Spring system inside an accelerating box

In summary, when the mass is initially at equilibrium, it is 3.00 inches closer to the bottom of the box than it is after the box has accelerated.
  • #1
physics517
7
0

Homework Statement



A mass m is resting at equilibrium suspended from a vertical spring of natural length L and spring constant K inside a box.

The box begins accelerating upward with acceleration a. How much closer does the equilibrium position of the mass move to the bottom of the box?

Homework Equations



f=kx
f=ma

The Attempt at a Solution



so this is my f=ma statement

kx-mg=ma

then i solved for x to get x= m(a+g) / k

but this answer is wrong and the correct answer is

ma / k
 
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  • #2
You are looking for the change in the mass' position from the initial equilibrium position, after the box has begun accelerating. What was its initial position?
 
  • #3
gneill said:
You are looking for the change in the mass' position from the initial equilibrium position, after the box has begun accelerating. What was its initial position?

In other words, what was the equilibrium position of the mass before the box was accelerated?

 
  • #4
hmm the problem doesn't specify that location.

Cant we just assume it to be at location 0
 
  • #5
physics517 said:
hmm the problem doesn't specify that location.

Cant we just assume it to be at location 0

Sure. You can assign your zero position to that location. Just keep in mind that it isn't the same location as the end of the relaxed, unloaded spring.

Recapping what's happening:

1. Box at rest. Spring unloaded, natural length L.
2. Box at rest. Spring loaded with mass M, settles at equilibrium.
3. Box accelerating. Spring stretches more, assumes new equilibrium.

The question is (essentially) asking for the amount of stretching that takes place between items 2 and 3.
 
  • #6
gneill said:
Sure. You can assign your zero position to that location. Just keep in mind that it isn't the same location as the end of the relaxed, unloaded spring.

Recapping what's happening:

1. Box at rest. Spring unloaded, natural length L.
2. Box at rest. Spring loaded with mass M, settles at equilibrium.
3. Box accelerating. Spring stretches more, assumes new equilibrium.

The question is (essentially) asking for the amount of stretching that takes place between items 2 and 3.


I think my problem was that i assumed L to be the spring's length when it was loaded. But what you stated makes sense now and it leads to the right answer.

excellent help. you gave me info but just enough to make me think and understand it

thank you
 

1. How does the spring system inside an accelerating box work?

The spring system inside an accelerating box works by utilizing the principle of Hooke's Law, which states that the force applied to a spring is directly proportional to the amount of stretch or compression of the spring. As the box accelerates, the spring stretches or compresses, creating a force that opposes the acceleration and maintains the equilibrium of the system.

2. What happens to the spring inside an accelerating box when the acceleration changes?

When the acceleration changes, the spring inside the box will either stretch or compress more, depending on the direction of the change in acceleration. If the acceleration increases, the spring will stretch more to oppose the increase, and if the acceleration decreases, the spring will compress more to oppose the decrease.

3. How does the mass of the spring affect the system inside an accelerating box?

The mass of the spring itself does not have a significant effect on the system inside an accelerating box. However, the mass of the object attached to the spring can affect the overall behavior of the system. A heavier object will require a stronger spring to maintain equilibrium, while a lighter object will require a weaker spring.

4. Can the spring system inside an accelerating box be used to measure acceleration?

Yes, the spring system inside an accelerating box can be used to measure acceleration. By measuring the amount of stretch or compression of the spring, the force and therefore the acceleration can be calculated using Hooke's Law. This principle is commonly used in devices such as accelerometers.

5. How does the spring constant affect the behavior of the system inside an accelerating box?

The spring constant, which is a measure of the stiffness of the spring, affects the behavior of the system inside an accelerating box by determining how much force is required to stretch or compress the spring. A higher spring constant means a stiffer spring, which will require more force to stretch or compress, while a lower spring constant means a more flexible spring that will require less force.

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