# Energy Question - Chain sliding off frictionless table

• dsy5037
In summary, the problem involves a frictionless chain of length 2.00m, with 20.0% of its length hanging over the edge of a table, being released to determine its speed when the entire chain comes off the table. The equations used were PE = KE and mgh = 1/2mv^2, with 80% of the total mass of the string used for PE and 1.6m as the remaining length of the string that still needed to fall. However, the correct approach is to consider the difference in potential energy between the initial state, when 20% of the chain is hanging over the edge, and the final state, when the entire chain is hanging over the edge. This results
dsy5037
Energy Question -- Chain sliding off frictionless table

## Homework Statement

Here is the problem that's confusing me: A frictionless chain of length 2.00m is held with 20.0% of its length hanging over the edge of a table. The chain is then released. Determine its speed the moment the entire chain comes off the table. (Answer = 4.34m/s)

## Homework Equations

I assumed this was an energy related question and that the equations for Kinetic/Potential energy would apply.

## The Attempt at a Solution

I wasn't really sure how to approach this problem but I tried

PE = KE since at first the chain is just hanging and then at the end the entire thing is off the table.

so I tried PE = KE
mgh = 1/2mv^2
(.8*mass)(g)(1.6m) = 1/2(1*mass)(v^2)

I thought you might only need to use 80% of the total mass of the string for the PE since 20% is hanging off. And I thought h would only be 1.6 since that's the remaining length of the string that still needed to fall. But I didn't get the correct answer and I'm probably totally wrong with my thinking but this question confused me :s Any help would be appreciated.

Assume the table is more than 2.00 meters above the ground. It may help to choose a reference height for potential energy, such as the surface of the table. (GPE below the table surface would be negative). What is the initial potential energy when 20% of the chain is hanging over the edge, and what is the potential energy of the chain when all of it is "hanging over the edge"?

Note this problem ignores chain fountain effect (you can find video examples of this by looking for "chain fountain" on youtube).

Last edited:
hi dsy5037! welcome to pf!
dsy5037 said:
I thought you might only need to use 80% of the total mass of the string for the PE since 20% is hanging off. And I thought h would only be 1.6 since that's the remaining length of the string that still needed to fall. But I didn't get the correct answer and I'm probably totally wrong with my thinking but this question confused me :s Any help would be appreciated.

your method is correct, but you've not thought out the PE difference correctly …

not all the 1.6 m of string has fallen 1.6 m, has it?

Isn't it related to density?

By setting the equation:

$$Ma=m(t)g$$

solve for $m(t)$, we have
$$m(t)=\frac{1}{5}\left[e^{\sqrt{\frac{g}{L}}t}+e^{-\sqrt{\frac{g}{L}}t}\right]=\frac{1}{5}\left[e^{\sqrt{5}t}+e^{-\sqrt{5}t}\right]$$ if we solve $m(t)=M[\itex] for t, the result would include density Last edited: Gzyousikai said: Isn't it related to density? By setting the equation: $$Ma=m(t)g$$ solve for [itex]m(t)$, we have
$$m(t)=\frac{1}{5}\left[e^{\sqrt{\frac{g}{L}}t}+e^{-\sqrt{\frac{g}{L}}t}\right]=\frac{1}{5}\left[e^{\sqrt{5}t}+e^{-\sqrt{5}t}\right]$$ if we solve [itex]m(t)=M[\itex] for t, the result would include density
You don't need to use density.

As tim (That's a tiny "tim".) points out, you haven't calculated Δ(P.E.) correctly.

How far does the (center of mass of the) 80% of the chain initially on the table top fall?

How far does the other 20% of the chain fall ?

SammyS said:
As tim (That's a tiny "tim".) …

™ !

rcgldr said:
Note this problem ignores chain fountain effect (you can find video examples of this by looking for "chain fountain" on youtube).
It is not a good problem because it ignores the fact that the chain will (non-uniformly) acquire horizontal speed, which means that when the trailing end leaves the table the airborne chain describes some arc, so its centre of mass has not descended as far as if it had remained vertical and straight.
However, this is not the "chain fountain" effect. As the video shows, the chain travels above the edge of the container without touching it along the way. The explanation comes from considering how the individual links are lifted from rest. If a uniform rod lying on a horizontal surface is snatched up at one end, there is an upward impulse, of half the magnitude, from the floor at the other end. This converts some the energy that would otherwise have gone into rotating the chain links (to be lost later) into upward movement. Note that this is quite different from the OP set-up, where the chain on the table is initially straight and taut.

## 1. What is the definition of energy?

Energy is the ability to do work or cause change. It can exist in many forms such as heat, light, motion, and chemical potential.

## 2. How is energy related to the chain sliding off a frictionless table?

In this scenario, the energy is converted from potential energy (stored energy due to the height of the chain) to kinetic energy (energy of motion) as the chain slides down the table. The absence of friction means that there is no loss of energy due to frictional forces.

## 3. What factors affect the amount of energy in this system?

The height of the table, the mass of the chain, and the acceleration due to gravity are the three main factors that affect the amount of energy in this system. The higher the table, the more potential energy the chain has. The heavier the chain, the more energy it will have as it falls. And the stronger the gravitational pull, the more energy the chain will have as it accelerates downwards.

## 4. Can energy be created or destroyed in this scenario?

No, according to the law of conservation of energy, energy cannot be created or destroyed. It can only be converted from one form to another. In the case of the chain sliding off a frictionless table, the potential energy of the chain is converted into kinetic energy as it slides down.

## 5. What would happen if there was friction on the table?

If there was friction on the table, some of the energy of the chain would be converted into heat due to the frictional forces. This means that the chain would have less kinetic energy as it slides down the table and would not reach the same speed as it would on a frictionless table.

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