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System shown in figure is released from ...

System shown in figure is released from rest . Pulley and spring is mass less and friction is absent everywhere. The speed of `5 kg` block when `2 kg` block leaves the constant of with ground is (force constant of spring `k = 40 N//m and g = 10 m//s^(2))`

A

`sqrt2 ms^(-1)`

B

`2 sqrt2 ms^(-1)`

C

`2 m s^(-1)`

D

`3 sqrt2 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
Let x be the extension in the string when 2 kg block leaves the contact with ground. Then tension in the spring should be equal to weight of 2 kg block,
`kx=2g` or `x=(2g)/(k) = (2 xx 10)/(40)= (1)/(2)` m
Now from conservation of mechanical energy, we get,
`mgx = (1)/(2)kx^(2) + (1)/(2) mv^(2)`
`rArr v= sqrt(2gx-(kx^(2))/(m)) = sqrt(2 xx 10 xx (1)/(2) - (40)/( 4xx5) ) = 2 sqrt2 m s^(-1)`.
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