Question

A force F=-k/x_2(x!=0) acts on a particle in x-direction. Find the work done by this force in displacing the particle from. x = + a to x = 2a. Here, k is a positive constant.

Answer 1

Given:

A variable force , F = -k/x² acting in x direction.

initial point  = x = a

Final point = x = 2a

To Find:

The work done by this force in displacing the particle.

Solution:

• Work done = ∫F.dx

Therefore ,

• Work done = $\int\limits^ {2a}_a {-k/x^{2} } \, dx$ = $\frac{k}{x}$ a to 2a = k/2a - k/a = -k/a  + k/2a = -k/2a
• Therefore work done by the force = - k/2a J