A loaded bus descends the winding ghat road from Mahabaleshwar at a steady, careful speed. At the bottom the smell of hot brakes hangs in the air. Drivers are told to use a low gear rather than the brakes alone. The bus lost hundreds of metres of height but gained no speed. The energy had to go somewhere.
When a non-conservative force does work
The last lesson assumed that no non-conservative force does work. Now remove that assumption. Friction, drag, the push of a hand and the pull of an engine are all non-conservative forces. Keep their total work \(W_{nc}\) in the derivation of the last lesson. The result is the most useful equation of the chapter.
\[ W_c + W_{nc} = \Delta K \]
Why. the work-energy theorem: the work of all forces equals the change in kinetic energy.
\[ W_c = -\Delta U \]
Why. the conservative forces are counted through their potential energy.
\[ -\Delta U + W_{nc} = \Delta K \]
Why. substitute step 2 into step 1.
\[ W_{nc} = \Delta K + \Delta U = \Delta E \]
Read it. move \(\Delta U\) to the right; the sum of the two changes is the change in mechanical energy.
The total work done by the non-conservative forces on a system equals the change in its mechanical energy: \(W_{nc} = \Delta E = (K_f + U_f) - (K_i + U_i)\).
The sign of \(W_{nc}\) tells the story. Friction and drag always oppose the motion, so their work is negative and the mechanical energy falls. A push from a hand, or the pull of a lift's cable, can do positive work and raise the mechanical energy. When \(W_{nc} = 0\), the equation becomes the conservation law of the last lesson.
A lift of mass \(500\,\text{kg}\) is pulled up \(10\,\text{m}\) by its cable at a steady speed. Take \(g = 10\,\text{m/s}^2\). What happens to the mechanical energy of the lift and the Earth?
Friction's bill: f times the distance slid
Kinetic friction of size \(f\) always points against the sliding. Over each small step of length \(ds\) it does work \(-f\,ds\), whatever the direction of the step. Add the steps. If \(f\) is constant, friction does work \(-fs\), where \(s\) is the total distance slid, not the displacement. A block that slides forward and then back pays for both legs.
So the mechanical energy falls by exactly \(fs\), and exactly that much heat appears in the block and the surface. Nothing is missing. The total energy, mechanical energy plus heat, stays constant. In the language of the last lesson, mechanical energy is not conserved, but energy is.
Follow the account in Fig 4.1. The block starts with \(K = \tfrac12 \times 2 \times 4^2 = 16\,\text{J}\). After sliding \(1\,\text{m}\) it reaches the spring with \(12\,\text{J}\), having made \(4\,\text{J}\) of heat. It compresses the spring by \(0.5\,\text{m}\): then \(U = \tfrac12 \times 80 \times 0.5^2 = 10\,\text{J}\) and the heat is \(6\,\text{J}\). The spring pushes it back and it leaves with \(8\,\text{J}\). It then slides \(2\,\text{m}\) and stops. It slid \(4\,\text{m}\) in all, so the heat is \(4 \times 4 = 16\,\text{J}\): every joule it started with.
A \(2\,\text{kg}\) block is released from rest at the top of an incline of \(37°\) and length \(5\,\text{m}\), with \(\mu_k = 0.25\). At the bottom it slides onto a floor with the same \(\mu_k\). Take \(g = 10\,\text{m/s}^2\), \(\sin 37° = 0.6\) and \(\cos 37° = 0.8\). Find the speed at the bottom of the incline and the distance it slides on the floor.
The incline is \(5 \times 0.6 = 3\,\text{m}\) high. On the incline \(N = mg\cos 37° = 16\,\text{N}\), so \(f = 0.25 \times 16 = 4\,\text{N}\). On the floor \(N = 20\,\text{N}\) and \(f = 5\,\text{N}\).
On the incline: \(mgh - fs = \tfrac12 mv^2\). On the floor: \(\tfrac12 mv^2 = f\,d\).
\(2 \times 10 \times 3 - 4 \times 5 = 60 - 20 = 40\,\text{J}\), so \(\tfrac12 \times 2 \times v^2 = 40\) and \(v = \sqrt{40} \approx 6.3\,\text{m/s}\). On the floor \(40 = 5\,d\), so \(d = 8\,\text{m}\).
The block reaches the bottom at about \(6.3\,\text{m/s}\) and slides \(8\,\text{m}\) on the floor. In all, friction turns the full \(60\,\text{J}\) of potential energy into heat: \(20\,\text{J}\) on the incline and \(40\,\text{J}\) on the floor.
The \(s\) in \(fs\) is the distance slid, which only grows. A block that slides \(3\,\text{m}\) forward and \(1\,\text{m}\) back has a displacement of \(2\,\text{m}\). But it has slid \(4\,\text{m}\), and friction makes heat for all four metres.
