4.2 Work Done by a Variable Force

4.2 Work Done by a Variable Force

In many cases, the force does not remain constant during the process of doing work. For example, as a rocket moves away from the Earth, work is done against the force of gravity, which varies as the inverse square of the distance from the Earth's centre. Similarly, the force exerted by a spring increases with the amount of stretch. How can we calculate the work done in such situations?

Figure 4.5 shows the path of a particle in the $x$-$y$ plane as it moves from point $P$ to point $Q$. The path has been divided into $n$ short intervals of displacements $\Delta\mathbf{d}_{1}$, $\Delta\mathbf{d}_{2}$, $\ldots$, $\Delta\mathbf{d}_{n}$ and $\mathbf{F}_{1}$, $\mathbf{F}_{2}$, $\ldots$, $\mathbf{F}_{n}$, are the forces acting during these intervals, respectively.

During each small interval, the force is supposed to be approximately constant. So, the work done for the first interval can then be written as:

and in the second interval:

and so on. The total work done in moving the object can be calculated by adding all these terms.

or

We can examine this graphically by plotting $F\cos\theta$ versus $d$ as shown in Fig. 4.6. The displacement $d$ has been sub-divided into the same $n$ equal intervals. The value of $F\cos\theta$ at the beginning of each interval is indicated in the figure.

Now the $i$th shaded rectangle has an area $F_{i}\cos\theta_{i}\,\Delta d_{i}$ which is the work done during the $i$th interval. Thus, the work done given by Eq. 4.4 equals the sum of the areas of all the rectangles. If we sub-divide the distance into a large number of intervals so that each $\Delta d$ becomes very small, the work done given by Eq. 4.4 becomes more accurate. If we let each $\Delta d$ to approach zero, then we obtain an exact result for the work done, such as:

If this limit $\Delta d$ approaches zero, the total area of all the rectangles (Fig. 4.6) approaches the area between the $F\cos\theta$ verses $d$ curve and $x$-axis from $P$ to $Q$ as shown shaded in Fig. 4.7.

Thus, the work done by a variable force in moving a particle between two points is equal to the area under the $F\cos\theta$ verses $d$ curve between the two points $P$ and $Q$ as shown in Fig. 4.7.

Example 4.1

A force $F$ acting on an object varies with distance $d$ as shown in Fig. 4.8. Calculate the work done by the force as the object moves from $d = 0$ to $d = 6\,\mathrm{m}$.

Solution

The work done by the force is equal to the total area under the curve from $d = 0$ to $d = 6\,\mathrm{m}$. This area is equal to the area of the rectangular section from $d = 0$ to $d = 4\,\mathrm{m}$, plus the area of triangular section from $d = 4\,\mathrm{m}$ to $d = 6\,\mathrm{m}$.

Hence

Work done represented by the area of rectangle $= 4 \times 5\,\mathrm{N} = 20\,\mathrm{Nm} = 20\,\mathrm{J}$

Work done represented by the area of triangle $= \tfrac{1}{2} \times 2\,\mathrm{m} \times 5\,\mathrm{N} = 5\,\mathrm{Nm} = 5\,\mathrm{J}$

Therefore, the total work done = $25\,\mathrm{J}$