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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Work, Power and Energy

BITSAT 2015 · Physics · Work, Power and Energy

An elastic string of unstretched length L and force constant k is...

An elastic string of unstretched length L\mathrm{L} and force constant k\mathrm{k} is stretched by a small length x\mathrm{x}. It is further stretched by another small length y. The work done in the second stretching is

Options

  1. A1/2Ky21 / 2 \mathrm{Ky}^{2}
  2. B1/2Ky(2x+y)1 / 2 \mathrm{Ky}(2 \mathrm{x}+\mathrm{y})
  3. C1 / 2 $\mathrm{ K} \left(\mathrmx^{2}+\mathrmy^{2}\right)
  4. D1/2k(x+y)21 / 2 \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}

Correct answer

B. 1/2Ky(2x+y)1 / 2 \mathrm{Ky}(2 \mathrm{x}+\mathrm{y})

Step-by-step solution

In the string elastic force is conservative in nature. ∴W=−ΔU\therefore \mathrm{W}=-\Delta \mathrm{U} Work done by elastic force of string, W=−(UF−Ui)=Ui−UFW=12kx2−k2(x+y)2=12kx2−12k(x2+y2+2xy)=12kx2−12ky2−12kx2−12k(2xy)=−kxy−12ky2\begin{aligned}\mathrm{W}&=-\left(\mathrm{U}_{\mathrm{F}}-\mathrm{U}_{\mathrm{i}}\right)=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{F}}\\ \mathrm{W}&=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{\mathrm{k}}{2}(\mathrm{x}+\mathrm{y})^{2}\\ &=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{1}{2} \mathrm{k} \left(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xy}\right)\\ &=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{1}{2} \mathrm{k} \mathrm{y}^{2}-\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}(2 \mathrm{xy})\\ &=-\mathrm{k} \mathrm{x} \mathrm{y}-\frac{1}{2} \mathrm{k} \mathrm{y}^{2} \end{aligned}Therefore,thethework done against elastic force \mathrm{W}_{\text{external }}=-\mathrm{W}=\frac{\mathrm{K} y}{2}(2 \mathrm{x}+\mathrm{y})
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