How Far Will the Spring Compress When a Block is Dropped?

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SUMMARY

The discussion focuses on calculating the maximum compression of a spring when a block of mass 2.0 kg is dropped from a height of 55 cm onto it, using a spring constant of 1960 N/m. The conservation of energy principle is applied, leading to the equation mgh = -(1/2)kx², where m is the mass, g is the acceleration due to gravity, h is the height, k is the spring constant, and x is the compression of the spring. The challenge presented is solving the quadratic equation for x, which represents the maximum distance the spring compresses.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with quadratic equations
  • Knowledge of spring mechanics and Hooke's Law
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the derivation of the conservation of energy equation in mechanical systems
  • Learn how to solve quadratic equations using the quadratic formula
  • Explore the applications of Hooke's Law in real-world scenarios
  • Investigate the effects of varying spring constants on compression distances
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Students in physics, engineers working with mechanical systems, and anyone interested in understanding the dynamics of spring compression and energy conservation.

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Homework Statement


A block of mass m = 2.0 kg is dropped from height h = 55 cm onto a spring of spring constant k = 1960 N/m. Find the maximum distance the spring is compressed.

Homework Equations


[tex]\Delta K+ \Delta U_G+ \Delta U_S=0[/tex] \change in kinetic energy+change in gravitational energy+change in spring energy=0
[tex]\Delta K=0[/tex]
[tex]U_{Gf}-U_{Gi}+U_{Sf}-U_{Si}=0[/tex]
[tex]U_{Si}=0[/tex] can be assumed
so
[tex]U_{Sf}=U_{Gi}-U_{Gf}[/tex]

The Attempt at a Solution


resumed from above:
[tex]\frac {kx_f^2}{2}=mgy_i-mgy_f[/tex]
so: [tex]\frac {kx_f^2}{2}=mgy_i-mgx_f[/tex]
With this I have tried over and over to solve for [tex]x_f[/tex] but I cannot find any way of getting [tex]x_f[/tex] by itself.
edit:LaTeX formatting
 
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conservation of energy, [tex]\Delta[/tex]U(g)+ [tex]\Delta[/tex]U(spring) = 0 <=== look, all the energy lost from one object goes to the other, so the sum is 0. Remember! conservation of energy. Think about it.

[tex]\Delta[/tex]U(g)=-[tex]\Delta[/tex]U(spring) <===== moved that equation around a it... just algebra.

mgh2-mgh1 = (1/2)kx^2 <===== look what we have here if we make it look more detailed with what we know.

Or simply mgh=-(1/2)kx^2
 
Last edited:
[tex]\frac {kx_f^2}{2}=mgy_i-mgx_f[/tex]
With this I have tried over and over to solve for [tex]x_f[/tex] but I cannot find any way of getting [tex]x_f[/tex] by itself.
It's a quadratic equation.
 

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