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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/s2 )

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√2m/sLet x be the extension in the spring when 2Kg mass leaves contact with ground So, kx=mg x=mgk =12 m Using conservation of energy mgx=v2m2+kx22 v=√2gx−kx2m v=√2×10×12−405×14 =2√2 m/s Therefore, the speed will be 2√2 m/s.

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