Solving for Spring Compression: A 0.10 kg Wooden Block

In summary: JIn summary, the problem involves two identical blocks, one sitting on a vertical massless spring and the other dropped from a height of 1.00 m above the first block, resulting in a perfectly inelastic collision. The task is to determine the compression of the spring, neglecting air drag and friction. The solution involves using conservation of linear momentum to find the combined velocity of the blocks and then applying conservation of energy to find the compression in the spring. For the second part of the conversation, the discussion shifts to a different homework problem involving a block of mass 1.50 kg attached to a horizontal spring with a stiffness constant of 200
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Homework Statement



A 0.10 kg flat wooden block is sitting on a vertical massless spring of spring constant 2000 N/m. An identical block is dropped from a height of 1.00 m above the first block. The collision between the two blocks is perfectly inelastic. How far down is the the spring compressed? Neglect air drag and friction in the spring.


Homework Equations



Conservation of Energy? Energy absorbed by spring = (1/2)(kX^2)


The Attempt at a Solution



Do I need to be concerned with potential energy? So far, I set up the equation:

(1/2)(2000)y2 - (1)(9.8)y + (1)(9.8) = 0 but that didnt give me the correct answer...Im confused as to why!

Any suggestions would be awesome, thanks in advance...
 
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  • #2


ops...
 
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  • #3


When the dropped block collides with the stationary block inelastically, find the combined velocity of the blocks by using law of conservation of linear momentum. Now apply the law conservation of energy to find the compression in the spring.
 
  • #4


Where does the 1+x come from? Because of the 1m height?

Or was that not correct Bright Wang, I saw you removed it...

Trying out rl.bhat's method now...
 
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  • #5


ops I was learning conservation of elastic energy in school today.. we never learned momentum in school so I didn't see that...

Ok first you find the velocity of the object 2. (the one that falls 1m). It hits object 1perfectly inelastic, so energy is not conserved but momentum always is!

gh=1/2*v^2
=> 9.8*0.1=1/2*v^2 => v=4.43m/s

Then P=P' => 0.1*4.43=(0.1+0.1)*V
V=2.21m/s

Then use conservation of energy...

1/2*(M1+M2)*V^2+(M1+M2)gh=1/2*k*x^2
then find x

Hope that's right!
 
  • #6


Wow that helps out a lot...makes sense:

I came up with:

(1/2)(.2)(2.21) + (.2)(9.8)(1m) = (1/2)(2000)x^2

x = .047m

That look right to you guys? I really appreciate the fast help!
 
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What is this (.2)(9.8)(1m) ?
 
  • #8


Combined mass of the 2 blocks * Gravity * The height the block is dropped from...

Or should it not be where the block is dropped from since they are moving as one object now?

Im thinking that's my potential energy (mgy) ?

Or instead of 1m I should have x on that side of the equation?
(1/2)(.2)(2.21) + (.2)(9.8)(x) = (1/2)(2000)x^2

x=.016m
 
  • #9


Now it makes sense.
 
  • #10


Thanks so much for yalls help...I really appreciate it! It makes sense...

I was able to most all of my HW and get the correct answers but I am not very sure about this last one I did...(Ya know, if you get bored and want to help) :)

A block of mass 1.50 kg is attached to one end of a horizontal spring, the other end of which is fixed to a vertical wall. The spring has a stifffness constant of 2000 N/m. The block slides without friction on a horizontal table, set close to the wall. Find the work done by the spring force if the block moves (a) from the equilibrium position till the spring is stretched by 10.0 cm, (b) from this last position till the sring is compressed by 3.00 cm.

For part A) I said

(1/2)*(2000)*(-.1m)^2

Or should I be looking at integrating from 0 to 10 of the Fspring?

I did it by spring potential = (1/2)(k)(x^2)
 

1. How do you calculate the spring constant in this experiment?

The spring constant, or k value, can be calculated by dividing the force applied to the spring by the resulting change in length. In this experiment, the force can be calculated by multiplying the mass of the wooden block by the acceleration due to gravity.

2. What is the purpose of solving for spring compression in this experiment?

The purpose of this experiment is to determine the relationship between the force applied to a spring and the resulting compression. This can help us understand the behavior of springs and their use in various applications.

3. How does the mass of the wooden block affect the spring compression?

The mass of the wooden block does not directly affect the spring compression, but it does affect the force applied to the spring. A heavier block will result in a greater force and therefore a greater compression.

4. Can the spring constant change over time?

The spring constant is a characteristic of a specific spring and does not change over time unless the spring is physically altered. However, the compression of the spring may change over time due to factors such as material fatigue or temperature changes.

5. How can this experiment be applied in real-world situations?

The concept of spring compression is applied in various real-world situations, such as in shock absorbers, trampolines, and car suspension systems. Understanding the relationship between force and compression can help engineers design and improve these systems.

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