Determination of Work by Variable Force using Integral Calculus

Determination of work by variable force using integral calculus:

Suppose, a variable force is acting along x-axis on a body. Magnitude of the force depends on distance travelled by the body i.e., F is the function of distance x. In the figure a graph has been plotted showing the variation of F(x) for different values of x.

Total displacement consists of equally divided n number of exceedingly small displacement ∆x, where ∆x is the displacement between x₁ and x₁ + ∆x. For this small change of displacement the magnitude of force is considered constant and the value of force is F₁. So work done by this force tin this small segment is,

∆W₁ = F₁ ∆x

Similarly, second segment is extended between x₁ + ∆x to x₁ + 2∆x and the small displacement is ∆x. For this small segment constant force is F₂. So work done by the force in the second segment is ∆W₂ = F₂ ∆x. So, total work done by the force F(x) to move the body from position x_i to x_f is given by,

W = ∆W₁ + ∆W₂ + ∆W₃ + … … …+ ∆W_N

= F₁∆x + F₂∆x + F₃∆x + … … … + F_N∆x

= ^N∑_K=1 F_K ∆x … … … (1)

When ∆_x becomes smaller and smaller, that means the value of N becomes very large, calculated value of work done will be more correct. We can get correct value of work done by the force F(x) it ∆x approaches zero and N becomes infinity. Then correct value will be,

W = lim_∆x→0 ^N∑_K=1 F_K ∆x … … … (2)

But in the language of calculus, the quantity, lim_∆x→0 ^N∑_K=1 F_K ∆x; is written as:

^xf∫_xi F(x) dx; which indicates the integration between x_i and x_f.

So, equation (2) becomes,

W = ^xf∫_xi F(x) dx

Numerically this quantity is the area subtended by the curved line of force and the displacement between x₁ and x_f. So, by integration work and area can be determined.