The system shown in figure is initially at rest. Neglecting friction determine the force ‘P’ required if the velocity of the collar B is 5 m/s after 2 sec and corresponding tension in the cable.[Figure: a horizontal smooth rod is fixed to a vertical…

Given: \(m_B = 3\) kg, \(m_A = 1.5\) kg, system starts from rest, \(v_B = 5\) m/s after \(t = 2\) s, all surfaces and pulleys frictionless.

1. Constraint (kinematic) relation. Let \(x_B\) = distance of collar B from the wall and \(y_A\) = depth of A below the fixed pulley. The cable has two horizontal segments whose length is \(x_B\) each (wall to B's pulley, and B's pulley back to the fixed pulley) and one vertical segment \(y_A\):

\(2x_B + y_A = \text{constant} \Rightarrow 2a_B + a_A = 0 \Rightarrow |a_A| = 2|a_B|\)

When B moves right, A moves up.

2. Acceleration of B (uniform acceleration from rest): \(v = u + at \Rightarrow 5 = 0 + a_B(2) \Rightarrow a_B = 2.5\ \text{m/s}^2\ (\rightarrow)\). Hence \(a_A = 5\ \text{m/s}^2\ (\uparrow)\).

3. Block A (FBD: T up, \(W_A = 1.5 \times 9.81 = 14.715\) N down), \(\sum F = ma\) upward:

\(T - 14.715 = 1.5 \times 5 \Rightarrow T = 22.215\ \text{N}\)

4. Collar B (horizontal forces: P to the right; the pulley on B carries two cable segments, each pulling B to the left with T):

\(P - 2T = m_B a_B \Rightarrow P - 2(22.215) = 3 \times 2.5 \Rightarrow P = 44.43 + 7.5 = 51.93\ \text{N}\)

Answer: P = 51.93 N (to the right); tension in the cable T = 22.22 N.