Why does Hooke's law not work here?

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SUMMARY

The discussion centers on the application of Hooke's Law in calculating the spring constant for a 1200-kg car that strikes a coiled spring after traveling at 85 km/h. The initial attempt to use average acceleration and force resulted in an incorrect spring constant of 69000 N/m. The correct approach involves applying the conservation of energy principle, equating kinetic energy (KE) to potential energy (PE) of the spring, which accurately reflects the variable force exerted by the spring as it compresses.

PREREQUISITES
  • Understanding of Hooke's Law (F = -kx)
  • Knowledge of kinetic energy formula (KE = (1/2)mv^2)
  • Familiarity with potential energy in springs (PE = (1/2)kx^2)
  • Concept of conservation of energy in physics
NEXT STEPS
  • Study the integration of variable forces in mechanics
  • Learn about energy conservation principles in mechanical systems
  • Explore advanced applications of Hooke's Law in real-world scenarios
  • Investigate the effects of non-constant acceleration on motion equations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to clarify misconceptions regarding Hooke's Law and energy principles.

MightyDogg
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Homework Statement


  1. A 1200-kg car moving on a horizontal surface has speed v = 85 kmh when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

Homework Equations


F=-kx
KE=(1/2)mv^2
PE(spring)=(1/2)kx^2

The Attempt at a Solution


I tried to find the average acceleration to slow the car from 85kmh to 0. I used the formula vfinal^2=vinitial^2 + 2ax, where velocity initial is 23.6m/s and x is 2.2m. This gave me an acceleration of -126m/s^2. Then I multiplied the acceleration by the mass to find the average force. This gave me -151200N. Then, I plugged that into Hooke's law with x being 2.2m. This gave the spring constant being 69000N/m.

However, I am supposed to use the conservation of energy principle where KE=PE. This gives the correct answer. Why does my method not work?
 
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MightyDogg said:

Homework Statement


  1. A 1200-kg car moving on a horizontal surface has speed v = 85 kmh when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

Homework Equations


F=-kx
KE=(1/2)mv^2
PE(spring)=(1/2)kx^2

The Attempt at a Solution


I tried to find the average acceleration to slow the car from 85kmh to 0. I used the formula vfinal^2=vinitial^2 + 2ax, where velocity initial is 23.6m/s and x is 2.2m. This gave me an acceleration of -126m/s^2. Then I multiplied the acceleration by the mass to find the average force. This gave me -151200N. Then, I plugged that into Hooke's law with x being 2.2m. This gave the spring constant being 69000N/m.

However, I am supposed to use the conservation of energy principle where KE=PE. This gives the correct answer. Why does my method not work?

I does not work because the acceleration is not constant. Hook's Law works, but not your expression ##v_f^2 = v_i^2 + 2ax##.
 
Last edited:
MightyDogg said:
Then I multiplied the acceleration by the mass to find the average force. This gave me -151200N. Then, I plugged that into Hooke's law with x being 2.2m. This gave the spring constant being 69000N/m.

By Hooke's law I assume you mean ##F=kx## where ##F## is the magnitude of the force and ##x## is the distance?

In that formula ##F## is not the average force. It's the magnitude of the force when the spring is stretched (or compressed) a distance ##x##.

If the force were constant, that would work, but the force is not constant. You could integrate the force, or use energy concepts.
 
Oh, okay it makes sense now. Thank you both very much.
 

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