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

• Lebombo
In summary: I could not follow the essence of the comment quoted below. 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, there may be experimental back-up or references.
Lebombo

## 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?

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

## The Attempt at a Solution

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

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.

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?

## 1. What is the relationship between the force and energy of a block on a spring as it compresses?

As the block on a spring compresses, the force exerted by the spring increases and the potential energy stored in the spring also increases. This is because the spring is being compressed, which requires work to be done and thus increases the potential energy.

## 2. What factors affect the force and energy of a block on a spring?

The force and energy of a block on a spring are affected by the spring constant, the displacement of the spring, and the mass of the block. A stiffer spring (higher spring constant) will exert a greater force and store more potential energy when compressed. A larger displacement will also result in a higher force and more potential energy, while a heavier block will require more force to compress the spring.

## 3. How does the force and energy of a block on a spring change as the spring is compressed?

As the spring is compressed, the force exerted by the spring will increase due to the increase in displacement. This increase in force will also result in an increase in the potential energy stored in the spring. As the spring is released and the block moves back to its original position, the force and energy will decrease accordingly.

## 4. What is the equation for calculating the force exerted by a spring on a block?

The force exerted by a spring on a block can be calculated using the equation F = -kx, where F is the force, k is the spring constant, and x is the displacement of the spring.

## 5. How does the energy of a block on a spring change when the spring is compressed at different rates?

The energy of a block on a spring will increase as the spring is compressed at a faster rate. This is because a faster compression means a greater displacement, resulting in a higher force and more potential energy stored in the spring. Similarly, if the spring is compressed at a slower rate, the energy of the block will increase at a slower rate.

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