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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 1 meter.A force of 25 Newtons stretches the spring by $\large\frac{1}{4}$ of a meter.Determine how much work is done by stretching the spring from a length of 1.5 meters to 2.5 meters.

$\begin{array}{1 1}100J\\250J\\140J\\150J\end{array} $

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1 Answer

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Answer : 100J
We first determine the spring constant,$k$.Because the force is 25N when $x$ is 0.25m,we can use Hooke's law to determine $k$
$f(x) =kx$
$25N=k\big(\large\frac{1}{4}$$m\big)$
$\therefore k =100\large\frac{N}{m}$
$f(x) =100x$
The work done by stretching the spring from a length of 1.5m to 2.5m:
$W=\int\limits_a^b f(x) dx$
$\;\;\;\;=\int\limits_{0.5}^{1.5} 100xdx$
$\;\;\;\;=\large\frac{100}{2}$$x^2\bigg|_{0.5}^{1.5}$
$\;\;\;\;=50(1.5)^2-50(0.5)^2$
$\;\;\;\;=100Nm$
$\;\;\;\;=100J$
answered Sep 1, 2014 by sreemathi.v
 

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