In many Indian homes a motor pumps water every morning from an underground sump to a tank on the roof. Then the motor stops. For the rest of the day the taps run without any pump at all. The water in the tank is at rest, yet it clearly holds something that the motor put there.
Energy with nowhere to be seen
In the chapter on Work, Kinetic Energy and Power, work was the transfer of energy by a force. The energy of a moving body is its kinetic energy \(K = \tfrac12 mv^2\), where \(m\) is the mass and \(v\) the speed. Now lift a brick slowly from the floor onto a shelf. You do work on it, but at the end the brick is at rest. Its kinetic energy is zero.
The energy has not vanished. Let the brick fall and it reaches the floor with exactly the speed that this energy would give it. So the energy was stored while the brick sat on the shelf. It was stored in the arrangement, in the brick being high above the Earth. Energy stored in an arrangement is called potential energy, written \(U\).
A stretched catapult, a wound clock spring and a compressed air cushion all store energy in the same way. Nothing in them moves and nothing is hot. The energy is in the shape: how far the parts are from where they would like to be.
How a change in potential energy is measured
The Earth pulls on the brick with its weight \(mg\), where \(g = 10\,\text{m/s}^2\) is the acceleration due to gravity. When the brick rises, gravity does negative work on it. When the brick falls, gravity does positive work and speeds it up. We define the change in potential energy as the negative of the work done by this force.
A system changes from one arrangement to another. The change in its potential energy is the negative of the work \(W\) done by the force that stores the energy: \(\Delta U = U_f - U_i = -W\).
The minus sign carries the meaning. When gravity does positive work, as in a fall, the potential energy decreases and the body gains speed. When gravity does negative work, as in a lift, the potential energy increases. Only some forces can store energy this way; the next lesson shows which. Gravity and the force of a spring are two of them.
While a coconut falls from a tree, gravity does \(+12\,\text{J}\) of work on it. What is the change in the potential energy of the coconut and the Earth?
Gravitational potential energy
Lift a body of mass \(m\) through a height \(h\) near the Earth's surface. The weight \(mg\) points down and the displacement points up, so gravity does work \(-mgh\). The change in potential energy is therefore \(\Delta U = -(-mgh) = +mgh\). The path does not matter: gravity's work depends only on the change in height.
Now choose a level and call it \(U = 0\). This level is the reference level. A body at a height \(h\) above it then has \(U = mgh\). A body below it has a negative \(h\), and so a negative \(U\). A negative potential energy is not a mistake. It only means the body is below the level you chose.
Try the three choices in Fig 1.1. With the floor as zero, \(U\) is \(15\,\text{J}\) at A and \(5\,\text{J}\) at B. With the table top as zero, it is \(10\,\text{J}\) and \(0\). With the ceiling as zero, both values are negative, \(-5\,\text{J}\) and \(-15\,\text{J}\). In all three cases the ball loses \(10\,\text{J}\) of potential energy in the fall. That \(10\,\text{J}\) is what becomes kinetic energy, so the speed at B is the same for every choice.
The zero of potential energy may be placed anywhere. No experiment measures \(U\) itself; every measurable result depends only on a change \(\Delta U\), which is the same for every choice of zero.
Choose the reference level once, at the start of a problem, and write it down. Measuring one height from the floor and another from the table top in the same problem gives a wrong \(\Delta U\).
A motor lifts \(1000\,\text{kg}\) of water from a sump \(2\,\text{m}\) below the ground to a tank \(10\,\text{m}\) above the ground. Take \(g = 10\,\text{m/s}^2\). How much potential energy does the water gain? Does the answer depend on the reference level?
Take the ground as the reference level. The sump is at \(h_i = -2\,\text{m}\) and the tank at \(h_f = +10\,\text{m}\).
\(U_i = mgh_i\) and \(U_f = mgh_f\), so \(\Delta U = mg\,(h_f - h_i)\).
\(U_i = 1000 \times 10 \times (-2) = -20\,\text{kJ}\). \(U_f = 1000 \times 10 \times 10 = 100\,\text{kJ}\). So \(\Delta U = 100 - (-20) = 120\,\text{kJ}\).
The tank holds \(120\,\text{kJ}\) more than the sump, whatever level is called zero. The taps spend this store during the day.
A \(2\,\text{kg}\) school bag is lifted from the floor onto a shelf \(1.5\,\text{m}\) high. Take \(g = 10\,\text{m/s}^2\). By how much does the potential energy increase, in joules?
A ball is held \(2\,\text{m}\) above a table that stands \(1\,\text{m}\) above the floor. Asha takes the floor as the reference level, and Ravi takes the table top. The ball is dropped onto the table. On what do they disagree?
The energy in a spring
A spring stretched by \(x\) from its natural length pulls back with a force of size \(kx\). Here \(k\) is the spring constant, in N/m. The force always points towards the natural length, so we write it as \(F = -kx\). In the chapter on Work you found the work done by this force from \(x = 0\) to \(x\). It is the area of a triangle under the force-extension graph, \(-\tfrac12 kx^2\).
