Work Done by a Variable ForceThe magnitude of the force acting on a body may not always remain constant. Suppose a rocket is fired upwards from the earth. The force required to lift it and make it move away from the earth must be sufficient to overcome the gravitational attraction of the earth which decreases inversely as the square of the distance of the rocket from the earth.
Thus the force necessary to keep the rocket moving away from the earth is not constant, it depends upon the position of the rocket with respect to the earth. Similarly, the force required to stretch a spring will depend on the amount of stretching. In such cases the work by force is calculated as follows.
Suppose a variable force F acts on a body in the x-direction. The force F is a function of x, i.e. F depends on x. Suppose the force moves the body from a position x1 to a position x2. What is the work done by this variable force F in moving the body from x1 to x2?
We plot F versus x. Let Fig. represent the graph of F versus x. we divide the total displacement into a large number of small elements, each equal to dx, as shown in Fig.
Finding work done by a variable force
dW = F dx
Likewise, during another small displacement dx from (x1 + dx) to (x1 +2dx), the force F is nearly constant and the work done by it is dW = F dx, where F is the value of the force at x1 + dx. The total work done by F is displacing the body from x1 to x2 is clearly equal to the sum of a large number of terms like that in eq. in which F has a different value for each term. Thus
The smaller the value of dx, the better will be the result of our calculation of W. Hence the exact result is
In the language of integral calculus, this relation is expressed as
Numerically, this quantity is equal to the area between the force curve and the x-axis between the values x1 and x2. This area is shown shaded in figure.
If the body is moved from x1 =0 to x2 =x, a displacement of x, then the work done by the variable force F to cause a displacement x is given by
A force F acts on a body in the x-direction. Figure (a) shows F as a function of x. find the work done by the force in moving the body (i) from x = 0 to x = 1 m, (ii) from x = 1 m to x = 3 m, (iii) from x = 3 m to x = 4 m, and (iv) from x = 0 to x = 4 m.
The work done by a variable force F to move a body from x = x1 to x = x2 is equal to the area of the force curve and the x-axis between the values x1 and x2.
(i) Referring to Fig. the work done by the force in moving the body from x = 0 to x = 1 m is given by
W1 = area of triangle OAD
(ii) The work done in moving the body from x = 1 m to x = 3 m is
W2 = area of rectangle ABED
= 5 N (3m - 1m) = 5 N 2 m = 10 J
(iii) The work done in moving the body from x = 3 m to x = 4 m is
W3 = area of triangle BEC
(iv) Since work is a scalar quantity, the total work done by the force in moving the body from x = 0 to x = 4 is equal to the algebraic sum of W1, W2 and W3, i.e.
W = W1 + W2 +W3
= 2.5 + 10 + 2.5 = 15 J