Physics problem, involving Hookes Law?

In summary, the problem asks to find the distance the spring was compressed by a lab partner, given a 0.500 kg mass resting on a horizontal spring with a constant of 45 N/m. The mass was released and reached a velocity of 3.4 m/s at the equilibrium point (x=0). Using the equations for elastic potential energy and kinetic energy, it can be determined that the elastic potential energy at the equilibrium point is equal to the kinetic energy at release. Therefore, the compression of the spring is 0.358 m. It is possible that the book answer is incorrect or there is a friction component not mentioned in the problem.
  • #1
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Homework Statement


A 0.500 kg mass is resting to a horizontal spring constant of 45 N/m. Your lab partner pulls the spring back and releases it when you are not looking. When the spring reaches its equilibrium point (x=0) the velocity of the mass is 3.4 m/s.

Find how far your partner compressed the spring.


Homework Equations


Ep = 1/2 k x ^2
k = spring constant
Ek = 1/2 m v ^2



The Attempt at a Solution


Not really sure how to do this, since there is more than one unknown.

Ep1 = Ep2 + Ek2

They never stated the spring stops at x=0, and I'm willing to bet it doesn't. However, when I assume that Ep1 = Ek2 and assume it stops, I get 0.30 m (rounded) when the answer should be 0.19m

Am I doing this right if I just set Ep1 to Ek2? It doesn't make sense to do it like this, because it should be

Ep1 = Ep2 + Ek2

But then we don't have enough info to solve the equation
 
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  • #2
What is the elastic potential energy at the equilibrium point (x=0)?
Check the first sentence of your post. Something is missing. ehild
 
  • #3
what you're really saying is that at x=0 the total energy of the system is just the KE because the spring is at its equilibrium position and hence there's zero PE and initially there's no KE and since energy is conserved.

PE1 + KE1 = PE2 + KE2 where KE1=0 and PE2=0 so PE1 = KE2 seems right.

I calculated the compression to be 0.358 m so I'm thinking maybe the book answer is wrong or there's some assumed friction component.
 

FAQ: Physics problem, involving Hookes Law?

1. What is Hooke's Law?

Hooke's Law is a fundamental law of physics that describes the relationship between the force applied to an object and the resulting deformation of that object. It states that the force applied to an object is directly proportional to the amount of deformation it undergoes, as long as the object remains within its elastic limit.

2. How is Hooke's Law expressed mathematically?

Hooke's Law can be expressed as F = -kx, where F is the force applied, k is the spring constant (a measure of the stiffness of the object), and x is the amount of deformation.

3. What is the elastic limit in Hooke's Law?

The elastic limit is the maximum amount of deformation an object can undergo and still return to its original shape once the force is removed. Beyond the elastic limit, the object will undergo permanent deformation or even break.

4. How is Hooke's Law used in real-world applications?

Hooke's Law is commonly used in the design and analysis of springs, rubber bands, and other elastic materials. It also plays a role in fields such as materials science, engineering, and biomechanics.

5. What are the limitations of Hooke's Law?

Hooke's Law is only valid for objects that remain within their elastic limit. Once an object is deformed beyond this limit, the relationship between force and deformation is no longer linear. Additionally, Hooke's Law only applies to objects that exhibit elastic behavior, meaning they can return to their original shape after the force is removed.

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