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Initially spring is in its natural length. Now block of mass $0.25 \mathrm{~kg}$ is released, then find out maximum force by system on the floor.
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The correct answer is:
$25 \mathrm{~N}$
By using work - energy theory on $(0.25) \mathrm{kg}$
$\begin{aligned} & \frac{1}{2} k x^2=m g x, \text { where } m=2 \mathrm{~kg} \\ & k x=2 m g=2 \times 0.25 \times 10=5 \mathrm{~N} \\ & N=k x+2 \times 10=5+20=25 \mathrm{~N}\end{aligned}$
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