In many cases the force does not remain constant during the process of doing work. Fro example, as a rocket moves away from the earth, work is done against the force of gravity, which varies as the invers square of the distance spring increase with the amount of sttetch. How can we calculate the work done in such a situation?
Fig. 4.4 shows the path of a particle in the x-y plane as it moves from point a to point b. the path has been divided into n short intervals of displacements and are the forces acting during these intervals.
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 so on. The total work done in moving the object can be calculated by adding all these terms. We can examine this graphically by plotting f cosθ verses d as shown in Fig. 4.5. the displacement d has been subdivided into n equal intervals. The value of F cos0 at the beginning of each interval is indicated in the figure.
Now the ith shaded rectangle has an area which is the work done during the ith interval. Thus, the work done given by Eq. 4.2 equals the sum of the areas of all the rectangles. Of we subdivide the distance into a large number of intervals so that each ∆d becomes very small, the work done given by Eq. 4.2 becomes more accurate. If we let each ∆d to approach zero then we obtain an exact result for the work done, such as
Thus, the work done by a variable force in moving a particle between two points is equal to the area under the F cos θ verses d curve between the two points a and b as shown in Fig. 4.6.
Example: 4.1: a force F acting on an object varies with distance x as shown in Fig. 4.7. calculate the work done by the force as the object moves from x = 0 to x = 6 m.
Solution: The work done by the force is equal to the total area under the curve from x = 0 to x = 6 m. this area is equal to the area of the rectangular section from x = 0 to x = 4 m, plus the area of triangular section from x = 4 m to x =6 m. Hence
Work done represented by the area of rectangle = 4m x 5 N
= 20 Nm = 20 J
Work done represented by the area of triangle = ½ x 2m x 5N
= 5 N m= 5 J
Therefore, the total work done = 20 J + 5 J = 25 J