Question

A force $F = Kx ^{2}$ acts on a particle at an angle of $60^{\circ}$ with the $x$-axis. the work done in displacing the particle from $x_{1}$ to $x_{2}$ will be -

• A. $\frac{ kx ^{2}}{2}$
• B. $\frac{ k }{2}\left( x _{2}^{2}- x _{1}^{2}\right)$
• C. $\frac{ k }{6}\left( x _{2}^{3}- x _{1}^{3}\right)$
• D. $\frac{ k }{3}\left( x _{2}^{3}- x _{1}^{3}\right)$

Answer 1

Answer: (C)

$d W= kx ^{2} dx \cos 60^{\circ}$
$\therefore WD =\frac{ k }{2} \int\limits_{ x _{1}}^{ x _{2}} x ^{2} d x =\frac{ k }{6}\left( x _{2}^{3}- x _{1}^{3}\right)$