Question

A block of mass m is suspended through a spring of
spring constant k and is in equilibrium. A sharp blow
gives the block an initial downward velocity v. How far
below the equilibrium position, the block comes to an
instantaneous rest?



please solve my question. I m new here ​

Answer 1

Because, the block was already hanging suspended in air, it would have reached it's equilibrium point which will be at a distance such that the forces in vertical direction are balanced , that is

mg = kx

or,

at a distance x = mg/k

Now,

Since, the initial kinetic energy of mass m when the sharp blow is given to it = 1/2mv^2

Now, the already potential energy stored in the spring block system will be 1/2k(mg/k)^2 = 1/2k m^2g^2

and,

At the moment when it's speed become zero it's kinetic energy is also zero and the extension be mg/k + A

such that A can also be called the amplitude in this case because that's the maximum distance the block is going from its mean position.

So, now conserving energy of the spring block system we get,

1/2mv^2 + m^2g^2/2k = 1/2 k ( mg/k + A)^2

Or,

mv^2 / k + m^2g^2 / k^2 = (mg/k + A)^2

or,

A = (√ mv^2 / k + m^2g^2 / k^2 ) - mg/k

So, this is the final answer.

Hope this helps you !