A boy pulls back the rubber of his gulel, a catapult, to shoot a pebble at a mango. The first few centimetres of the pull are easy. The last few centimetres take all his strength. The force he needs grows with every centimetre of stretch. The rule \(W = Fd\) needs one value of \(F\), and here there is no single value to use.
A constant force is a rectangle
Plot a constant force \(F\) against the position \(x\) of the body it acts on. The graph is a horizontal line at height \(F\). Over a displacement from \(x_1\) to \(x_2\), the region under the line is a rectangle, \(F\) tall and \(x_2 - x_1\) wide. Its area, \(F(x_2 - x_1)\), is exactly the work. So for a constant force, the work is the area under the \(F\)-\(x\) graph. The same idea works when \(F\) changes.
Strips under a changing force
Now let the force change with position. Cut the displacement into narrow strips, each of width \(\Delta x\). Across one strip the force hardly changes, so the work over that strip is nearly \(F\,\Delta x\), the area of a thin rectangle. Adding the strips adds their areas. As the strips are made narrower, the rectangles fit the curve better and better. In the limit their sum becomes the exact area under the graph.
For a force along the line of motion, the work done between two positions equals the area under the \(F\)-\(x\) graph between them. Area above the axis counts as positive work, and area below it as negative work.
Area below the axis
A force can reverse partway through the motion. The graph then dips below the \(x\) axis, and there the force points against the motion. On that stretch \(\cos\theta = -1\), so every strip gives negative work. The area below the axis therefore counts with a minus sign. The net work is the area above the axis minus the area below it.
Follow the readout in Fig 2.1. Over the first \(2\,\text{m}\) the force is a steady \(5\,\text{N}\), a rectangle of \(10\,\text{J}\). From \(2\,\text{m}\) to \(4\,\text{m}\) it falls to zero, a triangle of \(5\,\text{J}\), so the work so far is \(15\,\text{J}\). From \(4\,\text{m}\) to \(6\,\text{m}\) the force points backward, down to \(-4\,\text{N}\) and back to zero. That is a triangle of area \(4\,\text{J}\) below the axis. The net work is \(15 - 4 = 11\,\text{J}\).
A force on a block grows steadily from \(0\) to \(12\,\text{N}\) while the block moves \(4\,\text{m}\) along the force. How much work does the force do?
A force along \(x\) is \(6\,\text{N}\) from \(x = 0\) to \(x = 4\,\text{m}\). It then falls steadily to zero at \(x = 7\,\text{m}\). Find the work done by the force from \(x = 0\) to \(x = 7\,\text{m}\), in joules.
The integral
The area under a curve is what an integral computes. Shrinking the strips turns the sum of \(F\,\Delta x\) into the integral of \(F\,dx\). For a force \(F(x)\) along the \(x\) axis, the work from \(x_1\) to \(x_2\) is
\[ W = \int_{x_1}^{x_2} F(x)\,dx \]
When \(F\) is constant, the integral returns \(F(x_2 - x_1)\), the rectangle. So the integral is not a new rule. It is the old rule applied to one thin strip at a time.
A force along the \(x\) axis is \(F(x) = (3x^2 - 2x + 4)\,\text{N}\), with \(x\) in metres. Find the work it does as a particle moves from \(x = 1\,\text{m}\) to \(x = 3\,\text{m}\).
The work is the integral of the force: \(W = \int_1^3 (3x^2 - 2x + 4)\,dx = \left[x^3 - x^2 + 4x\right]_1^3\).
At the upper limit, \(27 - 9 + 12 = 30\). At the lower limit, \(1 - 1 + 4 = 4\).
So \(W = 30 - 4 = 26\,\text{J}\).
Along a curved path
In two or three dimensions, both the force and the direction of motion may change along the path. Take a tiny step \(d\vec r\) along the path. Over that step the force is nearly constant, so the work is \(\vec F\cdot d\vec r = F_x\,dx + F_y\,dy + F_z\,dz\). Adding the steps from \(A\) to \(B\) along the path gives the line integral \(\int_A^B \vec F\cdot d\vec r\). When each component of the force depends only on its own coordinate, the line integral splits into ordinary integrals, one for each axis.
A force \(\vec F = (2x\,\hat i + 3y^2\,\hat j)\,\text{N}\) acts on a particle that moves from the origin to the point \((2\,\text{m}, 1\,\text{m})\). Find the work done by the force.
Here \(F_x\) depends only on \(x\) and \(F_y\) only on \(y\), so the line integral splits: \(W = \int_0^2 2x\,dx + \int_0^1 3y^2\,dy\).
The first integral is \(\left[x^2\right]_0^2 = 4\,\text{J}\). The second is \(\left[y^3\right]_0^1 = 1\,\text{J}\).
So \(W = 5\,\text{J}\). This answer is the same for every path between the two points, because each integral depends only on where its coordinate starts and ends.