A \(0.5\,\text{kg}\) box slides across a floor with \(\mu_k = 0.4\) and stops after sliding \(2\,\text{m}\). Take \(g = 10\,\text{m/s}^2\). What was its initial speed, in m/s?
A block slides down a rough slope. On the way it loses \(50\,\text{J}\) of potential energy and reaches the bottom with \(30\,\text{J}\) of kinetic energy, starting from rest. Which statement is right?
Three stores in one account: the bungee jump
A bungee jumper falls from a platform tied to an elastic cord of natural length \(L\) and spring constant \(k\). For the first \(L\) metres the cord is slack and the jumper falls freely. Below that the cord stretches and pulls up. If air drag is small, no non-conservative force does work. Three stores then share one constant total: gravitational energy \(U_g\), kinetic energy \(K\) and the elastic energy of the cord.
A \(60\,\text{kg}\) jumper falls on a cord of natural length \(20\,\text{m}\) and constant \(120\,\text{N/m}\). The cord becomes taut after \(20\,\text{m}\) of fall, and the jumper stops \(40\,\text{m}\) down. Where is the jumper moving fastest?
The figure takes the zero of \(U_g\) at the lowest point. At the platform the jumper has \(U_g = mgH = 60 \times 10 \times 40 = 24\,\text{kJ}\) and nothing else. At the lowest point, \(40\,\text{m}\) down, the jumper is at rest and the cord is stretched by \(20\,\text{m}\). Its energy is \(\tfrac12 \times 120 \times 20^2 = 24\,\text{kJ}\): every joule has moved from height into the cord.
A \(60\,\text{kg}\) jumper falls from rest on a cord of natural length \(L = 20\,\text{m}\) and constant \(k = 120\,\text{N/m}\). Ignore drag and take \(g = 10\,\text{m/s}^2\). Find the total fall \(H\), the speed when the cord becomes taut and the greatest speed.
System: jumper, cord and Earth. The jumper is at rest at the start and at the lowest point. Measure the fall \(y\) downward from the platform.
At the lowest point \(mgH = \tfrac12 k(H - L)^2\). When the cord becomes taut \(\tfrac12 mv^2 = mgL\). The greatest speed is at \(k(y - L) = mg\).
\(600H = 60(H - 20)^2\) gives \(H^2 - 50H + 400 = 0\), so \(H = 40\,\text{m}\) or \(10\,\text{m}\). Only \(H = 40\,\text{m}\) is more than \(L\). Taut: \(v = \sqrt{2 \times 10 \times 20} = 20\,\text{m/s}\). Fastest at \(y = 20 + 600/120 = 25\,\text{m}\). There \(\tfrac12 \times 60\,v^2 = 600 \times 25 - \tfrac12 \times 120 \times 5^2\), which is \(13\,500\,\text{J}\), so \(v \approx 21.2\,\text{m/s}\).
The jumper falls \(40\,\text{m}\), is moving at \(20\,\text{m/s}\) when the cord tightens and at \(21.2\,\text{m/s}\) at its fastest, \(25\,\text{m}\) down. The cord's greatest pull, at the bottom, is \(120 \times 20 = 2400\,\text{N}\), four times the jumper's weight.
A \(50\,\text{kg}\) jumper uses the same cord: natural length \(20\,\text{m}\), \(k = 120\,\text{N/m}\). Ignore drag and take \(g = 10\,\text{m/s}^2\). How far below the platform is the lowest point, in metres?
Now return to the bus on the ghat. Suppose a \(10\,000\,\text{kg}\) bus descends \(500\,\text{m}\) of height at a steady speed. Its kinetic energy does not change, so \(W_{nc} = \Delta U\). That is \(-10\,000 \times 10 \times 500\), or \(-5 \times 10^7\,\text{J}\). Fifty megajoules become heat, much of it in the brakes. That is enough to bring more than a hundred litres of water to the boil. So drivers use a low gear and let the engine share the load.
India's best-known bungee jump is near Rishikesh, from a platform about \(80\,\text{m}\) above a river gorge. The operators choose a cord for each jumper's weight. The energy account of this lesson decides that choice. The cord must store all of \(mgH\) before the jumper reaches the rocks. Its greatest pull must stay within a few times the body weight.
- Using the displacement instead of the distance slid in \(fs\).Friction makes heat on every metre slid, forward or back. Use the total path length.
- Saying the energy lost to friction has disappeared.It is heat in the surfaces. Total energy is conserved; mechanical energy is not.
- Taking the fastest point of a bungee jump to be where the cord becomes taut.The jumper gains speed until the cord's pull equals the weight, a little below that point.
- Keeping the root \(H < L\) of the bungee equation.The lowest point must be below the length of the slack cord.