Energy analysis of the system (leaking bucket from Morin's "Mechanics"

[Michael Korobov]

Homework Statement

Problem 5.17 from David Morin's "Introduction to classical mechanics"
At t = 0, a massless bucket contains a mass M of sand. It is connected to a wall by a massless spring with constant tension T (that is, independent of length). The ground is frictionless, and the initial distance to the wall is L. At later times, let x be the distance from the wall, and let m be the mass of sand in the bucket. The bucket is released, and on its way to the wall, it leaks sand at a rate dm/dx = M/L. In other words, the rate is constant with respect to distance, not time; and it ends up empty right when it reaches the wall. Note that dx is negative, so dm is also.
Q.
(a) What is the kinetic energy of the (sand in the) bucket, as a function of x? What is its maximum value?

Relevant Equations

Conservation of momentum

Hi,
Can you please help me understand how the formula of energy decreasing during a sand leaking is obtained?
One of possible solution to this problem, suggested in the textbook, states that when the bucket moves from x to x+dx (d is negative), there are two components responsible of energy change: one is due to the work done by the spring (-T)dx, and another which is proportional to dx/x, i.e. the energy change is E dx/x. Thus
$$dE=-Tdx+E\frac{dx}{x}$$.
How this component E dx/x is obtained?
Thanks a lot!

 

[haruspex]

How much sand is in the bucket when x from the wall?
What fraction is lost in the next dx?
What fraction of its KE is that?

 

[kuruman]

Without wishing to detract from the textbook's approach, I think it is more straightforward to write down Newton's second law as $T=m(\xi)a$ where $m(\xi)$ is the mass of the sand when the bucket is at distance $\xi$ from the wall and then note that $a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt}=v\dfrac{dv}{dx}=-v\dfrac{dv}{d\xi}.$ The resulting equation is separable.

 

[Tomy World]

To solve this problem, these are the key equations to be resolved.
$$\frac {dm} {dx} = \frac {M} {L} $$.... get m(x)
$$T=ma ⇒ \frac {dT} {dx} = 0 = m \frac {da} {dx} + a \frac {dm} {dx} $$.... get a(x)
$$a = \frac {dv} {dt} = \frac {dv} {dx} \frac {dx} {dt} = v \frac {dv} {dx} ⇒ a dx = v dv $$.... get v(x)

Finally,
$$\text{Kinetic Energy} = E =\frac {mv^2} {2} $$.... get E(x)

If you got the equation correctly, the graph should look similar to the attached image (plot using https://www.transum.org/Maths/Activity/Graph/Desmos.asp)

 

[Tomy World]

Attached image shows how to get
$$dE = -Tdx + E\frac{dx} {x}$$
This is interesting. But I'm not sure how this approach helps in solving the problem ;-)

 

[haruspex]

Attached image shows how to get
$$dE = -Tdx + E\frac{dx} {x}$$
This is interesting. But I'm not sure how this approach helps in solving the problem ;-)

Dividing through by x makes it directly integrable.