The problem is as follows:
The figure from below represents a block which is tied by means of a wire to a brass block over a frictionless jet air board incline. The brass block has a mass equal to $4m$ (assume $m=1\,kg$). While the block has a mass of $m$. It is known that the block descends $5\,m$. Keep in mind that the incline makes an angle with the horizontal of $37^{\circ}$. Using this information find the work done by the gravitational force acting on the body whose mass is $4m$ and also find the work acting on the block whose mass is $m$.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\textrm{-200 J and -30 J}\\ 2.&\textrm{200 J and -30 J}\\ 3.&\textrm{-200 J and 30 J}\\ 4.&\textrm{200 J and 30 J}\\ \end{array}$
Well from what I could spot is that to get the work acting on the block which is descending it would follow this: (For brevity purposes I'm omitting units but they are consistent with their respective dimentions)
$W=F\cdot d = (mg)\cdot h = 4\cdot 1 \cdot 10 \cdot 5 =200 \,J$
In this case the work it is positive due the force acting on the object is in the same direction as the displacement.
But the source of confusion comes with the block which is ascending on the air jet incline.
The thing is, that if the object descends 5,m then in the other part, it will also move 5,m isn't it?. However the force is not acting in the same direction as the slope. The force is acting in the $\textrm{y-direction}$.
Thus the displacement in the $\textrm{y-direction}$ for that block would be:
$d=5\sin 37^{\circ}$
Hence the work acting on the object must be:
$W=-F_{y}\cdot 5 \sin 37^{\circ}=-(40\cdot 5 \cdot \frac{3}{5})= -30 J$
Now the reason for which I'm using a negative here is because the block is moving in a direction contrary to the force applied. I mean the force is acting downwards but the block is ascending. In other words, by returning to the definition of work as $W=F\cos \omega$ in this case $\omega = \pi$.
And if I'm not mistaken, the answer should be the second option.
But is my analysis correct?. I need help here because I'm lost.
The thing is, I thought that I could change the direction of the force using trigonometry so that it acts in the same direction as the slope as follows:
$$W=F\csc \omega \cdot 5 = (40)\cdot \frac{5}{3} \cdot 5$$
But as it can be seen the result isn't exactly nice. The only thing which it does seem to make sense is that I can spot here that the Work is negative because the force is acting in a direction contrary to the displacement.
To better illustrate what I'm proposing the figure from below represents the way how I intended to decompose the force so it is in line with the displacement over the incline.
But as I mentioned lines above. What went wrong in this second analysis?. Is it just that it cannot be done in that way and that the force cannot be decomposed when it is in only one direction?.
Moreover, why I can't just multiply the force and the displacement which is $5\,m$?. I mean, aren't pulleys just machines which do transmit the force by means of tension through the wires passing through them?.
Can someone help me here?. As I mentioned I'm lost in the conceptual and in the mathematical intepretation. Please try to explain the concepts first so I can understand and why the trigonometrical manipulation didn't worked out as I supposed? Can someone help me to address these doubts?.