# Energy Conservation as an explanation for v=0 for spring mass system

• User1265
In summary, the conversation discusses the behavior of a block of mass m connected to a spring when it is released from its initial position. The displacement graph is crucial in predicting the block's behavior through time, and the gradient is an important factor. The graph shows that the block initially accelerates downwards due to the force of gravity, but as it continues to move, the restoring force from the spring increases and eventually equals the weight of the block, causing the net acceleration to approach zero. This results in the block reaching equilibrium and moving with a constant velocity. However, due to its inertia, the block continues past equilibrium, causing the restoring force to still increase and the velocity to decrease. Eventually, the velocity approaches zero and the curve on the graph
User1265
Homework Statement
I watched a video on a wesbite which explained the graph attached below for the following question:

A block of mass m is connected to the ceiling by a spring with spring constant k. Initially the spring has zero extension and the block is held in position by a support. At time t=0 the support is removed from the block and it falls vertically downwards. Given that the spring has no damping, and ignoring air resistance, draw a graph of the vertical position of the block as a function of time during its initial fall and oscillation.
Relevant Equations
F= -kx

My Solution:

For the displacement graph, the gradient is crucial to predict the behaviour of the displacement of the block through time.

At 1: System is released - velocity is zero, considering forces acting on block, kx < mg, as block is observed to move downwards, and object is accelerating in the negative direction, taking direction of mg to be -ve

At 2: Velocity is increasing so graphs gradient can be seen to steepen in the negative direction.

(Velocity is increasing at a decreasing rate - by virtue of springs extension, restoring force is increasing linearly in the opposite direction, hence net acceleration in direction of motion is decreasing (to zero) - thus there must come a time when resotring force is equal in size to the objects weight.)At 3 : Net acceleration in -ve direction approaches zero, kx = mg , then the block at equilbrium moves with a constant velocity , as seen by green straight line, representing velocity not changing.

At 4: Due to object's inertia , continues past equilbrium, and as block contunues to move below equilibrum, its restoring force is still increasing so now velocity in that direction decreasing as acceleration is now increasing in the opposite (+ve) direction as kx > mg

At 5: in negative diretcion the decreasing velocity is approaching zero, (as kx>>mg) hence curve flattens out

Part Where I am stuck

At 6: The block reaches a point of lowest displacement by virtue of the fact the velocity equals zero, at this point in time, (the extension of spring has reached its maximum with the given weight and so the net acceleration upwards must be at a maximum).

This was my reasoning since velocity is decreasing to zero in negative direction there cannot be any further displacement thereon from this point in time. Hence the block can be said to reach its maximum displacement.

I thought by reasoning this way it would at least be consistent with my considerations of velocity and acceleration in turn affecting velocity in the previous parts.

However the actual solution At 6 gives the reason of conversation of energy as reasoning for why v=0 - stating block moves downwards until a point where all Kinetic energy and gravitational potential energy transferred to elastic PE in the spring, so at this point of maximum displacement the block must have zero velocity.

Q: I don't understand why the argument has shifted to the conservation of energy at the critical point where velocity reaches zero at the maxmium displacement, as opposed to just carrying on the consideratons of velocity decreasing until it is about to change direction. Is it better, more intuitive explanation?

User1265 said:
Homework Statement: I watched a video on a wesbite which explained the graph attached below for the following question:A block of mass m is connected to the ceiling by a spring with spring constant k. Initially the spring has zero extension and the block is held in position by a support. At time t=0 the support is removed from the block and it falls vertically downwards. Given that the spring has no damping, and ignoring air resistance, draw a graph of the vertical position of the block as a function of time during its initial fall and oscillation.
Homework Equations: F= -kx

View attachment 253224
My Solution:

For the displacement graph, the gradient is crucial to predict the behaviour of the displacement of the block through time.

At 1: System is released - velocity is zero, considering forces acting on block, kx < mg, as block is observed to move downwards, and object is accelerating in the negative direction, taking direction of mg to be -ve

At 2: Velocity is increasing so graphs gradient can be seen to steepen in the negative direction.

