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

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SUMMARY

The discussion focuses on the mechanics of a block suspended by a spring and its behavior after receiving a sharp downward blow. The block's mass (m) and the spring constant (k) are critical in determining how far below the equilibrium position the block comes to an instantaneous rest. The total mechanical energy is calculated using the equation 1/2 mv² + (mg)²/2k, leading to the conclusion that the distance below equilibrium is given by h = v√(m/k). The potential energy of the spring remains positive regardless of whether it is compressed or elongated.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants (k)
  • Knowledge of gravitational potential energy (mgh)
  • Familiarity with the conservation of mechanical energy principles
  • Basic algebra for solving equations involving kinetic and potential energy
NEXT STEPS
  • Study the principles of Hooke's Law in detail
  • Learn about energy conservation in mechanical systems
  • Explore the dynamics of oscillatory motion in springs
  • Investigate the effects of damping on spring systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for examples of spring dynamics and potential energy concepts.

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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