Question #0e5ad

1 Answer
Nov 22, 2016

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......(1)#

Now in this equilibrium state before application of sharp blow the PE of the spting-block system is #=1/2kx^2#

The inital downward velocity of the suspended block is v.This means it has gained a KE #=1/2mv^2#

So initial toral mechanical energy just after the application of blow is
#E_1=1/2kx^2+1/2mv^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 #=1/2k(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.

So in this position the total energy is given by

#E_2=1/2k(x+y)^2-mgy#

By conservation of energy we have

#E_2=E_1#

#=>1/2k(x+y)^2-mgy=1/2kx^2+1/2mv^2#

#=>1/2k(x+y)^2-1/2kx^2-mgy=1/2mv^2#

#=>k(x+y)^2-kx^2-2mgy=mv^2#

#=>k(x+y+x)y-2mgy=mv^2#

#=>2kxy+ky^2-2mgy=mv^2#

#=>2mgy+ky^2-2mgy=mv^2#

#color(red)("by equation (1) "kx= mg)#

#=>ky^2=mv^2#

#=>y^2=m/kv^2#

#=>y=vsqrt(m/k)#