Solving this problem using the energy method

[Abhishek11235]

Homework Statement

This is the exercise from Morin's book on classical mechanics. The problem is as follows:
Consider a leaky bucket pulled by a string towards wall(see screenshot). The bucket leaks all the way which is constant (i.e $dm/dx = \lambda$ where m is the instantaneous mass of bucket. He asks to calculate kinetic energy as the function of distance

Relevant Equations

$dE/dx= F$

I attempted the solution using force method. I got correct. However, I was stuck at the alternative way to solve problem using energy method. As shown in screenshot 2, he tells that the energy of system changes due to 2 ways:
- The tension T
- Leaking of mass

As shown in screenshot 2 ,the leaking of mass contributes dx/x. My question is how this expression comes up? Please detailed justification and Thanks in advance

 

[jbriggs444]

As the mass moves an incremental distance x, it loses a fraction of its mass. If it is at distance 10 and moves incrementally by dx, it will lose dx/10 of its mass. If it is at distance x, it will lose a fraction, dx/x, of its mass.

If it loses dx/x of its mass, it loses dx/x of its energy.

Just as Morin says, the contribution to dE from this is E dx/x.

 

[haruspex]

My question is how this expression comes up

It is confusing because of the way x is defined.
Try obtaining the equation for m as a function of x. The leak contributes $\frac 12 v^2dm$ to dE. Substitute for $v^2$ using E.

 

[Abhishek11235]

As the mass moves an incremental distance x, it loses a fraction of its mass. If it is at distance 10 and moves incrementally by dx, it will lose dx/10 of its mass. If it is at distance x, it will lose a fraction, dx/x, of its mass.

If it loses dx/x of its mass, it loses dx/x of its energy.

Just as Morin says, the contribution to dE from this is E dx/x.

How can it lose dx/x fraction? The rate of leaking is constant. So $dE= 1/2 dm v^2 = 1/2 \lambda dx v^2 = E dx/L$ since $\lambda= M/L$. I want justification for dx/x.

 

[Abhishek11235]

It is confusing because of the way x is defined.
Try obtaining the equation for m as a function of x. The leak contributes $\frac 12 v^2dm$ to dE. Substitute for $v^2$ using E.

I got E dx/L as I showed above

 

[TSny]

The rate of leaking is constant. So $dE= 1/2 dm v^2 = 1/2 \lambda dx v^2 = E dx/L$.

The last equality is not correct. It looks like you assumed that $E = \frac{1}{2}Mv^2$. But the mass of sand in the bucket is not $M$ (except at the initial point of release).

 

[Abhishek11235]

The last equality is not correct. It looks like you assumed that $E = \frac{1}{2}Mv^2$. But the mass of sand in the bucket is not $M$ (except at the initial point of release).

Ok. So, I tried doing the dE. However,to get correct answer I have to assume dm/dx= M/x. Otherwise I get L everytime. Can you elaborate your answer please?

 

[haruspex]

Ok. So, I tried doing the dE. However,to get correct answer I have to assume dm/dx= M/x. Otherwise I get L everytime. Can you elaborate your answer please?

As @TSny writes, you substituted E/M for ½v², but E depends on the current mass, m, not the initial mass, M