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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2014
  • Medium
  • MCQ

When a rubber-band is stretched by a distance \(x\), it exerts restoring force of magnitude  \(F=ax+bx^2\) where \(a\) and \(b\) areconstants. The work done in stretching the unstretched rubber-band by \(L\) is

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Answer & Solution

Correct Answer

(B) \(\;\frac{{a{L^2}}}{2} + \frac{{b{L^3}}}{3}\)

Step-by-step Solution

Detailed explanation

Work done in stretching the \(rubber-band\) by a distance \(dx\) is

\(dW =F dx =\) \(\left( {ax + b{x^2}} \right)dx\)

Integrating both sides,

\(W = \int\limits_0^L {axdx} \, + \,\int\limits_0^L {b{x^2}\,dx = \frac{{a{L^2}}}{2}}  + \frac{{b{L^3}}}{3}\)

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