Maximum Compression for a Block on Spring

Click For Summary
The discussion focuses on the discrepancy between using Newton's laws and conservation of energy to determine maximum compression of a spring by a block. The initial approach using Newton's laws only accounts for the equilibrium position, where gravitational and elastic forces balance, leading to an incorrect answer. In contrast, conservation of energy correctly considers the block's kinetic energy, allowing for maximum compression to be calculated accurately. The key takeaway is that maximum compression occurs after the block passes the equilibrium point, where the elastic force exceeds gravitational force, resulting in oscillation. The exercise is best approached using energy conservation principles for accurate results.
Abu
Messages
39
Reaction score
1
Homework Statement
A spring of negligible mass has force constant k = 800 N/m. You place the spring vertically with one end on the floor. You then lay a 1.6 kg block on top of the spring and release the book from rest. Find the maximum distance the spring will be compressed
Relevant Equations
F = kx
Hi everyone, just a quick question..

I tried this problem using Newtons laws, not conservation of energy, and I got an answer exactly half of what the correct answer is, and I'm not sure why. Here is what I did:

Net force = zero once the spring is compressed, therefore

mg - kx = 0
mg = kx
mg/k = x
1.6(9.8)/800 = x
x = 1.96 cm

The actual answer is 3.92 cm however, and I don't know why.

If I do this problem using conservation of energy, I get the correct answer but I can't explain why my Newtons laws method is wrong.

Thanks for your consideration!
 
Physics news on Phys.org
When you let go of the block and it starts compressing the spring, the spring will start oscillating. The equation you were using is at the equilibrium position where gravity and elastic force cancel. However, once block reaches this position, it will have non-zero velocity, so it will move past this point and compress the spring further, until it's velocity drops to zero. At this point, elastic force will be bigger than gravity, so it will push the block up, hence it will oscillate.

Therefore, since the energy is conserved, in order to get the maximum compression, you need the equation where kinetic energy is zero, so by substituting correct potential energy for gravity and elastic force, you get the right answer, as you did. In order to get that same point via Newton's law, you'd need to search for the point where elastic force is maximal, not the point where it cancels with gravity. This would prove more complicated than using energy conservation in my opinion, so that's why this exercise is most likely intended to be done using energy conservation.

Hope that helps.
 
  • Like
Likes Abu
Antarres said:
When you let go of the block and it starts compressing the spring, the spring will start oscillating. The equation you were using is at the equilibrium position where gravity and elastic force cancel. However, once block reaches this position, it will have non-zero velocity, so it will move past this point and compress the spring further, until it's velocity drops to zero. At this point, elastic force will be bigger than gravity, so it will push the block up, hence it will oscillate.

Therefore, since the energy is conserved, in order to get the maximum compression, you need the equation where kinetic energy is zero, so by substituting correct potential energy for gravity and elastic force, you get the right answer, as you did. In order to get that same point via Newton's law, you'd need to search for the point where elastic force is maximal, not the point where it cancels with gravity. This would prove more complicated than using energy conservation in my opinion, so that's why this exercise is most likely intended to be done using energy conservation.

Hope that helps.
Thank you so much for your fast reply. Your response helped me a lot, I really appreciate it! Thanks!
 
You're welcome!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 22 ·
Replies
22
Views
2K
Replies
24
Views
4K
Replies
8
Views
6K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K