Question

A force $F=K{x}^2$ acts on a particle at an angle of ${60}^o$ with the work done in displacing the particle from $x_1$ to $x_2$ will be:

• A. $\frac{k{x}^2}{2}$
• B. $\frac{k}{2}(x_2^2-x_1^2)$
• C. $\frac{k}{6}(x_2^3-x_1^3)$
• D. $\frac{k}{3}(x_2^3-x_1^3)$

Answer 1

The correct option is C $\frac{k}{6}(x_2^3-x_1^3)$

Given, $F=k{x}^2,Angle={60}^{\circ }$

We know that the work-done by the force is:

$W={\int }_{x_1}^{x_2}k{x}^{2}dx=k[\frac{x^3}{3}{]}_{x_1}^{x_2}\cos {60}^{\circ }$

$W=\frac{k}{6}[x_2^3-x_1^3]$ Since $\cos \theta =\frac{1}{2}$