoning about directions: a body is pushed towards lower potential energy, downhill on the graph, and the steeper the graph the harder the push.

Two instances cover almost everything in this course. Near the ground, $F = - m g$ taking up as positive, so $U = m g y$ , with the zero of height chosen wherever it is convenient. For a spring, $F = - k x$ gives $U = \frac{1}{2} k x^{2}$ , measured from the natural length.

The arbitrary constant is not a defect. Only differences in $U$ appear in any physical result, so choosing the floor, the table top or sea level as the zero changes every value of $U$ and no answer. It is also worth being careful about where the energy lives. Potential energy is not stored in the body; it belongs to the configuration of the interacting system, the book and the Earth together. Saying that a raised book has potential energy is shorthand, and the shorthand becomes misleading in the lesson on gravitation, where the two bodies are of comparable importance.

Conservation of mechanical energy

Now put the two halves together. The work energy theorem says $W_{net} = Δ K$ . If every force doing work is conservative, then $W_{net} = - Δ U$ , so $Δ K + Δ U = 0$ , which means the sum

E = K + U = \frac{1}{2} m v^{2} + U (x)

does not change as the body moves. This is conservation of mechanical energy, and it is a consequence of the second law rather than an addition to it.

Its practical value is that it relates speed to position directly, skipping the trajectory. A ball dropped 40 m arrives at $\sqrt{2 g h} = \sqrt{2 (9.81) (40)} = 28.0$ m s⁻¹, and so does a ball that slid down a frictionless curved chute of the same height, or one that swung down on a string, because the equation contains only the height. That indifference to the path is what makes it powerful, and it is the same indifference that makes it silent about direction and about time.

Example. A roller coaster car starts from rest at the top of a 40 m drop and runs on a frictionless track into a vertical loop of radius 12 m. How fast is it at the bottom, how fast at the top of the loop, and does it make it round?

Take the bottom of the track as the zero of height. From the start to the bottom, $m g h = \frac{1}{2} m v^{2}$ gives $v = 28.0$ m s⁻¹, independent of mass, which is why coasters do not need to be weighed. At the top of the loop the car is $2 \times 12 = 24$ m up, so $\frac{1}{2} m v^{2} = m g (40 - 24)$ and $v = \sqrt{2 (9.81) (16)} = 17.7$ m s⁻¹. The previous lesson's condition for staying on the track is $v \geq \sqrt{g r} = \sqrt{9.81 \times 12} = 10.85$ m s⁻¹, so 17.7 is comfortable. Note that the two lessons were needed together: energy gave the speed, and circular dynamics said what speed was enough.

Now you. The same car starts from rest at the top of a 25 m drop, on a frictionless track. What is its speed at the bottom, and could it get round a loop of radius 12 m?

Answer

At the bottom, $v = \sqrt{2 (9.81) (25)} = 22.1$ m s⁻¹. At the top of the loop, 24 m up, only 1 m of drop is left, giving $v = \sqrt{2 (9.81) (1)} = 4.4$ m s⁻¹, well below the 10.85 m s⁻¹ needed. The car would leave the track before reaching the top.

Energy diagrams

Draw $U (x)$ against $x$ and add a horizontal line at the total energy $E$ . Since $K = E - U$ and kinetic energy cannot be negative, the body is confined to the regions where the curve lies below the line, and the points where the curve meets the line are turning points, where the speed is zero and the motion reverses.

The picture answers questions that would otherwise take a calculation. A dip in $U$ with the line cutting both sides gives motion trapped between two turning points, which is bound: a pendulum, a mass on a spring, a planet in orbit. Raise the line above the lip on one side and the body escapes that way, which is how escape velocity is defined in a later lesson. Where the curve is flat, $d U / d x = 0$ and the force is zero, which is equilibrium: stable at a minimum, since a displacement produces a force back towards it, and unstable at a maximum, since a displacement produces a force away. A ball in a valley and a ball balanced on a hilltop both have zero force on them, and the second derivative is what tells them apart.

That last observation is the seed of a later lesson. Near a minimum, every smooth potential looks like a parabola, and a parabolic potential is exactly a spring, so every stable system oscillates the same way when disturbed gently. Reading it off the graph costs nothing; deriving it takes a Taylor expansion, which is done when the oscillation lesson arrives.

