Work


The physics idea of Work

In physics, the word "Work" has a very precise definition, which has little to do with sitting at a desk and doing mental computations. Rather, work is defined as follows
\[ {\rm Work}~=~{\rm Force}~ \times~ {\rm Distance} \]

If we use meters to measure distance, and seconds to measure time, then the units of force are the Newton (which is 1 kg m/sec2), and the units of work are Joules. One Joule is the same as 1 Newton-meter, which is also the same as 1 kg m2/sec2). Work has the same units as energy, and when work is done on an object (for example when a book is lifted from the table to a higher shelf), that object gains Potential Energy relative to its initial state. (Ofcourse, if the book then falls off the shelf, that potential energy can get converted to kinetic energy.. but enough said.)

We often have to resort to calculations involving integrals when we want to compute the work done on an object. However, the first, and simplest example will illustrate how work is calculated for the case of a constant force.


Example 1: Calculating Work for a Constant force

How much work does it take to lift a 1 kg box up to a 10 m height?

The force needed must overcome the force of gravity, which is a constant force (at least close to the surface of the Earth; see example 3 for a case when this is not true !) . Indeed, we know that

\[ F~ =~ m ~g \]

where $ g=9.8 m /sec^2 $ is the acceleration due to gravity, and $ m $ is the mass of the object, in this case 1 kg. The work done is thus

\[ W ~=~ m ~g~d = 1~ (9.8) ~10 = 98 ~{\rm Joules} \]

So in this case, finding work done is easy, since the force is constant throughout the motion. In the next example, we will need to take some complications into account.



Example 2: Pulling up a Chain (Work as an integral)

A chain of length $ L $ and mass $ M $ hangs down over the roof of a new construction site. The chain must be pulled up onto the roof. How much work is done in pulling the chain up?

In this problem, the amount of chain that still has to get pulled up is changing. This is equivalent to moving an object whose mass changes, so the simple calculation we did above will not quite do it. However, we could think of this chain as being made up of many many little bits, and figure out how much work is required to pull up each of the bits.
We are given that the mass of the chain is $ M $ , and that its length is $ L $ . This means that the mass per unit length is $ M/L $ . Let us focus attention on a small piece of the chain at a distance $ x_k $ from the top of the roof, and suppose that the size of each of the little bits is $ \Delta x $ . Then the mass of this piece is:

\[ \Delta m =~\frac{\Delta x}{L}~M \]
Once we get used to the idea of tiny tiny bits, which are called "infinitesimally small" pieces, we will also get comfortable with a notation commonly used by mathematicians,

\[ d m =~\frac{d x}{L}~M \]

(This is called differential notation. It just means that we are already mentally prepared for the fact that, coming up ahead will be a step in which we will let $ \Delta x $ get infinitesimally small.)

The force of gravity acting on the little piece of chain is
\[ F = \Delta m~ g~ = ~g~\frac{M}{L}~\Delta x \]

This implies that the work done in carrying just that small piece up to the top of the roof is
\[ \Delta W = F~x_k~= \Delta m~ g~ = ~g~\frac{M}{L}~x_k~\Delta x \]

Now we are to add up the total work done for each one of the pieces, located all the way down the length of the chain, from the piece right next to the roof, all the way down to the piece at the bottom of the chain (which is a distance $ L $ from the top). This sum is:
\[ W = ~\sum_0^n ~ F~x_k~= ~\sum_0^n ~g~\frac{M}{L}~x_k~\Delta x \]

We recognize that when we make the pieces very small, (i.e. take a limit as the number of pieces, $ n \to \infty $ ), this sum turns into the definite integral


\[ W = ~\int_0^L ~ F~x dx~= ~\int_0^L ~g~\frac{M}{L}~x~ dx \]

We can compute this integral as follows:
\[ W = ~g~\frac{M}{L}~ ~\int_0^L ~x~ dx \]

Simplifying, we get
\[ W = ~g~\frac{M}{L}~ ~\frac{x^2}{2}~|_0^L = ~g~\frac{M}{L}~ \frac{L^2}{2} = ~~\frac{g~ M~ L}{2}~ \]

Example 3: Sending a rocket to space
How much work does it take to send a rocket of mass $ m $ from the surface of the earth to a distance $ r $ above the surface ?

The force of gravitation is not constant over large distances away from the surface of the earth, and we need to use the inverse square law from physics to describe how this force changes with distance. This law says that the force of gravitational attraction between two objects is
\[ F(x) = ~G~ \frac{M~ m}{x^2} \]


where $ M, m $ are the masses of the objects (in this case the mass of the rocket and of the earth), $ G $ is the universal gravitation constant, and $ x $ is the distance between the centers of mass of the objects. (In this case, the distance $ x $ is the distance between the center of the earth and the rocket.)
The work done to move the rocket by a very small distance, from $ x $ to $ x + dx $ above the surface of the earth is would be
\[ dW = F(x) dx \]

The force in this expression is not constant, as we have seen above, so that the total work is expressed as an integral of the form

\[ W = \int_{x=R}^{x=R+r}~ F(x)~ dx = G M m ~ \int_{R}^{R+r} ~\frac{1}{x^2}~ dx \]

Note that the integration is carried out from the surface of the earth (where is $ -1/x $ , obtaining
\[ W = G M m ~ [-\frac{1}{x}]~|_R^{R+r} = G M m ~ [\frac{r}{R(R+r)}] \]

Some of the details of the calculation have been left for you to fill in.


For your consideration:

  • (1) Explain why the antiderivative of $ 1/x^2 $ is $ -1/x $ .

  • (2) Fill in any missing steps in the last calculation above.

  • (3) Assume that the radius of the earth is $ R \approx 6,400 $ and that the mass of the rocket is 500 kg. Use the fact that when the rocket is at the surface of the earth, the force due to gravity acting on it is $ F = m g $ to determine the magnitude of the constants $ G m M $ in the expression for the gravitational force.

  • (4) Determine how much work is done in lifting the rocket to a height 300 km above the surface of earth.