Question

Question #0e5ad

Answer 1

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)`