At an amusement park such as Imagica near Mumbai, a chain pulls the roller-coaster car slowly to the top of the first hill. Then the chain releases the car. For the rest of the ride no engine pushes the car. It plunges, climbs, loops and twists, and it still reaches the end. Everything it does comes from that first climb.
Kinetic plus potential
The mechanical energy \(E\) of a system is the sum of its kinetic energy and its potential energy, \(E = K + U\). Here \(U\) includes every kind of potential energy in the system: gravitational, elastic, or both. Mechanical energy is the energy of motion and of arrangement together.
Recall the work-energy theorem from the chapter on Work, Kinetic Energy and Power. The total work done on a body equals the change in its kinetic energy. Split that work into two parts. The conservative forces do work \(W_c\), and the non-conservative forces do work \(W_{nc}\). Now suppose \(W_{nc} = 0\), and follow the derivation below.
\[ W_c + W_{nc} = K_f - K_i \]
Why. the work-energy theorem: the work of all the forces equals the change in kinetic energy.
\[ W_c = -(U_f - U_i) \]
Why. the definition of potential energy, \(\Delta U = -W_c\), from lesson 1.
\[ -(U_f - U_i) = K_f - K_i \]
Why. put \(W_{nc} = 0\): no non-conservative force does work.
\[ K_i + U_i = K_f + U_f \]
Read it. collect the initial terms on one side and the final terms on the other.
If only conservative forces do work on a system, its mechanical energy \(E = K + U\) stays constant. Energy moves between the kinetic and the potential forms, but their sum does not change.
A condition about work, not about force
Read the condition carefully. It does not say "only conservative forces act". It says "only conservative forces do work". A block on a smooth curved track feels the normal force of the track all the way down. That force is always perpendicular to the track, and so to the motion. It does no work, and mechanical energy is still conserved. The same holds for the tension in the string of a swinging pendulum.
A \(0.5\,\text{kg}\) ball is dropped from rest from a height of \(20\,\text{m}\). Ignore air resistance and take \(g = 10\,\text{m/s}^2\). Find its speed just before it reaches the ground.
Only gravity does work, so mechanical energy is conserved. Take the ground as the reference level.
At the start \(K_i = 0\) and \(U_i = mgh = 0.5 \times 10 \times 20 = 100\,\text{J}\). At the ground \(U_f = 0\) and \(K_f = \tfrac12 \times 0.5 \times v^2\).
\(\tfrac12 \times 0.5 \times v^2 = 100\) gives \(v^2 = 400\), so \(v = 20\,\text{m/s}\).
Notice that \(m\) appears on both sides of \(\tfrac12 mv^2 = mgh\) and cancels: \(v = \sqrt{2gh}\) for any mass. No equation of motion was needed.
A bead slides without friction along a wire, starting from rest \(5\,\text{m}\) above the lowest level. One wire has a dip and a tall hump; the other is a smooth S. On which wire is the bead moving faster when it reaches the lowest level?
The bead in the figure has a mass of \(0.2\,\text{kg}\) and starts \(5\,\text{m}\) up, so \(E = mgh = 10\,\text{J}\). Press play on each wire. The \(U\) bar shrinks when the bead drops and grows when it climbs, and the \(K\) bar does the opposite. The total bar never changes. Near the top of the hump the bead has almost stopped, with \(v\) about \(2.8\,\text{m/s}\). At the lowest level both wires deliver \(10\,\text{m/s}\).
This is the great saving of the energy method. The normal force of the wire changes direction at every point, and the acceleration of the bead changes with it. Solving the motion with Newton's second law would be hard. Energy conservation links the start and the end directly, and the shape of the track drops out. Only the drop in height survives.
Energy gives the speed, not the time. On two smooth tracks with the same drop, the speed at the bottom is the same, but the times are different. A track that drops steeply at first gains speed early and usually arrives first. Energy conservation says nothing about time.
A stone is thrown from the top of a \(20\,\text{m}\) high cliff at \(15\,\text{m/s}\), at an angle above the horizontal. Ignore air resistance and take \(g = 10\,\text{m/s}^2\). With what speed does it reach the ground below, in m/s?
Three smooth water slides start at the same height and end at the same pool. One is straight, one is steep at first, and one is shallow at first. Compare the speeds at the bottom and the times taken.
The pendulum and the nail
A pendulum is the same idea in a swing. The tension in the string is always perpendicular to the path of the bob, so only gravity does work. The bob therefore rises on the far side to exactly the height from which it was released. Galileo went further and asked what happens if the string catches on a nail below the pivot.
