The force acting on a particle of mass m moving along the x-axis is given by F(x) = Ax2 – Bx. Which one of the following is the potential energy of the particle ?
2Ax – B
$ -rac{x^2}{6} (2Ax – 3B) $
$ Ax^3 – Bx^2 $
Zero
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-2 – 2017
$U(x) = -\int (Ax^2 – Bx) dx = – \left( A \int x^2 dx – B \int x dx \right) = – \left( A \frac{x^3}{3} – B \frac{x^2}{2} \right) + C$
$U(x) = -\frac{A}{3}x^3 + \frac{B}{2}x^2 + C$.
If we set the integration constant $C=0$, the potential energy is $U(x) = -\frac{A}{3}x^3 + \frac{B}{2}x^2$. Let’s rewrite Option B: $ -\frac{x^2}{6} (2Ax – 3B) = -\frac{2Ax^3}{6} + \frac{3Bx^2}{6} = -\frac{A}{3}x^3 + \frac{B}{2}x^2 $. This perfectly matches the derived potential energy expression with $C=0$.
– To find $U(x)$ from $F(x)$, integrate: $U(x) = -\int F(x) dx$.