Work Power and Energy 1 Question 3
4. 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 unstretched rubber band by $L$ is
(2014 Main)
(a) $a L^{2}+b L^{3}$
(b) $\frac{1}{2}\left(a L^{2}+b L^{3}\right)$
(c) $\frac{a L^{2}}{2}+\frac{b L^{3}}{3}$
(d) $\frac{1}{2} \frac{a L^{2}}{2}+\frac{b L^{3}}{3}$
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Answer:
Correct Answer: 4. (c)
Solution:
- Thinking Process We know that change in potential energy of a system corresponding to a conservative internal force as
$$ U _f-U _i=-W=-\int _i^{f} F \cdot d r $$
Given,
$$ F=a x+b x^{2} $$
We know that work done in stretching the rubber band by $L$ is $|d W|=|F d x|$
$$ \begin{aligned} |W| & =\int _0^{L}\left(a x+b x^{2}\right) d x \\ & ={\frac{a x^{2}}{2}}^{L}+{\frac{b x^{3}}{3}} _0 \\ & =\frac{a L^{2}}{2}-\frac{a \times(0)^{2}}{2}+\frac{b \times L^{3}}{3}-\frac{b \times(0)^{3}}{3} \\ & =|W|=\frac{a L^{2}}{2}+\frac{b L^{3}}{3} \end{aligned} $$
