Box on a string, find displacement when first at rest

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SUMMARY

The discussion centers on calculating the displacement of a block attached to a spring when it first comes to rest after being displaced. The block has a mass of 3.57 kg, a spring constant of 165.7 N/m, and a coefficient of kinetic friction of 0.239. The initial displacement from equilibrium is 0.216 m. The correct approach involves equating the initial spring energy to the energy dissipated by friction and the remaining spring energy when the block stops, leading to the equation: (Initial Spring Energy) = (Energy Dissipated) + (Spring Energy When Block Stops).

PREREQUISITES
  • Understanding of Hooke's Law and spring potential energy
  • Knowledge of kinetic friction and its effects on motion
  • Familiarity with energy conservation principles in mechanical systems
  • Basic skills in solving algebraic equations
NEXT STEPS
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  • Study the dynamics of oscillatory motion in spring systems
  • Explore the equations of motion for blocks on surfaces with friction
  • Investigate the relationship between thermal energy and mechanical work
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Students studying physics, particularly those focusing on mechanics, energy conservation, and oscillatory motion. This discussion is beneficial for anyone tackling problems involving springs and frictional forces.

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Homework Statement


A block with mass m = 3.57 kg is attached to a spring with spring constant k = 165.7 N/m and negligible mass. The coefficient of kinetic friction between the block and the table is µk = 0.239. The box is displaced from its equilibrium position to the left by 0.216 m. If the block is released from rest, calculate the block's displacement when it first comes to rest again.
(Picture attached)

Homework Equations


Us = 1/2kΔL²
Wfriction = μmgdcosθ


The Attempt at a Solution



This is what I tried:
Us = Wfriction
1/2(k)(xf)² - 1/2(k)(xi)² = μmgdcosθ
I then said xf = d - xi and subbed in this for xf
1/2(165.7)(d-0.216)² - 1/2(165.7)(0.216)² = 0.239(3.57)(9.8)d(cos180°)
solving for d I get d = 0.33 m but this is incorrect
I'm not sure what I'm doing wrong, suggestions?
 

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When the block first comes to rest, does the spring exert sufficient force to move the block in the opposite direction. If so, then friction hasn't yet dissipated all the energy initially stored in the spring, and your energy approach won't work. Try solving the equations of motion and calculating the time at which the block comes to rest.
 
The reason I'm doing it with energy is that this problem is under the heading: Thermal energy and power. How I see it is if the ground were frictionless then the energy stored in the spring at xi would be equal to that stored at xf, and energy is conserved. However with friction energy will be lost to heat therefore the difference in energy stored in the spring at xf and xi should be equal to the energy lost to heat (friction).
 
"The reason I'm doing it with energy is that this problem is under the heading: Thermal energy and power."

Got it. You'll have to account for the energy stored in the spring when the block comes to rest:

(Initial Spring Energy) = (Energy Dissipated) + (Spring Energy When Block Stops)

= μmgd + k(d - xi)^2

Actually, that's easier than solving the equations of motion!
 

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