Two blocks connected over a pulley

Given: $m_A = 40$ kg, $m_B = 25$ kg, $\mu = 0.20$, $g = 9.81$ m/s². The string is inextensible, so both blocks have the same acceleration $a$. B moves down and A moves toward the pulley.

   +-----+        ___
   |  A  |-------(   )     pulley
 ==+-----+==========\ \====
   friction <--      \ \ T
                      [B] (down, a)

Free-body equations

For A (horizontal): friction $F = \mu m_A g = 0.20 \times 40 \times 9.81 = 78.48$ N.

$$T - 78.48 = 40a$$

For B (vertical): $m_B g = 25 \times 9.81 = 245.25$ N.

$$245.25 - T = 25a$$

(a) Acceleration

Adding the two equations:

$$245.25 - 78.48 = 65a \;\Rightarrow\; a = \frac{166.77}{65} = 2.566\ \text{m/s}^2$$

(b) Tension

$$T = 78.48 + 40(2.566) = 181.1\ \text{N}$$

Check from B: $T = 25(9.81 - 2.566) = 181.1$ N. It agrees.

(c) Speed and distance after 2 s

Starting from rest with constant $a$:

$$v = at = 2.566 \times 2 = 5.13\ \text{m/s} \qquad s = \tfrac12 a t^2 = 5.13\ \text{m}$$

(d) Load on the pulley support

The two straight parts of the string pull on the pulley with force $T$ horizontally (toward A) and $T$ vertically downward (toward B). The pulley is light and smooth, so the load it passes to its support is the resultant of these two pulls:

$$R = \sqrt{T^2 + T^2} = \sqrt{2}\,T = 256.1\ \text{N}$$

This load acts at $45^\circ$ below the horizontal, directed toward the A side (down and back toward the table). The support pushes on the pulley with an equal and opposite force.

Answer: a = 2.57 m/s²; T = 181.1 N; v = 5.13 m/s and s = 5.13 m at 2 s; pulley load = 256.1 N at 45° below the horizontal, toward A.