Connected Particles with a Slack String
Block A of mass 3 kg rests on a rough horizontal table and is connected by a light inextensible string over a smooth pulley at the table edge to block B of mass 2 kg, which hangs vertically. The coefficient of friction between A and the table is 0.25. Block B is initially 1.2 m above the floor, and A is initially 3 m from the pulley. The system is released from rest. The table is long enough that A does not reach the pulley before B lands. Use g = 9.8 m/s^2. A student calculates one acceleration for the whole motion and assumes the string remains taut after B hits the floor. The teacher asks them to separate the motion into two stages and to decide whether A reaches the pulley.
1. Find the acceleration of the system and the tension in the string while B is descending.
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Let a be the acceleration of A towards the pulley and of B downwards; the string is taut. For B: 2g-T=2a. For A: T-F=3a, where F=μR=0.25(3g)=0.25(29.4)=7.35 N.
Adding: 2g-7.35=(2+3)a, so 19.6-7.35=5a, giving a=12.25/5=2.45 m/s^2.
Then T=2g-2a=19.6-4.9=14.7 N. The same tension follows from T=3a+7.35=7.35+7.35=14.7 N.
2. Find the time taken for B to reach the floor and its speed just before impact.
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B falls 1.2 m from rest with a=2.45 m/s^2. Using s=ut+(1/2)at^2: 1.2=0.5(2.45)t^2, so t^2=2.4/2.45=0.9796 and t=0.990 s.
Speed just before impact: v=at=2.45(0.990)=2.43 m/s. This is also obtained from v^2=2as=2(2.45)(1.2).
3. After B lands, the string goes slack. Find the further distance A moves before stopping and determine whether A reaches the pulley.
Read the worked answer
After B lands, the string is slack, so T=0. A moves on the rough table under friction only. Its acceleration is a=-F/m=-7.35/3=-2.45 m/s^2, i.e. 2.45 m/s^2 opposing motion.
Its speed at the start of this stage is 2.43 m/s. Stopping distance: s=v^2/(2*2.45)= (2.43)^2/4.9 =1.20 m.
Before B landed, A had moved 1.2 m. Total distance moved by A is 2.4 m. Since A started 3 m from the pulley, it stops 0.6 m short and does not reach the pulley. The common error is to keep the string tension acting after B lands; it does not, because the string is no longer taut.