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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A spring with a spring constant of 250 N/m has a length of 0.5 meters at rest. What magnitude of force is needed to stretch the spring so that is 0.75 meters long?

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Answer: 62.5 N
Per Hooke's Law, in an ideal spring, the spring force F$_s$ is proportional to x where x is the displacement of the block from the equilibrium position
$F_s = -kx$.
$x = $ displacement of the block, i.,e, $x =0.75m - 0.5m = 0.25m$; It is given that the spring constant is 250Nm$^{-1}$
$\Rightarrow F_s = -k x = 250 Nm^{-1} \times 0.25\;m = 62.5 N$
answered Aug 25, 2014 by balaji.thirumalai

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