Two friends climb from the same village to the same temple on a hilltop near Dharamshala. One takes the short, steep trail. The other takes the long trail that winds gently around the hill. Both reach the temple with the same weight on their backs. The one on the long trail is more tired, but is that because of gravity?
The path test
Carry a body from a point A to a point B along some path. A force acting on it does some work \(W_{AB}\) along the way. Now carry it from A to B along a different path. For some forces the work is the same on every path. It depends only on where the body starts and where it ends. Such a force is a conservative force.
Gravity near the Earth is the standard example. Its work is \(-mg\,\Delta h\), where \(\Delta h\) is the rise in height. A steep path and a gentle path between the same two points have the same \(\Delta h\). So gravity does the same work on both. The spring force is another example: its work depends only on the stretch at the start and at the end.
Play the route "up, across, down" in Fig 2.1 and watch the gravity readout. It falls to \(-50\,\text{J}\) at the top of the climb. On the way down it rises again to \(-30\,\text{J}\). Gravity returns on the descent the extra it took on the climb. The drag readout only ever falls, by \(2\,\text{J}\) for every metre moved.
The round-trip test
Now choose "to B and back". The body ends where it started, so the path is a closed loop. Gravity's work over the round trip is zero: it took \(30\,\text{J}\) on the way up and returned \(30\,\text{J}\) on the way down. The drag took \(10\,\text{J}\) on each leg, \(20\,\text{J}\) in all. It returned nothing.
A force is conservative if the work it does on a body moving between any two points is the same for every path joining them. Equivalently, the work it does around every closed path is zero.
The two tests always agree. Suppose the work from A to B is the same on path 1 and on path 2. Go out along path 1 and return along path 2. The return trip reverses every displacement, so its work is the negative of the work along path 2. The round trip then gives \(W_1 - W_2 = 0\). Read backwards, a zero round trip forces \(W_1 = W_2\).
A \(3\,\text{kg}\) box is pushed up a \(10\,\text{m}\) long ramp onto a platform \(4\,\text{m}\) high. Take \(g = 10\,\text{m/s}^2\). What is the work done by gravity on the box, in joules?
A box is dragged across a rough floor from the door to a cupboard and then back to the door. What is the total work done on it by gravity, and by kinetic friction?
Why the test decides everything
A potential energy \(U\) is a single number attached to each point. Between two points its change \(U_B - U_A\) has one value. Lesson 1 set that change equal to \(-W_{AB}\). This is possible only if \(W_{AB}\) also has one value, the same on every path. So a force can have a potential energy only if it is conservative.
A force that fails the test is a non-conservative force. Kinetic friction always opposes the sliding, so it takes energy on every metre in every direction. Air drag does the same. A longer route costs more, and no number attached to the points could track that. So friction and drag have no potential energy. Their work must be written out in full in every calculation.
A \(5\,\text{kg}\) crate is dragged across a level floor with \(\mu_k = 0.2\) from a point A to a point B \(6\,\text{m}\) away. Take \(g = 10\,\text{m/s}^2\). Find the work of friction on the straight route, on a detour \(10\,\text{m}\) long, and on a straight round trip to B and back.
Friction has a constant size \(f = \mu_k mg = 0.2 \times 5 \times 10 = 10\,\text{N}\), always against the motion.
Straight route: \(W = -10 \times 6 = -60\,\text{J}\). Detour: \(W = -10 \times 10 = -100\,\text{J}\).
Round trip: \(W = -10 \times 12 = -120\,\text{J}\), not zero. Gravity does no work on any of these routes, because the floor is level.
Gravity works like a bank locker: whatever you deposit on the climb, you withdraw in full on the descent. Friction works like a toll road: it charges for every kilometre, in both directions, and never refunds a rupee.
Where the lost energy goes
A puck slides across a table and stops. Its kinetic energy is gone, but nothing was destroyed. The rubbing surfaces warm up slightly. Friction has turned the ordered motion of the puck into the random jiggling of the molecules of the puck and the table. That random motion is heat, or thermal energy. The total energy is unchanged.
We still call it a loss, for a practical reason. The heat is spread over countless molecules moving in all directions. No mechanical process can collect it into the ordered motion of the puck again. The energy is still there, but it can no longer do the job we want.
A puck slides across a rough table and stops. It had \(40\,\text{J}\) of kinetic energy. Which statement is right?
Central forces
A central force always points along the line joining the body to a fixed point O, towards O or away from it. Its size depends only on the distance \(r\) from O. The Sun's pull on a planet is a central force. So is the electric force of a fixed charge. So is the pull of a spring tied to a fixed peg on a smooth table.
Every central force is conservative, and the proof needs only one idea. Move the body along an arc of a circle about O. The displacement along the arc is perpendicular to the radius, and the force lies along the radius. So the force does no work on any arc. On a short step straight towards or away from O, the work is \(F(r)\,dr\).
\[ d\vec s = dr\,\hat r + r\,d\theta\,\hat\theta \]
Why. any small step splits into a part along the radius and a part along the arc. Here \(\hat r\) points away from O and \(\hat\theta\) along the arc.
\[ dW = \vec F \cdot d\vec s = F(r)\,\hat r \cdot (dr\,\hat r + r\,d\theta\,\hat\theta) \]
Why. the force lies along \(\hat r\) and has size \(F(r)\), positive when it points away from O.
\[ dW = F(r)\,dr \]
Why. \(\hat r \cdot \hat r = 1\) and \(\hat r \cdot \hat\theta = 0\); the arc part of the step does no work.
\[ W_{AB} = \int_{r_A}^{r_B} F(r)\,dr \]
Read it. add the steps. Only the change in \(r\) enters, so the result depends on \(r_A\) and \(r_B\) alone, whatever the path.
Any path can be approximated by such a staircase of arcs and radial steps, as finely as we like. The arcs contribute nothing. The radial steps add up to \(\int F(r)\,dr\) between the same two radii on every staircase. So the work depends only on the distances of A and B from O. That is the path test, passed. Every central force has a potential energy that depends on \(r\) alone.
A planet moves along a short arc of its orbit at a nearly constant distance from the Sun. What work does the Sun's gravity do on it over that arc?
A puck on a smooth table is tied to a fixed peg by a spring of natural length \(0.5\,\text{m}\) and constant \(100\,\text{N/m}\). The puck moves along a winding path from a point \(0.6\,\text{m}\) from the peg to a point \(0.8\,\text{m}\) from the peg. Find the work done by the spring, in joules.
Return now to the two trekkers. Gravity is conservative, so it does the same work, \(-mg\,\Delta h\), on both trails. The trekker on the long trail is more tired for other reasons. The muscles do non-conservative work on every step, and much of it becomes heat in the body. Gravity charged both of them the same.
In the 1840s James Prescott Joule, the son of a brewer near Manchester, let falling weights turn a paddle wheel inside a tank of water. The paddles churned the water, and friction in the water stopped them. Joule measured a tiny rise in the water's temperature. He found that the same loss of potential energy of the weights always gave the same amount of heat. Energy that friction seemed to destroy had become heat, in a fixed exchange. The unit of energy is named after him.
- Calling a force conservative because it is constant in size.A drag of constant size still does more work on a longer path. Apply the path test or the round-trip test.
- Saying friction destroys energy.Friction turns mechanical energy into heat. Total energy is conserved.
- Taking friction's work over a round trip to be zero.Friction opposes the motion on every leg, so its negative works add.