So the potential energy stored in the spring is \(U = +\tfrac12 kx^2\), with \(U = 0\) at the natural length. The square has a plain meaning. A compression \(x = -0.2\,\text{m}\) gives the same energy as a stretch \(x = +0.2\,\text{m}\). The spring stores energy for being deformed, and it does not care in which direction.
The last chip of Fig 1.2 shows the square at work. Doubling the stretch doubles the force, and the force acts over twice the distance. So the stored energy grows four times. Tripling the stretch would store nine times the energy.
A spring stores \(5\,\text{J}\) when it is compressed by \(3\,\text{cm}\). How much does it store when it is stretched by \(6\,\text{cm}\)?
Springs in series and in parallel
Two springs joined end to end are springs in series. In the chapter Objects in Equilibrium you saw that one force \(F\) runs through both. So each spring stores \(U = F^2/2k\). The softer spring, with the smaller \(k\), stretches more and stores more energy. For two springs in series, \(U_1/U_2 = k_2/k_1\).
Two springs side by side, both fixed to the wall and to the block, are springs in parallel. They share one stretch \(x\), and their forces add: \(F = k_1x + k_2x\). So the pair acts as one spring of constant \(k = k_1 + k_2\). Each spring stores \(\tfrac12 k x^2\) with its own \(k\), so now the stiffer spring stores more, and \(U_1/U_2 = k_1/k_2\).
Cutting a spring
Cut a spring of natural length \(L\) and constant \(k\) into a piece of length \(L_1\). Under one force, every centimetre of the spring stretches by the same amount. So a piece stretches in proportion to its length, and a shorter piece is stiffer: \(k_1 L_1 = kL\). A spring cut in half has twice the spring constant.
A spring of constant \(200\,\text{N/m}\) and natural length \(60\,\text{cm}\) is cut into pieces of \(20\,\text{cm}\) and \(40\,\text{cm}\). Find the constant of each piece. The two pieces are then joined in parallel. Find the constant of the pair, and the energy stored in the short piece when it alone is stretched by \(5\,\text{cm}\).
(a) \(k_1 = kL/L_1 = 200 \times 60/20 = 600\,\text{N/m}\), and \(k_2 = 200 \times 60/40 = 300\,\text{N/m}\). Both pieces are stiffer than the whole spring.
(b) In parallel the constants add: \(600 + 300 = 900\,\text{N/m}\).
(c) \(U = \tfrac12 k_1 x^2\), so \(U = \tfrac12 \times 600 \times (0.05)^2 = 0.75\,\text{J}\). Convert \(5\,\text{cm}\) to \(0.05\,\text{m}\) before squaring.
Springs of \(200\,\text{N/m}\) and \(600\,\text{N/m}\) are joined in series and hold a \(3\,\text{kg}\) block at rest. Take \(g = 10\,\text{m/s}^2\). What is the total energy stored in the two springs, in joules?
A spring of constant \(k\) is cut into two equal halves. Which statements are true?
Whose energy is it?
We often say "the potential energy of the ball". That is a convenient short form. The energy is stored in the separation between the ball and the Earth, and it takes both of them to store it. Neither body owns it. The same is true of a spring joining two blocks: the energy is in the spring between them, not in either block.
The short form is harmless near the Earth for one reason. When a ball falls, the Earth also moves towards the ball, but its enormous mass keeps its motion tiny. Almost all the released energy goes to the ball. Two bodies of similar mass, such as two stars or two blocks joined by a spring, share the energy. Then the potential energy must be assigned to the pair.
When a \(1\,\text{kg}\) ball and the Earth pull each other together, their momenta stay equal and opposite. Kinetic energy is \(p^2/2m\), so the two kinetic energies are in the inverse ratio of the masses. The Earth receives about \(1/(6 \times 10^{24})\) of the energy the ball receives, far too little to measure.
Two blocks on a smooth floor are joined by a compressed spring and held still. When they are released, both blocks move off. Where was the energy stored?
In 1853 the Scottish engineer William Rankine wrote a paper on how energy changes form. He needed a word for energy that sits in a body because of its position or shape, ready to become motion. He called it "potential energy", and he called the energy of motion "actual energy". The second name did not last; Kelvin's "kinetic energy" replaced it. Rankine's term stayed, and it is the \(U\) of this chapter.
A pumped-storage power station has two lakes, one high on a hill and one low in the valley. When electricity is cheap and plentiful, pumps lift water to the upper lake and store it as potential energy. When the demand is high, the water runs down through turbines and makes electricity again. India is building such stations, for example at Tehri in Uttarakhand, to store solar power made during the day for use at night.
- Measuring different heights from different levels in one problem.Fix one reference level at the start. Every \(h\) is measured from it, upward positive.
- Treating a negative potential energy as impossible.A negative \(U\) only means the body is below the chosen zero. Only changes in \(U\) matter.
- Using \(U = \tfrac12 kx\) or measuring \(x\) from the wall.\(U = \tfrac12 kx^2\), and \(x\) is the change in length from the natural length.
- Taking a cut piece to be softer than the whole spring.A shorter piece is stiffer: \(k_1 = kL/L_1\).