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

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Homework Help Overview

The problem involves a block of mass m suspended by a spring with spring constant k, which is initially at equilibrium. After receiving a sharp blow that imparts an initial downward velocity v, the question is how far below the equilibrium position the block comes to an instantaneous rest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of energy, considering the potential energies involved at different states of the block's motion. Questions arise regarding the calculation of energy at the moment of instantaneous rest and the sign of the spring potential energy during elongation.

Discussion Status

Some participants have provided insights into the energy calculations immediately after the blow and at the point of instantaneous rest. There is an ongoing exploration of the correct formulation of the spring potential energy and its implications for the problem.

Contextual Notes

Participants are navigating the definitions and assumptions related to potential energy in the context of spring mechanics, particularly regarding the reference points for gravitational and elastic potential energy.

RandomGuy1
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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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