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Question

System shown in figure is released from rest. pulley and spring is massless and friction is absent everywhere. The speed of $$5\ kg$$ block when $$2\ kg$$ block leaves the contact with ground is :
(Take force constant of spring $$k=40\ N/m $$ and $$g=10\ m/s^2$$ ) 
1060480_9c3494d3c5344eec8fec286c47488304.PNG


A
2m/s
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B
22m/s
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C
2m/s
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D
42m/s
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Solution

The correct option is B $$2\sqrt 2 \,m/s$$

Let x be the extension in the spring when 2Kg mass leaves contact with ground

So,  $$ kx=mg $$

 $$ x=\dfrac{mg}{k} $$

 $$ =\dfrac{1}{2}\ m $$

Using conservation of energy

  $$ mgx=\dfrac{{{v}^{2}}m}{2}+\dfrac{k{{x}^{2}}}{2} $$

 $$ v=\sqrt{2gx-\dfrac{k{{x}^{2}}}{m}} $$

 $$ v=\,\sqrt{2\times 10\times \dfrac{1}{2}-\dfrac{40}{5}\times \dfrac{1}{4}} $$

 $$ =2\sqrt{2}\ m/s $$

Therefore, the speed will be $$2\sqrt2 \ m/s$$.

 


Physics

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