With the nail, the far side of the swing is a much tighter arc, as Fig 3.1 shows. Yet the bob stops at the same height, now at about \(80°\) from the vertical below the nail. The tension still does no work. So the kinetic energy at the bottom, \(10\,\text{J}\) for every kilogram, must all become height again: the same \(1\,\text{m}\). The shape of the path has dropped out again.
A pendulum bob on a \(2\,\text{m}\) string is released from rest with the string at \(60°\) to the vertical. Take \(g = 10\,\text{m/s}^2\). Find its speed at the lowest point, in m/s.
In his last book, the Two New Sciences of 1638, Galileo described a pendulum whose thread catches on a nail below the pivot. The bob then swings on a shorter arc, yet it always rises to the level from which it started. Galileo argued from this that a falling body gains exactly enough speed to rise again to its starting height, by any path. It was an energy argument, two hundred years before the word energy took its modern meaning.
Draw the boundary first
Potential energy exists only when both interacting bodies are inside the system. Take the ball alone as the system. Then the Earth is outside, and gravity is an outside force that does work on the ball. Take the ball and the Earth together. Then there is no outside force at all, and the same energy appears as \(U\) inside the system.
Both accounts give the same answer. For the falling ball of the example: with the ball alone, gravity does \(+100\,\text{J}\) of work, so \(K\) rises by \(100\,\text{J}\). With the ball and the Earth, \(U\) falls by \(100\,\text{J}\) and \(K\) rises by \(100\,\text{J}\). The error to avoid is mixing the two: writing both the work of gravity and the change in \(mgh\) counts gravity twice.
So every energy problem begins with four steps:
- Draw the boundary of the system and name what is inside it.
- List the forces. Mark which are conservative and inside the system, and which do work from outside or are non-conservative.
- Choose the reference level for each potential energy.
- Write \(K_i + U_i = K_f + U_f\), adding the work of any non-conservative force if one does work.
A \(0.5\,\text{kg}\) block is pressed against a spring of constant \(200\,\text{N/m}\) on a smooth floor, compressing it by \(0.2\,\text{m}\). When released, the block leaves the spring and slides up a smooth ramp. Take \(g = 10\,\text{m/s}^2\). Find the speed of the block as it leaves the spring and the height it reaches on the ramp.
System: block, spring and Earth. The normal forces do no work, and there is no friction, so mechanical energy is conserved. Take the floor as the reference level.
At release \(E = \tfrac12 kx^2\). On leaving the spring \(E = \tfrac12 mv^2\). At the highest point \(E = mgh\).
\(\tfrac12 \times 200 \times 0.2^2 = 4\,\text{J}\). Then \(\tfrac12 \times 0.5 \times v^2 = 4\) gives \(v = 4\,\text{m/s}\), and \(0.5 \times 10 \times h = 4\) gives \(h = 0.8\,\text{m}\).
The block leaves the spring at \(4\,\text{m/s}\) and rises \(0.8\,\text{m}\) up the ramp, whatever the angle of the ramp.
In which of these is the mechanical energy of the system (body and Earth, with any spring included) conserved?
A block slides down a smooth curved track. The track pushes on the block the whole way. Why does that push not change the block's mechanical energy?
The push of a smooth track is the normal force, and it is always perpendicular to the surface. The block moves along the surface, so the normal force is perpendicular to every small displacement and does zero work. Since no non-conservative force does work, the mechanical energy stays constant even though a force from outside acts.
The first drop of a roller coaster is \(45\,\text{m}\) high. Ignore friction and air resistance and take \(g = 10\,\text{m/s}^2\). Estimate the top speed of the car at the bottom of the drop, in km/h.
Now return to the roller coaster at the start of the lesson. After the chain releases the car, gravity is the only force doing work if friction is small. So at every point of the ride, \(\tfrac12 mv^2 + mgh\) equals the energy at the top of the first hill. The speed at any point follows from the height alone: \(v = \sqrt{2g(h_{\text{top}} - h)}\), whatever loops and twists lie between.
- Adding the work of gravity and the change in \(mgh\) in the same equation.Count gravity once: either as potential energy inside the system, or as work from outside.
- Thinking a force that acts must spoil conservation.Only forces that do work matter. A normal force on a fixed smooth surface does none.
- Using energy to find the time of a motion.Energy gives speeds at positions. For times, use the equations of motion.