For most forces the line integral does depend on the path. Friction on a box pushed round a long route takes more energy than on a short route between the same two points. Some forces do the same work on every path, like the force in the last example, and like gravity. Such forces have a special name and a stored energy of their own. The next chapter sorts forces into these two kinds.
A force along the \(x\) axis is \(F(x) = 4x^3\,\text{N}\), with \(x\) in metres. Find the work it does as a particle moves from \(x = 0\) to \(x = 2\,\text{m}\), in joules.
Stretching a spring slowly
A spring of spring constant \(k\), stretched by \(x\) from its natural length, pulls back with a force \(kx\). This is Hooke's law, from the chapter on Forces and Newton's Laws. To stretch the spring slowly, a hand must pull with a force equal to \(kx\) at each extension. This pull starts at zero and grows in proportion to \(x\). Its \(F\)-\(x\) graph is a straight line through the origin, and the area under it is a triangle.
\[ F_{\text{hand}}(x') = kx' \]
Why. in a slow stretch the spring stays in balance, so the hand's pull equals the spring's pull at every extension \(x'\).
\[ W = \int_0^{x} kx'\,dx' \]
Why. the pull changes with position, so the work is the integral of the pull over the stretch.
\[ W = \left[\tfrac12 kx'^2\right]_0^{x} = \tfrac12 kx^2 \]
Read it. the integral of \(x'\) is \(\tfrac12 x'^2\). This is the area of the triangle, half of the base \(x\) times the height \(kx\).
The half in \(\tfrac12 kx^2\) is there because the pull grows from zero. Its average over the stretch is half its final value, \(\tfrac12 kx\). In Fig 2.2 the final pull is \(100\,\text{N}\) over \(0.2\,\text{m}\), and the work is \(\tfrac12 \times 0.2 \times 100 = 10\,\text{J}\). The formula gives the same: \(\tfrac12 \times 500 \times 0.2^2 = 10\,\text{J}\).
The spring pulls on the hand the other way. Its force points back towards the natural length while its end moves away from it. So the spring does negative work, \(-\tfrac12 kx^2\), during the stretch. Stretching from an extension \(x_1\) to a larger \(x_2\) takes the area of a trapezium, \(\tfrac12 k(x_2^2 - x_1^2)\). A compression works in the same way, with \(x\) measured on the other side of the natural length.
Do not multiply the final force by the stretch. That uses \(kx\) as if it had acted from the start, and gives \(kx^2\), twice the right answer. The spring force is a variable force; its work is the area of a triangle, not of a rectangle.
A spring has \(k = 800\,\text{N/m}\). Find the work needed to stretch it (a) from its natural length to \(5\,\text{cm}\), (b) from \(5\,\text{cm}\) to \(10\,\text{cm}\).
(a) \(W = \tfrac12 kx^2 = \tfrac12 \times 800 \times 0.05^2 = 1\,\text{J}\).
(b) \(W = \tfrac12 k(x_2^2 - x_1^2) = \tfrac12 \times 800 \times (0.10^2 - 0.05^2) = 400 \times 0.0075 = 3\,\text{J}\).
The second \(5\,\text{cm}\) cost three times as much as the first, because the pull was larger all the way through it.
Now return to the gulel. Its rubber behaves roughly like a spring. The last few centimetres of the pull are hard because the force is largest there, and they add the tallest strips to the area. Doubling the stretch multiplies the work by four. That is why a small extra pull of the rubber sends the pebble much faster, since the rubber later hands that work to the pebble.
A spring with \(k = 200\,\text{N/m}\) is stretched slowly from its natural length by \(10\,\text{cm}\). Find the work done by the hand, in joules.
A spring is stretched from \(2\,\text{cm}\) to \(4\,\text{cm}\). How does this work compare with the work to stretch it from its natural length to \(2\,\text{cm}\)?
A spring is stretched by \(10\,\text{cm}\). It is then moved slowly through its natural length to a compression of \(10\,\text{cm}\). What total work does the spring do over this move?
Archers measure the pull a bow needs at each length of draw and plot it as a draw-force curve. The area under that curve is the work the archer does in one draw. Most of it passes to the arrow when the string is released. A modern compound bow has cams that make the pull peak in the middle of the draw and drop near full draw. The archer can then hold the bow steady at full draw, while the area under the curve stays large.
- Using \(W = Fd\) with one value of a force that changes.For a changing force, find the area under the \(F\)-\(x\) graph, or integrate \(F\,dx\).
- Adding all the areas as positive.Area below the \(x\) axis is negative work. The net work is the signed area.
- Writing \(kx^2\) for the work of a spring.The pull grows from zero, so the work is the triangle, \(\tfrac12 kx^2\).
- Using centimetres in \(\tfrac12 kx^2\) with \(k\) in N/m.Convert the extension to metres before squaring it.