Which would require less energy to rub the black board? I have been thinking about a problem that would give teachers and students that rubs blackboard the optimum way to rub the chalk off the blackboard.
The problem is as follows:
"A man is going to rub the chalk off a blackboard, he is going to choose a way to rub off the chalk in two ways 


*

*Starting from the upper left corner of the board and moving horizontally to the right and moving slightly down and then again moving to the right and then when it reaches the left corner and so on 

*The second way is given in the following: He starts at the top and rubs down and then up again and so on.

Assumption: I have assumed that the blackboard is a square! with side length $d$ also there is NO friction between the duster and the board. Only hand motion is considered
He is going to choose the way that is less tiring in his arms. i.e., less work done by his muscles!
Somehow from vague intuitive notion I am inclined to agree that it would be less tiring to rub off the chalk if he moves the duster beginning at the top and rubbing horizontally and gradually decreasing the height when one reach the corners till everything is off. 
I tried to calculate the muscle work by considering the following ideas(please correct me if I'm wrong):


*

*The nerves inside our muscle has to fire its signal continuously throughout the entire interval of the process of rubbing and its has to oppose the gravitational force keeping the hand up the air.

*When the hand is at the highest point on the blackboard, the muscle has to work against gravity $U=mgd$, as the muscle continues to work against gravity throughout the interval from left to right (the horizontal path of the first case) I couldn't find the total work withstood by our muscle.



Designing a method of approach.
In the above diagram, we move by horizontally with a velocity $v$ to the right. But due to gravity the hand falls a distance $dy$ as we move $dx$. 
Therefore the time it takes to fall by a distance $dy$ is given by $dt=\frac{dx}{v}$. 
If $dx=\frac{S}{n}$   as $n\to \infty$, then from the laws of motion
$dy=\frac{1}{2}g\frac{(dx)^2}{v^2}$
Therefore for the whole journey from left to right our muscle has to give a beat up at an amount equal to the gravity force. So, the work supplied by the muscle for one trip would be
$$\int mgdy=\int_{0}^S  mg \frac{1}{2}g\frac{(dx)^2}{v^2}$$
Please help how to resolve this.
Thanks!
 A: I think you are confusing physical work (as measured by physics) with physiological work (energy expended to make your body move or keep muscles extended). See Why does holding something up cost energy while no work is being done? and Does energy expenditure require movement?.
As far as physics is concerned, no  work is being done by the man in the 1st scenario. In fact the man gains energy (work is done on him) as his arm is lowered and its gravitational potential energy is reduced. This work is $mgh$ where $m$ is the mass of the arm and $h$ is the vertical distance between the highest and lowest positions of the centre of mass of the arm. (If the man's arm is assumed to start and end at the same position, the net work done on or by his arm is zero.)
In the 2nd scenario work is done by the man when going up the board and work is done on the man when going down the board. If his arm acted like a spring the energy gained when going down could be used to go back up again. If he starts at the top and finishes at the bottom he gains the same net amount of energy as in the 1st scenario. 
However, muscles do not act like springs : they do not store energy. They use energy going up; they also use a small amount of energy coming down, to slow the arm and prevent it from "free falling" and crashing into something.
As far as biophysics/biochemistry is concerned, it is very difficult to calculate the energy expended in the man's muscles to keep his arms raised at a certain level for some time, or to move them sideways or even down. This is not a problem which physics can solve. Even a biochemist would find it very difficult.
Probably the 1st scenario is the more energy-efficient, but I do not think it is possible to prove it using classical dynamics.
A: Ok so you are quite mistaken about one thing. The nerve signals actually can be ignored and the muscle strain only be considered or we would be double counting . Technically I would suggest that we do it diagonally and for a good reason too. Here are the reasons :
1) It would take less time.
2) In the vertical way you are gaining maximum potential energy repeatedly and therefore subjugating to repeated strain and unstrain which may weaken your bones and muscles and therefore create unnecessary friction. 
2) In the horizontal way which is better than the vertical one, you are certainly decreasing the potential energy , but you may have to stand in awkward positions in order to completely rub the board.
3) Diagonal is good considering you have enough place to move around.
edit: Horizontal is better than diagonal as you have a steady decline of potential and it is almost at par with circular.
A: Qualitatively, and negating friction, I would hazard a guess at horizontal, As there is generally only a reactive force preventing the duster from falling, whereas the vertical method involves considerable work being done, as a force must be exerted to not only resist the downward movement of the duster but also provide the work that lifts it to a state of higher gravitational energy, as per @Sammy Gerbil's comment.
Not sure on peoples fuss on diagonal dusting?? Sure you would be able to get a few longer runs which improves effective movement to ineffective movement (moving to new position to begin new run, whilst not erasing anything). However, there would also be a considerable number of shorter runs that would be horrible inefficient relative to a regular movement.
The efficiency thing was originally thought up for a rectangle board, in which case the longer and less runs would be more efficient. 
A: Assuming that all motions require a uniform amount of energy (ie. gravity and human musculature is irrelevant), it would be whatever technique happens to work out to the least amount of overlap (the eraser erasing part that was already erased) while still erasing the board entirely. So you'd need to know the dimensions of the eraser and chalkboard to figure this out.