(Velocity is increasing at a decreasing rate - by virtue of springs extension, restoring force is increasing linearly in the opposite direction, hence net acceleration in direction of motion is decreasing (to zero) - thus there must come a time when resotring force is equal in size to the objects weight.)At 3 : Net acceleration in -ve direction approaches zero, kx = mg , then the block at equilbrium moves with a constant velocity , as seen by green straight line, representing velocity not changing.

At 4: Due to object's inertia , continues past equilbrium, and as block contunues to move below equilibrum, its restoring force is still increasing so now velocity in that direction decreasing as acceleration is now increasing in the opposite (+ve) direction as kx > mg

At 5: in negative diretcion the decreasing velocity is approaching zero, (as kx>>mg) hence curve flattens out

Part Where I am stuck

At 6: The block reaches a point of lowest displacement by virtue of the fact the velocity equals zero, at this point in time, (the extension of spring has reached its maximum with the given weight and so the net acceleration upwards must be at a maximum).

This was my reasoning since velocity is decreasing to zero in negative direction there cannot be any further displacement thereon from this point in time. Hence the block can be said to reach its maximum displacement.

I thought by reasoning this way it would at least be consistent with my considerations of velocity and acceleration in turn affecting velocity in the previous parts.

However the actual solution At 6 gives the reason of conversation of energy as reasoning for why v=0 - stating block moves downwards until a point where all Kinetic energy and gravitational potential energy transferred to elastic PE in the spring, so at this point of maximum displacement the block must have zero velocity.

Q: I don't understand why the argument has shifted to the conservation of energy at the critical point where velocity reaches zero at the maxmium displacement, as opposed to just carrying on the consideratons of velocity decreasing until it is about to change direction. Is it better, more intuitive explanation?

Your explanation was good up to the point where you said you got stuck. Why not continue with this line of reasoning?

In general, you can look at a problem like this through analysis of forces and/or energy conservation. In this case, looking at forces seems to me to be the best way to explain the different phases of the motion.

It depends what you are trying to explain. If asking why it reaches a lowest point you cannot answer that it is because the velocity becomes zero. You would then have to prove that it does so.
The conservation of energy argument doesn’t really work either. It supposes an initial finite total of energy then reasons that if it were to descend far enough then the elastic energy would hold it all, leaving nothing for KE. To see that this flawed, imagine gravity increasing as the mass descends, maybe towards a black hole.

## 1. Why does the spring mass system have a velocity of 0?

According to the principle of energy conservation, the spring mass system has a velocity of 0 because all of the potential energy stored in the spring has been converted into kinetic energy and then dissipated as heat due to friction. This results in a net loss of energy, causing the system to come to a complete stop.

## 2. How does energy conservation explain the velocity of 0 in a spring mass system?

Energy conservation states that the total energy of a closed system remains constant. In the case of a spring mass system, the initial potential energy stored in the spring is converted into kinetic energy as the mass moves. However, due to friction and other dissipative forces, this kinetic energy is eventually converted into heat, resulting in a decrease in the total energy of the system and causing the mass to come to rest at a velocity of 0.

## 3. Can energy conservation be applied to any system with a velocity of 0?

Yes, energy conservation can be applied to any system where the velocity is 0. This is because in order for an object to have a velocity of 0, all of its kinetic energy must have been converted into other forms of energy, such as potential or thermal energy, in accordance with the principle of energy conservation.

## 4. How does the mass of the object affect its velocity in a spring mass system?

The mass of the object does not directly affect its velocity in a spring mass system. However, a heavier mass will require more energy to be stored in the spring in order for it to reach the same velocity as a lighter mass. This is because the greater mass will have a higher inertia, making it more difficult to accelerate.

## 5. Is energy conservation the only explanation for a velocity of 0 in a spring mass system?

No, there could be other factors at play that contribute to the velocity of 0 in a spring mass system. These could include external forces acting on the system, such as air resistance or friction, or any other dissipative forces that may affect the transfer and conservation of energy. However, energy conservation is a fundamental principle that can be applied to explain the behavior of the system in most cases.

Replies
24
Views
2K
Replies
22
Views
796
Replies
5
Views
349
Replies
29
Views
2K
Replies
3
Views
765
Replies
5
Views
2K
Replies
2
Views
1K
Replies
3
Views
774
Replies
2
Views
1K
Replies
1
Views
5K