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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 20 cm. A 40 N force is required to stretch (and hold the spring) to a length of 30 cm. How much work is done in stretching the spring from 35 cm to 38 cm?

$\begin{array}{1 1}1.98J\\1.78J\\1.38J\\1.48J\end{array} $

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Answer : 1.98J
A force of 40N is required to stretch the spring $30cm-20cm=10cm=0.10m$ from its natural length.
Using Hooke's law which tells us that the force required to stretch a spring a distance of $x$ meters from its natural length is $F(x) =kx$ where $k > 0$ is called the spring constant.
Using Hooke's law we have, $40=0.10k$ $\rightarrow k=400$
Therefore, the force required to hold this spring $x$ meters from its natural length is $F(x)=400x$.
We want to know the work required to stretch the spring from 35cm to 38cm. Doing that gives us distances of 35-20 = 15cm = 0.15 m and 38-20 = 18 cm = 0.18m.
The work is then, $W=\int\limits_{0.15}^{0.18} 400xdx$ $=200x^2\big|_{0.15}^{0.18}=1.98J$
answered Sep 1, 2014 by sreemathi.v
edited Sep 2, 2014 by balaji.thirumalai
 

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