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Let the spring is elongated by a length x at equilibrium condition when a block of mass m is hanged from its end.
Here force exerted by the weight of the block is
F =mg
Again force constant of the spring being k,then by Hook's law F = kx
So
kx = mg
Now in this equilibrium state before application of sharp blow the PE of the spting-block system is
= 1/2 k x^2
The inital downward velocity of the suspended block is v.This means it has gained a KE
= 1/2 m *v^2
So initial toral mechanical energy just after the application of blow is
E1 = 1/2 k *x^2 +1/2 m*v^2
Now let the maximum elongation from equilibrium position be y after the application of sharp blow. Then in this conditon total elongation becomes x+y and the PE of the spring becomes
= 1/2 k*(x+y)^2
But the mass m has decreased its height by y.So there occurs a decrease in PE by the amount = mgy
.When the block comes to an instantaneous rest its KE will be zero.
By cinservation of energy
We get,
y = v (m/k)^(1/2)
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