How Does Spring Compression Vary with Different Methods of Loading?

In summary, when suddenly dropping a 5.0 kg block onto a vertical spring with a spring constant of 490 N/m, the spring will compress by approximately 0.2m. However, if the block is slowly lowered to the point where it can be removed without disturbing the spring, the displacement will be different due to the net force being 0. This is because Hooke's law states that the force exerted by the spring is directly proportional to its displacement from equilibrium. Therefore, when the block is dropped, there is a net force and the spring will oscillate, while when it is slowly lowered, the net force is 0 and the spring will not oscillate.
  • #1
phazei
9
0
A vertical spring with k = 490 N/m is standing on the ground. You are holding a 5.0 kg block just above the spring, not quite touching it.

A)How far does the spring compress if you let go of the block suddenly?

B)How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it?


A) This part was simple enough,
U(potential) = U(spring)
where the h in potential is the height from where the spring would compress to...
mg(deltaS) = 1/2 k*(deltaS)^2
plug #'s and solve for deltaS. 0.2m

B)I don't know where to start this any differently. 0.2m isn't the correct answer. I just don't know how this differs...
help?

Thanks
 
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  • #2
Hint: If you can move your hand without disturbing it, what must be the net force on the block?
 
  • #3
the net force must be 0.

So the force upward by the spring = mg

But how to i relate that to figure out what the displacement is?
 
  • #4
How does the spring force depend on its displacement from equilibrium? (What is Hooke's law?)
 
  • #5
oh...

mg=-kdeltaS
5(-9.81)=-490deltaS
:biggrin: :biggrin:


so they're different because it occilates when dropped and not when lowered, right?
 
  • #6
phazei said:
oh...

mg=-kdeltaS
5(-9.81)=-490deltaS
Right! (Here's a tip: Don't get hung up on the signs. The minus sign in Hooke's law just means that the spring always exerts a force opposite to its displacement from equilbrium. And g is always a positive number, by the way.)


so they're different because it occilates when dropped and not when lowered, right?
Exactly right!
 

Related to How Does Spring Compression Vary with Different Methods of Loading?

1. What is a block placed on a spring?

A block placed on a spring refers to a physical system in which a block or object is attached to a spring and can move freely in one dimension.

2. What is the purpose of placing a block on a spring?

The purpose of placing a block on a spring is to study the behavior of a simple harmonic oscillator and analyze the relationship between the displacement of the block and the force exerted by the spring.

3. How does the mass of the block affect the motion of the spring?

The mass of the block affects the motion of the spring by changing the period and frequency of the oscillation. A heavier block will have a longer period and lower frequency compared to a lighter block on the same spring.

4. How does the spring constant affect the motion of the block?

The spring constant, also known as the stiffness of the spring, affects the motion of the block by determining the amount of force needed to stretch or compress the spring. A higher spring constant will result in a stiffer spring and a larger force required to create the same displacement in the block.

5. Is there any real-world application for a block placed on a spring?

Yes, there are many real-world applications for a block placed on a spring. Some examples include shock absorbers in vehicles, door closers, and pogo sticks. The behavior of a block on a spring is also used in seismology to understand and predict earthquake motion.

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