When a rubber band is stretched by a distance ‘X ’, it exerts a restoring force of magnitude F = ax + bx^2, where ‘a’ and ‘b’ are constants. The work done in stretching the unstretched rubber band by ‘L’ is

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$\frac{{aL}^{3}}{3} + \frac{{bL}^{2}}{2}$

$\frac{{aL}^{3}}{2} + \frac{{bL}^{2}}{3}$

$\frac{{aL}^{2}}{2} + \frac{{bL}^{3}}{3}$

$aL + {bL}^{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} F = ax + {bx}^{2} dw = F \cdot dx \\ dw = (ax + {bx}^{2}) dx \Rightarrow W = \int_{0}^{L} dW = \int_{0}^{L} (ax + {bx}^{2}) dx = \frac{{aL}^{2}}{2} + \frac{{bL}^{3}}{3} \end{array}$

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