9.5: Work

A fundamental concept in classical physics is work: If an object is moved in a straight line against a force \(F\) for a distance \(s\) the work done is \(W=Fs\).

The force due to gravity on a 10 pound weight is 10 pounds at the surface of the earth, and it does not change appreciably over 5 feet. The work done is \(W=10\cdot 5=50\) foot-pounds.

In reality few situations are so simple. The force might not be constant over the range of motion, as in the next example.

This is the same problem as before in different units, and we are not specifying a value for \(D\). As before \[W=\int_{r_0}^{D} {k\over r^2}\,dr= -\left.{k\over r}\right|_{r_0}^{D}=-{k\over D}+{k\over r_0}. \nonumber \] While "weight in pounds'' is a measure of force, "weight in kilograms'' is a measure of mass. To convert to force we need to use Newton's law \(F=ma\). At the surface of the earth the acceleration due to gravity is approximately 9.8 meters per second squared, so the force is \(F=10\cdot 9.8=98\). The units here are "kilogram-meters per second squared'' or "kg m/s\(^2\)'', also known as a Newton (N), so \(F=98\) N. The radius of the earth is approximately 6378.1 kilometers or 6378100 meters. Now the problem proceeds as before. From \(F=k/r^2\) we compute \(k\): \(98=k/6378100^2\), \(k= 3.986655642\cdot 10^{15}\). Then the work is: \[W=-{k\over D}+6.250538000\cdot 10^8\quad\hbox{Newton-meters.} \nonumber \] As \(D\) increases \(W\) of course gets larger, since the quantity being subtracted, \(-k/D\), gets smaller. But note that the work \(W\) will never exceed \(6.250538000\cdot 10^8\), and in fact will approach this value as \(D\) gets larger. In short, with a finite amount of work, namely \(6.250538000\cdot 10^8\) N-m, we can lift the 10 kilogram object as far as we wish from earth.

Next is an example in which the force is constant, but there are many objects moving different distances.

Here we have a large number of atoms of water that must be lifted different distances to get to the top of the tank. Fortunately, we don't really have to deal with individual atoms---we can consider all the atoms at a given depth together. To approximate the work, we can divide the water in the tank into horizontal sections, approximate the volume of water in a section by a thin disk, and compute the amount of work required to lift each disk to the top of the tank. As usual, we take the limit as the sections get thinner and thinner to get the total work.

Figure 9.5.1. Cross-section of a conical water tank.

At depth \(h\) the circular cross-section through the tank has radius \(r=(10-h)/5\), by similar triangles, and area \(\pi(10-h)^2/25\). A section of the tank at depth \(h\) thus has volume approximately \(\pi(10-h)^2/25\Delta h\) and so contains \(\sigma\pi(10-h)^2/25\Delta h\) kilograms of water, where \(\sigma\) is the density of water in kilograms per cubic meter; \(\sigma\approx 1000\). The force due to gravity on this much water is \(9.8\sigma\pi(10-h)^2/25\Delta h\), and finally, this section of water must be lifted a distance \(h\), which requires \( h9.8\sigma\pi(10-h)^2/25\Delta h\) Newton-meters of work. The total work is therefore \[W={9.8\sigma\pi\over 25} \int_0^{10} h(10-h)^2\,dh={980000\over3}\pi\approx 1026254\quad\hbox{Newton-meters.} \nonumber \]

A spring has a "natural length,'' its length if nothing is stretching or compressing it. If the spring is either stretched or compressed the spring provides an opposing force; according to Hooke's Law the magnitude of this force is proportional to the distance the spring has been stretched or compressed: \(F=kx\). The constant of proportionality, \(k\), of course depends on the spring. Note that \(x\) here represents the change in length from the natural length.