A boy fits a small stone into the leather pad of a gulel, the rubber catapult of every Indian village. He pulls the pad back slowly, holds it, and lets go. The stone flies off far faster than any throw of his arm. While he held the pad still, nothing moved, yet the stretched rubber held the energy that the stone then carried away.
The work of a growing force
To stretch a wire you must pull it, and your pull does work. But the force is not the same all the way. At the start the wire is slack and needs no force at all. As it stretches, the force needed grows in proportion to the extension, \(F = k\,\Delta L\) with \(k = YA/L\). A force that changes cannot simply be multiplied by the distance.
In the chapter on Work, Kinetic Energy and Power, the work of a changing force was the area under its force-displacement graph. For a wire the graph is a straight line through the origin. The area under it, up to a force \(F\) at an extension \(\Delta L\), is a triangle of area \(\tfrac12 F\,\Delta L\).
Watch the readout in Fig 4.1. At half a millimetre the force is \(100\,\text{N}\) and the work is only \(0.025\,\text{J}\). At one millimetre the force is \(200\,\text{N}\) and the work is \(0.1\,\text{J}\). Twice the stretch needs four times the work, because the triangle doubles in both height and width.
The wire gives this work back when it is released, as long as it stays within its elastic limit. So the work is stored in the stretched wire as elastic potential energy, called strain energy. It is the same energy as \(\tfrac12 kx^2\) in a spring, written for a wire.
A steel wire \(1\,\text{m}\) long and \(1\,\text{mm}^2\) in section, with \(Y = 2 \times 10^{11}\,\text{Pa}\), is stretched by \(200\,\text{N}\). Find its extension and the energy stored.
The stiffness is \(k = YA/L = 2 \times 10^{11} \times 10^{-6}/1 = 2 \times 10^5\,\text{N/m}\), so \(\Delta L = 200/(2 \times 10^5) = 1\,\text{mm}\).
\(U = \tfrac12 F\,\Delta L = \tfrac12 \times 200 \times 10^{-3} = 0.1\,\text{J}\).
Check with the other form: \(\tfrac12 k(\Delta L)^2 = \tfrac12 \times 2 \times 10^5 \times 10^{-6} = 0.1\,\text{J}\).
A wire is stretched slowly until the force in it is \(200\,\text{N}\) and its extension is \(1\,\text{mm}\). Why is the work done \(0.1\,\text{J}\) and not \(200 \times 10^{-3} = 0.2\,\text{J}\)?
A wire stretched by \(0.5\,\text{mm}\) stores \(0.02\,\text{J}\). How much energy does the same wire store when stretched by \(1.5\,\text{mm}\), still within its elastic limit? Answer in joules.
Energy per unit volume
Divide the strain energy by the volume of the wire, \(AL\). The result is the strain energy density \(u\), the energy stored in each cubic metre of the metal. It depends only on how hard the metal is being stretched, not on the size of the wire.
\[ U = \tfrac12 F\,\Delta L \]
Why. the strain energy of the wire, from the area under its force-extension line.
\[ u = \dfrac{U}{AL} = \dfrac12 \times \dfrac{F}{A} \times \dfrac{\Delta L}{L} \]
Why. divide by the volume \(AL\), and give \(A\) to the force and \(L\) to the extension.
\[ u = \tfrac12\,\text{stress}\times\text{strain} \]
Read it. \(F/A\) is the stress and \(\Delta L/L\) is the strain.
Hooke's law, \(\text{stress} = Y \times \text{strain}\), turns this one form into two more. Try the step yourself below: replace the strain by stress divided by \(Y\).
\[ u = \tfrac12\,\text{stress}\times\text{strain} \]
Why. the strain energy density from the derivation above.
?
Your turn. Hooke's law gives the strain as stress divided by \(Y\). Which line is it?
\[ u = \dfrac{(\text{stress})^2}{2Y} \]
Read it. collect the two factors of stress.
A steel wire (\(Y = 2 \times 10^{11}\,\text{Pa}\)) \(1\,\text{m}\) long and \(1\,\text{mm}^2\) in section carries a stress of \(2 \times 10^8\,\text{Pa}\). Find the strain, the strain energy per unit volume and the total strain energy.
Strain \(= \text{stress}/Y = 2 \times 10^8/(2 \times 10^{11}) = 10^{-3}\).
\(u = \tfrac12 \times 2 \times 10^8 \times 10^{-3} = 10^5\,\text{J/m}^3\).
The volume is \(1 \times 10^{-6} = 10^{-6}\,\text{m}^3\), so \(U = 10^5 \times 10^{-6} = 0.1\,\text{J}\). This is the wire of the last example: \(2 \times 10^8\,\text{Pa}\) on \(1\,\text{mm}^2\) is \(200\,\text{N}\).
The last form, \((\text{stress})^2/2Y\), holds a surprise. At the same stress, a material with a smaller \(Y\) stores more energy, because it stretches further. Thin wires behave the same way. Two wires of one metal and one length carry the same load. The thinner wire has the larger stress, and its energy per volume grows as the square of that stress.
Two wires of the same material and length have radii \(r\) and \(2r\). They carry the same load. What is the ratio of the strain energy in the thin wire to that in the thick wire?
