When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude $\mathrm{F}=\mathrm{ax}+\mathrm{bx}^{2}$ where a and b are constants. The work done in stretching the unstretched rubber band by L is:

(A) $\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{bL}^{3}}{3}$

(B) $\frac{1}{2}\left(\frac{a L^{2}}{2}+\frac{b L^{3}}{3}\right)$

(D) $\frac{1}{2}\left(\mathrm{aL}^{2}+\mathrm{bL}^{3}\right)$

JEE Main · 2014 · 06 Apr 2014 · Offline · Work, Power & Energy

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:

(A) $\frac{aL ^{2}}{2} + \frac{bL ^{3}}{3}$ ✓ Correct
(B) $\frac{1}{2} (\frac{a L ^{2}}{2} + \frac{b L ^{3}}{3})$
(C) $a L^{2} + b L^{3}$
(D) $\frac{1}{2} (aL^{2} + bL^{3})$

Answer: (A)

Solution

The work stored in stretching is the force integrated over the extension,

W = \int_{0}^{L} (a x + b x^{2}) d x

. Carrying out the integral,

W = [\frac{a x ^{2}}{2} + \frac{b x ^{3}}{3}]_{0}^{L} = \frac{a L ^{2}}{2} + \frac{b L ^{3}}{3}

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option

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