This question was previously asked in

NDA (Held On: 10 Sept 2017) General Ability Test Previous Year paper

Option 2 : \(- \frac{{{x^2}}}{6}\left( {2Ax - 3B} \right)\)

Electric charges and coulomb's law (Basic)

41133

10 Questions
10 Marks
10 Mins

**CONCEPT****:**

- Force: The
**interaction**which**changes or tries to change the position**of anybody is called**force.** **SI unit**of**force**is**Newton (N).**- Mathematically force can be written as
**F = ma**

Where m = mass and a = acceleration of the object**.**

- Potential energy :The energy
**due to the position**of the particle is called**the potential energy**of the particle**.** - The SI unit of
**potential energy**is**Joule (J).** - The relation between
**force and potential energy**is given by: - \({\rm{Force\;}}\left( {\rm{F}} \right) = - \frac{{dU}}{{dx}}\)

Where **U is potential energy** and x is the position.

__EXPLANATION__:

Given- Force = F(x) = AX^{2}- Bx

\({\rm{F}} = - \frac{{{\rm{dU}}}}{{{\rm{dx}}}} = {\rm{A}}{{\rm{X}}^2} - {\rm{Bx}}\)

dU = - (AX^{2}- Bx) dx

Integrate on both sides of the above equation

\(\smallint dU = - \smallint ({\rm{A}}{{\rm{X}}^2} - {\rm{\;Bx}}){\rm{dx}}\)

\({\rm{Potential\;energy\;}}\left( {\rm{U}} \right){\rm{\;}} = {\rm{\;}} - \frac{{A{x^3}}}{3} + \frac{{B{x^2}}}{2} = - \frac{{{x^2}}}{6}\left( {2Ax - 3B} \right)\)