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

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

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

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$\frac{a L^{2}}{2} + \frac{b L^{3}}{3}$

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$a L^{2} + b L^{3}$

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Solution

The correct option is C
$\frac{a L^{2}}{2} + \frac{b L^{3}}{3}$

Work done is given by $W = \int F d x = \int (a x + b x^{2}) d x$ .
Integrating from $x = 0$ to $x = L$ , we get $W = \frac{a L^{2}}{2} + \frac{b L^{3}}{3}$ .

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When a rubber band is stretched by a distance x, it exerts a restoring force of magnitude F=ax+bx2, where a and b are constants. The work done in stretching the upstretched rubber band by L is

The correct option is C aL22+bL33 Work done is given by W=∫Fdx=∫(ax+bx2)dx. Integrating from x=0 to x=L, we get W=aL22+bL33.

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

The correct option is C
$\frac{a L^{2}}{2} + \frac{b L^{3}}{3}$

Work done is given by $W = \int F d x = \int (a x + b x^{2}) d x$ .
Integrating from $x = 0$ to $x = L$ , we get $W = \frac{a L^{2}}{2} + \frac{b L^{3}}{3}$ .