JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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
- A \(\frac{1}{2}\left( {a{L^2} + b{L^3}} \right)\)
- B \(\;\frac{{a{L^2}}}{2} + \frac{{b{L^3}}}{3}\)
- C \(\;\frac{1}{2}\) \(\left( {\frac{{a{L^2}}}{2} + \frac{{b{L^3}}}{3}} \right)\)
- D \(\;a{L^2} + b{L^3}\)
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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