Why is a gulel made of rubber and not of steel? Steel stretched to its elastic limit, near \(2.5 \times 10^8\,\text{Pa}\), stores about \(1.6 \times 10^5\,\text{J/m}^3\), some \(20\,\text{J}\) per kilogram. Rubber can be stretched to several times its length and give most of the work back. Per kilogram it stores many times more energy than ordinary steel, which is why catapults, and the bows of some archers, use it.
Lowered gently or let go
Hang a load \(m\) on a wire in two ways. First, lower it gently by hand until the wire holds it alone. At rest the wire pulls up with \(mg\), so the extension is \(x_0 = mg/k\). Now try the second way: hold the load with the wire just taut, and let go.
A \(2\,\text{kg}\) block hangs from a steel wire held so that the wire is just taut. Lowered gently, it would stretch the wire \(0.2\,\text{mm}\). If it is let go instead, how far does the wire stretch at the lowest point of the motion?
The figure above uses a steel wire \(2\,\text{m}\) long and \(1\,\text{mm}^2\) in section, so \(k = 10^5\,\text{N/m}\) and \(x_0 = 20/10^5 = 0.2\,\text{mm}\). Let the block go. It speeds up at first, because the weight beats the small pull of the wire. At \(x_0\) the forces balance, but the block is moving at its fastest. It overshoots and is stopped only at \(0.4\,\text{mm}\).
Energy gives the lowest point at once. There the block is at rest again, so all the work of gravity sits in the wire: \(mgx = \tfrac12 kx^2\), and \(x = 2mg/k = 2x_0\). The stress at that instant is twice the resting stress. A suddenly applied load strains a wire twice as hard as the same load applied gently.
Where half the energy goes
Follow the bars of the figure above in the "small losses" state. The block settles at \(x_0 = 0.2\,\text{mm}\). Gravity has done \(mgx_0 = 20 \times 2 \times 10^{-4} = 4\,\text{mJ}\) of work. The wire stores \(\tfrac12 k x_0^2 = 2\,\text{mJ}\). The other \(2\,\text{mJ}\), exactly half, has become heat in the wire and the air as the bobbing died away.
When the load is lowered gently instead, the same \(4\,\text{mJ}\) of gravity's work is shared differently. The wire still stores \(2\,\text{mJ}\). The hand that lowered the load did \(-2\,\text{mJ}\) of work, holding back and taking half the energy. Either way the wire ends with half of gravity's work.
A \(2\,\text{kg}\) block is fixed to the lower end of a steel wire \(2\,\text{m}\) long and \(1\,\text{mm}^2\) in section (\(Y = 2 \times 10^{11}\,\text{Pa}\)). It is held with the wire just taut, and released. Take \(g = 10\,\text{m/s}^2\). Find the greatest extension, and compare it with the extension at rest.
The wire has \(k = YA/L = 10^5\,\text{N/m}\).
At the lowest point \(mgx = \tfrac12 kx^2\), so \(x = 2mg/k = 2 \times 20/10^5 = 4 \times 10^{-4}\,\text{m} = 0.4\,\text{mm}\).
At rest \(kx_0 = mg\) gives \(x_0 = 0.2\,\text{mm}\). The released block stretches the wire twice as far, and the stress is momentarily doubled, from \(2 \times 10^7\) to \(4 \times 10^7\,\text{Pa}\).
A \(5\,\text{kg}\) block is let go from rest on the same wire, \(k = 10^5\,\text{N/m}\), held just taut. Take \(g = 10\,\text{m/s}^2\). What is the greatest extension of the wire, in mm?
A load is let go on a just-taut wire and finally comes to rest at the extension \(x_0 = mg/k\). Explain why exactly half of the work done by gravity has become heat.
Gravity does work \(mgx_0\) as the load falls through \(x_0\). At rest the wire stores \(\tfrac12 kx_0^2\). Since \(kx_0 = mg\), that is \(\tfrac12 mgx_0\), exactly half of gravity's work. The load ends at rest with no kinetic energy, so the other half must have left the mechanical energy: the swings were damped by the wire and the air, which turned it into heat.
A lift cable or a crane rope is chosen for the load it must carry, with a wide margin. The margin must cover more than a heavy load. A load that drops onto a slack rope strains it at least twice as hard as a load hung gently. The farther it falls, the harder the jerk. That is why a crane driver takes up the slack slowly before lifting.
Return to the gulel. The boy's slow pull did work against a force that grew from zero. The rubber stored it as strain energy, the area under its force-stretch curve. On release, the rubber gave that energy to the stone as kinetic energy. A wire does the same, with far less stretch and far more force.
- Writing the work of stretching as \(F\,\Delta L\).The force grows from zero, so the work is the triangle, \(\tfrac12 F\,\Delta L\).
- Thinking the strain energy is proportional to the extension.\(U = \tfrac12 k(\Delta L)^2\): three times the stretch stores nine times the energy.
- Giving the thicker wire more strain energy under the same load.Under one load, \(U = F^2L/(2AY)\): the thinner wire stretches more and stores more.
- Taking the extension of a suddenly released load to be \(mg/k\).It overshoots to \(2mg/k\) and settles at \(mg/k\) only after the swings die away.