A force (F) acting on a particle is such that F is inversely proportional to the distance covered. The work done by the force in moving the particle from point 'a' to point 'b' is proportional to-

The correct answer is option 4) i.e. \(ln(\frac{b}{a})\)

CONCEPT:

• Work is said to be done by a force when the force acting on it causes the object to displace.

Mathematically it is given by;

W = F.x = Fxcosθ 

Where F is the force acting on the object and x is the displacement caused.

• The force acting on an object and displacement is represented graphically as shown. 
  • Consider a varying force acting on the object. If we divide the region under the curve into infinitesimally small regions, the force would appear constant for that region which has caused a displacement of Δx. 
  • In such a case, the area of that small region = Force × displacement (Δx) = work done.

Therefore, work done by a variable force is given by:

​\(W = ∫ _{x_1} ^{x_2} F(x)dx\)

EXPLANATION:

Let x be the distance covered. Given that \(F \propto \frac{1}{x}\)

Work done, \(W = ∫ _{a} ^{b} F(x)dx\)

\(\Rightarrow W \propto ∫ _{a} ^{b} \frac{1}{x}dx\)

\(\Rightarrow W \propto [ln(x)]_a ^b\)

\(\Rightarrow W \propto ln(b)-ln(a)\)

\(\Rightarrow W \propto ln(\frac{b}{a})\)