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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A spring has a natural length of 18 inches and a force of 20 lbs is required to stretch and hold the spring to a length of 24 inches. What is the work required in ft-lbs to stretch the spring from a length of 21 inches to a length of 26 inches?

$\begin{array}{1 1}7.6389ft-lbs\\6.6389ft-lbs\\5.6389ft-lbs\\4.6389ft-lbs\end{array} $

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Answer : 7.6389ft-lbs
We are told that a force of 20lbs is needed to stretch the spring 24in-18in =6in=0.5ft from its natural length.
Then using Hooke's law we have, $20=k(0.5)$
$\Rightarrow k= 40$
Again,using Hooke's law we can see that the force function is $F(x) =40x$
For the limits of the integral we can see that we start with the spring at a length of 21-18=3in or $\large\frac{1}{4}$ feet and we end with a length of 26 - 18=8in or $\large\frac{2}{3}$ feet.
These are then the limits of the integral. The work is then, $W=\int\limits_{\large\frac{1}{4}}^{\large\frac{2}{3}} 40\;x\;dx$
$\;\;\;\;=20 x^2\bigg|_{\Large\frac{1}{4}}^{\Large\frac{2}{3}}$
$\;\;\;\;=\large\frac{275}{36}$
$\;\;\;\;=7.6389\;ft-lbs$
answered Sep 1, 2014 by sreemathi.v
edited Sep 2, 2014 by balaji.thirumalai
 

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