Example. A pendulum of length 2.0 m is released from rest at 40° to the vertical. How fast is the bob moving at the lowest point?

The tension does no work, being perpendicular to the motion throughout, so mechanical energy is conserved with gravity alone. The bob rises above its lowest point by $h = L (1 - \cos θ) = 2.0 (1 - \cos 40^{\circ}) = 2.0 (1 - 0.766) = 0.468$ m. Then $v = \sqrt{2 g h} = \sqrt{2 (9.81) (0.468)} = 3.03$ m s⁻¹. Nothing about the swing's shape or duration entered.

Now you. A pendulum of length 1.5 m is released from rest at 60° to the vertical. Find the speed at the lowest point.

Answer

$h = 1.5 (1 - \cos 60^{\circ}) = 1.5 \times 0.5 = 0.750$ m, so $v = \sqrt{2 (9.81) (0.750)} = 3.84$ m s⁻¹.

When friction is there anyway

Real problems have friction, and the method survives with one extra term. Split the forces into conservative and the rest, and the work energy theorem becomes

Δ K + Δ U = W_{other}

where $W_{other}$ is the work done by every non-conservative force, negative for friction and drag, positive for a motor or a person pushing. Mechanical energy is not conserved; it changes by exactly the work of the other forces, and no book-keeping is lost.

Example. A 2.0 kg block slides 5.0 m down a 30° incline with $μ_{k} = 0.25$ , starting from rest. How fast is it going at the bottom?

The height dropped is $5.0 \sin 30^{\circ} = 2.50$ m, so gravity releases $m g h = (2.0) (9.81) (2.50) = 49.05$ J. The normal force is $m g \cos 30^{\circ} = 16.99$ N, so friction is 4.248 N and removes $4.248 \times 5.0 = 21.24$ J over the slide. What is left is $49.05 - 21.24 = 27.81$ J of kinetic energy, giving $v = \sqrt{2 (27.81) / 2.0} = 5.27$ m s⁻¹, against the 7.0 m s⁻¹ a frictionless slide would have given.

Now you. A 3.0 kg block slides 4.0 m down a 25° incline with $μ_{k} = 0.35$ , from rest. Find its speed at the bottom.

Answer

The drop is $4.0 \sin 25^{\circ} = 1.690$ m, releasing $m g h = 49.75$ J. Friction is $0.35 (3.0) (9.81) \cos 25^{\circ} = 9.335$ N, removing 37.34 J. The remaining kinetic energy is 12.41 J, so $v = \sqrt{2 (12.41) / 3.0} = 2.88$ m s⁻¹.

Where the missing energy goes

The 21.24 J that friction removed did not cease to exist. It went into heating the block and the incline, and mechanics has nothing to say about that, because a temperature is not a mechanical variable and the sliding surfaces are not a point particle.

This is the honest boundary of the subject. Mechanical energy is conserved only when the non-conservative forces do no work; total energy is conserved always, but proving that requires counting the energy stored in the disordered motion of enormous numbers of molecules, which is where thermodynamics begins. The joule is the same unit in both subjects, and the reason the two fields use it is Joule's discovery, in the 1840s, that a fixed amount of mechanical work always produces a fixed amount of heating: 4.18 kJ per kilogram of water per kelvin, measured by letting falling weights turn a paddle in an insulated tank.

There is also an asymmetry worth stating plainly. The kinetic energy of an ordered stream of molecules can be turned entirely into disordered motion, as friction does effortlessly; the reverse conversion is limited by a law that mechanics does not contain. Classical mechanics is time reversible, and a film of a block sliding to a halt run backwards shows a block spontaneously cooling and accelerating, which no one has ever seen. The equations of this course permit it.

What one scalar equation cannot do

Energy conservation gives a single equation relating speed and position, and one equation cannot determine a velocity in two dimensions. That is a real limitation, and the case where it bites is the collision.

Two pucks meet on ice and separate. Energy conservation, if it held, would give one relation between the two outgoing speeds, leaving the directions undetermined. Worse, it usually does not hold: a collision that dents, deforms or sticks converts mechanical energy into heat and sound, and there is no way to know in advance how much.

What survives is a different quantity, and it survives precisely because the third law makes the internal forces cancel in pairs. It is a vector, so it carries direction, and it is conserved in every collision whether energy is or not. That quantity is momentum, and it is next.