What is the potential energy of a system with N cubes at equidistant heights?

[RubroCP]

Homework Statement

A system with $N$ small cubes (small enough to be considered point masses) are connected by an ideal cable (zero mass and inextensible), each cube is $a$ distance from the next. Suppose that $N_2$ of the $N$ cubes are arranged in a plane situated on a very high cliff ($Na$ much smaller than the height of the cliff) and that the others are initially suspended and guided by a channel, so that, while a cube is falling, its entire motion is parallel to the gravitational field. In a moment, the system is released and the squares start to fall in a straight line. Consider the gravitational acceleration constant and with modulus $g$. Disregard any dissipative effects. In an instant $t_0$, instantly after $N_3$ blocks are falling ($N_3>N_2$), what is the magnitude of the speed of the blocks? Use energy conservation to answer the question.

Relevant Equations

Those involving energy conservation.

I have tried to apply the conservation methods, but I am not understanding what the statement is asking for.

 

[kuruman]

The statement of the problem is clear. You have a linear chain of $N$ cubes on a table cliff). $N_2$ of these are on the table which means that $N-N_2$ are hanging over the edge. These are constrained by a tube to be vertical, i.e. don't worry about pendulum motion. The blocks are released and allowed to fall. After time $t_0$, the number of falling blocks is $N_3$. Find the speed of the blocks.

As per forum rules you have to make an honest effort and show what you have tried before receiving help.

 

[haruspex]

The condition $N_3>N_2$ is odd. Maybe they mean $N_3>N-N_2$.

 

[Delta2]

What is the total potential energy of $n$ cubes that are equidistant from each other with distance $a$ and the first cube is on height $h$ from the floor and the last cube is at height $(n-1)a+h$. You will have to calculate a sum of n terms, that is $$\sum_{i=1}^{n} a_i$$ where $a_i$ the gravitational potential energy of the i-th cube.
To this sum you will also have to add the potential energy of the cubes that are resting at the table, but this would be kinda easy cause they are all in the same height.

So to solve this problem you may follow this procedure:

• Calculate the potential energy $U_1$ of the system of $N-N_2$ cubes hanging and $N_2$ cubes resting at table
• Calculate the potential energy $U_2$ of the system of $N-N_3$ cubes hanging and $N_3$ at table.
• Apply conservation of energy. The difference between the potential energies $U_1-U_2$ equals the total kinetic energy of the system of cubes (assuming their initial kinetic energy is zero). It is easy to express the total kinetic energy of $N$ cubes since all are moving with the same speed V.
• Solve the above equation for the common speed V of all cubes.