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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A spring has a length of 15.4 cm and is hanging vertically from a support point above it.A weight with a mass of 0.200 kg is attached to the spring,causing it to extend to a length of 28.6cm.What is the value of the spring constant?

$\begin{array}{1 1}14.9N/m\\15.9N/m\\24.9N/m\\17.9N/m\end{array} $

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Answer : 14.9N/m
We place the origin of our coordinate system at the top of the spring,with the positive direction upward,as is customary.
Then $x_0=-15.4cm$ and $x=-28.6cm$.
According to Hooke's law,the spring force is $F_s=-k(x-x_0)$
Also,we know the force exerted on the spring was provided by the weight of the 0.200-kg mass :
$ F=-mg $  
$ \;\;\;=-(0.200 kg)(9.81m/s^2) = -1.962 N $
Again,the negative sign indicates the direction.
Now we can solve the force equation for the spring constant :
$ k =-\large\frac{F_s}{x-x_0}=-\large\frac{1.962N}{(-0.286m)-(-0.154m)} $
answered Sep 1, 2014 by sreemathi.v
edited Sep 2, 2014 by pady_1

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