Force and Energy of a block on a spring as it compresses.

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a 10kg block placed on a spring scale with a spring constant of 10N/m. The scenario includes calculating the compression of the spring, the energy stored in the spring when compressed, and the force exerted by the block on the spring at maximum compression. The context includes concepts of gravitational potential energy and spring potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations related to the spring's compression and the energies involved, with some expressing uncertainty about the correctness of their reasoning. Questions arise about the role of potential energy in the context of the problem and how to interpret the scenario of gently placing the block on the spring.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup and the implications of their calculations. Some guidance is offered regarding the analysis of forces and energy conservation, but there is no explicit consensus on the correct approach or interpretation.

Contextual Notes

Participants note potential ambiguities in the problem setup, particularly regarding the initial conditions of the block and spring interaction. There are mentions of the need for a free body diagram and considerations of energy dissipation in real springs.

Lebombo
Messages
144
Reaction score
0

Homework Statement



Suppose is a 10kg block sitting at rest on the ledge of a building 20 meters tall. The 10kg block applies a Force of F = ma = 10kg(10m/s^2) = 100N. The potential energy with reference to the ground would then be PE = mgh = 10kg(10m/s^2)(20m) = 2000J.

If the 10kg block is then gently placed on a 20m tall spring scale of spring constant k = 10N/M

How far does the spring compress?
Determine the amount of energy in the fully compressed spring.
How much force is the block applying to the spring when the spring is fully compressed?

Homework Equations



F*d = mgh
F*d = (1/2)kx^2
F = kx

The Attempt at a Solution



Answer to question 1)

F = kx
F/k = x
mg/k = x
(10kg)(10m/s^2) / (10N/m) = x
10 meters = x

So the spring compresses 10m when the 10kg block is gently placed on the spring.Answer to question 2)

Gravitational PE = F*d = mgh = (10kg)(10m/s^2) * 10m = 100N*10m/s^2 = 1000J

Spring PE = (1/2)kx^2 = (1/2)(10N/m)(10m)^2 = 500J
Answer to Question 3

1st guess:

kx = F = 100N2nd guess:

F*d = (1/2)kx^2
F(10m) = 500J
F = 500J/10m = 50N3rd guess:

(1/2)(base)(height) = (1/2)(Δx)(Δy) = (1/2)(Distance)(Force) = (1/2)(10m)(F) = 1000J
F = 2(1000J)/(10m) = 200NDon't know if any of this is correct. I don't fully understand the energy and force and whether or not potential energy even has any part in determining the energy and force. Or how spring potential energy determines the energy and force. Thank you for any reply.
 
Physics news on Phys.org
Lebombo said:
If the 10kg block is then gently placed on a 20m tall spring scale of spring constant k = 10N/M

Lebombo said:
Don't know if any of this is correct. I don't fully understand the energy and force and whether or not potential energy even has any part in determining the energy and force. Or how spring potential energy determines the energy and force. Thank you for any reply.

to attempt such 'events' draw a free body diagram of both situations and then decide if you wish to analyse the situation in detail or initial and final equilibrium states of the block and the spring.
1. initially at the ledge the body is in equilibrium- one does transfer it gently- without disturbing the equilibrium but now the spring can not support it as it is not compressed - as time advances the body will move down with some velocity under the action of force (downward weight- K.x) if x is the displacement.
2. naturally spring force will put it finally to zero velocity but at the point where mg=k.x , the body will have some K.E. left over so it will go further down and one can expect oscillations about a mean point of the spring and finally it will die out in real springs- so some part of energy is going to be dissipated .

3. if one takes the final position of the spring and adds spring energy 1/2 . k.x^2 + P.E. of the body at the new height = total energy and gets a value which is not equal to the initial potential energy - one may wonder where the energy is gone?
i think the above is a fair analysis of the physical situation but i may be corrected if something is overlooked.
 
Lebombo said:
If the 10kg block is then gently placed ...
I'm not certain how to interpret that. I can hold a block on top of an uncompressed spring, gently, then let it go, or I can let it down gently so that when I let go nothing moves. drvrm's interpretation (the former of those) is probably right, but it leads to the physically unrealistic situation that the spring can be compressed to nothing.
 
haruspex said:
can hold a block on top of an uncompressed spring, gently, then let it go, or I can let it down gently so that when I let go nothing moves

one can try 'gently' pushing the spring down without generating any perceptible velocity- but it will be a cumbersome process taking a lot of time and the losses in the spring can not be avoided for 'real springs'.
sorry, i could not follow the essence of the comment quoted below
haruspex said:
but it leads to the physically unrealistic situation that the spring can be compressed to nothing.
moreover as the problem has been set-up by the author -is there any experimental back -up /references?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
44
Views
7K
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
30
Views
2K
Replies
17
Views
2K
Replies
29
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K