Work Done By A Variable Force

A constant force is rare. It is the variable force, which is more commonly encountered. Fig. 1 is a plot of a varying force in one dimension. If the displacement ∆x is small, we can take the force F (x) as approximately constant and the work done is then 

∆W = F(x).∆x

This is illustrated in Fig. 1(a). 

Fig.1(a) (a) The shaded rectangle represents the work done by the varying force F(x), over the small displacement ∆x, ∆W = F(x) ∆x.

Adding successive rectangular areas in Fig. 1(a) we get the total work done as

$W \cong\sum_{x_i}^{x_f}F(x)\Delta x$

where the summation is from the initial position x$_i$ to the final position x$_f$. 

Fig.1(b) (b) adding the areas of all the rectangles we find that for ∆x → 0, the area under the curve is exactly equal to the work done by F(x).

If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 1(b). Then the work done is

$W =\lim_{x\rightarrow 0}\sum_{x_i}^{x_f}F(x)\Delta x$

$W =\int_{x_i}^{x_f}F(x)dx$

where ‘lim’ stands for the limit of the sum when ∆x tends to zero. Thus, for a varying force the work done can be expressed as a definite integral of force over displacement (see also Appendix 3.1).

Example 1

A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N. The total distance through which the trunk has been moved is 20 m. Plot the force applied by the woman and the frictional force, which is 50 N versus displacement. Calculate the work done by the two forces over 20 m.

Answer

Fig.2: Plot of the force F applied by the woman and the opposing frictional force f versus displacement.

The plot of the applied force is shown in Fig. 6.4. At x = 20 m, F = 50 N (≠ 0). We are given that the frictional force f is |f|= 50 N. It opposes motion and acts in a direction opposite to F. It is therefore, shown on the negative side of the force axis.

The work done by the woman is WF → area of the rectangle ABCD + area of the trapezium CEID

W$_F$ = ×+ +  100 × 10 + $\frac{1}{2}$(100 + 50) × 10 

 = 1000 + 750 = 1750 J

The work done by the frictional force is 

W$_f$ → area of the rectangle AGHI 

W$_f$  = (−50) × 20 = − 1000 J 

The area on the negative side of the force axis has a negative sign.