How Far Does the Block Travel Below Equilibrium After a Sharp Blow?

[RandomGuy1]

Homework Statement

A block of mass m is suspended through a spring of spring constant k[I/] and is in equilibrium. A sharp blow gives the block an initial downward velocity v. How far below the equilibrum position does the block come to an instantaneous rest?

Homework Equations

Potential energy of a compressed or extended spring: 1/2 kx²
Gravitational Potential energy: mgh

The Attempt at a Solution

When the block is at equilibrium, kx = mg.
This implies x = mg/k
Potential energy of the spring in this position = 1/2 kx² = (mg)²/2k
Taking the gravitational potential energy to be zero in this position,
total mechanical energy = 1/2 mv² + (mg)²/2k.

I can't figure out where to go from there.

 

[voko]

What is the energy immediately after the blow? What is it at the instantaneous rest?

 

[RandomGuy1]

Ah, I get your point.
Energy immediately after blow is 1/2 mv² + (mg)²/2k. It is at energy at instantaneous rest that I'm not sure. As the kinetic energy is zero, only elastic and gravitational potential energies have values. So, I get

1/2mv² + m²g²/2k = 1/2 k[(mg/k) + h²] - mgh

Solving that, I get h = v√(m/k).

Just one thing, how do I know if the spring potential energy is negative or positive? I mean, should the 1/2 k[(mg/k) + h²] term be negative as the spring is being brought from its natural state to an elongated state?

 

[voko]

Just one thing, how do I know if the spring potential energy is negative or positive? I mean, should the 1/2 k[(mg/k) + h²] term be negative as the spring is being brought from its natural state to an elongated state?

First of all, it is not 1/2 k[(mg/k) + h²] but 1/2 k[(mg/k) + h]². Since you got the correct answer above, I assume that was a typo.

From the equation it follows immediately that the potential energy, as a function of elongation, is always positive. This is logical, too. If the elongation is positive, that the spring is stretched, so it has some energy stored. If the elongation is negative, then the spring is compressed, again with energy in it. The lowest energy state of a spring is zero, when it is left alone.

 

[Nickie]

To find the distance the block will come to an instantaneous rest after the sharp blow, we need to use the principle of conservation of energy. At the instant the block comes to rest, all of its initial kinetic energy will be converted into potential energy of the spring. Therefore, we can set the total mechanical energy at this instant equal to the potential energy of the spring.

Using the equation for potential energy of a compressed or extended spring, we can set it equal to the total mechanical energy:

1/2 kx^2 = 1/2 mv^2 + (mg)^2/2k

From here, we can solve for x, which represents the distance the block will come to rest below the equilibrium position. This can be done by rearranging the equation and taking the square root of both sides:

x = √(mv^2/k - (mg)^2/k^2)

Therefore, the block will come to an instantaneous rest at a distance of x below the equilibrium position.