The system shown is initially at rest. Neglecting friction, determine the force P required if the velocity of collar B is 5 m/s after 2 s, and the corresponding tension in the cable. (asked in MU Dec 2025; also June 2025 KT)Figure (described in words…

Answer: \(P=51.93\) N; cable tension \(T=22.22\) N.

Kinematic relation. Let \(x_B\) be the distance of collar B from the wall and \(y_A\) the depth of block A below the fixed pulley. The cable has a run from the wall anchor to the pulley on B (length about \(x_B\)), a run from the pulley on B back to the fixed pulley (about \(x_B\)), and the hanging part \(y_A\):

\[2x_B+y_A=\text{constant}\Rightarrow v_A=2v_B,\quad a_A=2a_B\ (\text{A moves up when B moves right})\]

Accelerations (from rest, uniform acceleration): \(a_B=\dfrac{5-0}{2}=2.5\) m/s², so \(a_A=5\) m/s² upward (velocity of A after 2 s \(=10\) m/s).

Block A (1.5 kg), upward positive:

\[T-1.5(9.81)=1.5(5)\Rightarrow T=14.715+7.5=22.22\ \text{N}\]

Collar B (3 kg), to the right positive. The two cable runs attached to the pulley on B both pull to the left, so the cable force on B is \(2T\):

\[P-2T=3(2.5)\Rightarrow P=2(22.215)+7.5=51.93\ \text{N}\]

(The same result follows from the impulse–momentum principle over 2 s: for A, \((T-14.715)(2)=1.5(10)\); for B, \((P-2T)(2)=3(5)\).)