The block shown in figure has a mass M and descends with an acceleration a. the mass of the string below the point A is m. Find the tension of the string at the point A and at the lower end.

Explanation

Consider the block +the part of the string below A'' as the system . Let the tension at A be T. The forces acting on this system are
a. (M+mg)g, downward, by the earth
T, upward, by the upper part of the string.
The first is gravitatioN/Al and the second is electromagnetic. We do not have to write the force by the string on the block. This electromagnetic force is by one part of the system on the other part. Only the forces acting on the system by the objects other than the system are to be included.
The system is descending with an acceleration a. Taking the downward direction as the X-axis, the total force along the X-axis is `(M+m)g-T.` Using Newton's law along the X-axis is ` (M+mg)g-T=(M+m)a`.
or `T=(M+m)(g-a)`
we have omitted the free body diagram. This you can do if you can draw the free body diagram in your mind and write the equations correctly.
To get the tension T' at the lower end we can put `m=0 in i.`
Effectively we take the point A at the lower end. Thus, we get `T'=M(g-a)`.