Why does Hooke's law not work here?

In summary, the car has a force of -151200N when it strikes the spring and the spring has a stiffness constant of 69000N/m.
  • #1
MightyDogg
14
0

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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  • #2
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:
  • #3
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.
 
  • #4
Oh, okay it makes sense now. Thank you both very much.
 

Related to Why does Hooke's law not work here?

1. What is Hooke's law and why is it important?

Hooke's law is a principle in physics that states that the force needed to extend or compress a spring by some distance is directly proportional to that distance. It is important because it helps us understand the behavior of elastic materials and allows us to make predictions about their behavior.

2. In what situations does Hooke's law not work?

Hooke's law does not work when the material being stretched or compressed is not perfectly elastic. This means that the material does not return to its original shape and size when the force is removed. Additionally, Hooke's law may not apply when the forces involved are very large or when the material is being stretched beyond its elastic limit.

3. Can you give an example of when Hooke's law does not work?

One example of when Hooke's law does not work is when a rubber band is stretched too far. Initially, the rubber band may follow Hooke's law, but at some point, it will reach its elastic limit and no longer return to its original shape. This is because rubber bands are not perfectly elastic and have a limit to how much they can stretch.

4. How do we account for Hooke's law not working in certain situations?

In situations where Hooke's law does not work, we can use more complex equations or models to describe the behavior of the material. For example, we can use stress-strain curves to analyze the behavior of a material that is not perfectly elastic. We can also use different laws and principles, such as the Young's modulus or the shear modulus, to describe the behavior of non-ideal materials.

5. Is Hooke's law still useful even though it does not work in all situations?

Yes, Hooke's law is still useful because it provides a simple and intuitive way to understand the behavior of elastic materials. It also allows us to make predictions and approximations in situations where the forces involved are not too large and the material is not being stretched beyond its elastic limit. Additionally, Hooke's law serves as a fundamental principle in the study of elasticity and is the basis for more complex models and theories.

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