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

AI Thread Summary
A block suspended from a spring experiences a downward velocity after a sharp blow, leading to a discussion on how far it travels below the equilibrium position before coming to rest. The potential energy of the spring and gravitational potential energy are key to solving the problem, with the total mechanical energy expressed as the sum of kinetic and potential energies. The equation derived indicates that the distance below equilibrium, h, is given by h = v√(m/k). Clarification was provided on the correct form of the spring potential energy equation, confirming it remains positive regardless of whether the spring is compressed or extended. The discussion emphasizes understanding energy conservation principles in the context of spring mechanics.
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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.
 
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What is the energy immediately after the blow? What is it at the instantaneous rest?
 
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?
 
RandomGuy1 said:
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.
 
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