A particle A of mass 5kg rests on a smooth horizontal table. particle A is attached to one end of a light inextensible string which passes over a smooth pulley fixed to the edge of the table. The other end of the string is attached to particle B of mass 4kg which hangs freely below the pulley 1.4m above the ground. the system is released from rest with a string taut. particle A does not reach the pulley before B reaches the ground. [a] find the tension in the string before B hits the ground. [b] find the time taken for B to reach the ground.
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The tension of the string is 21.8N. Time taken for B to reach ground is 0.8sec
We need to write the force equations for the two particles A and B.
- For particle A,
- T - ma = 0, since Tension from the string T is the only force acting on it.
- T = 5a ---(1)
- for particle B,
- mg - T = ma, since the particle is moving down due to its weight,but tension due to the string is also acting on it.
- 4x9.8 - T = 4a---(2)
- Since , the string is taut, both particle moves with same acceleration.
- Hence, 4x9.8 - 5a = 4a;
a = 4x9.8/9 = 4.36m/s^2
(a) Hence the tension in the string T = 5a = 21.8
(b) time taken , we know by the equations of motion that,
- s = ut + 0.5xax(t^2)
Since it falls from rest, initial velocity, u =0.
- s = 1.4m, distance to reach the ground.
Therefore, t = square root(2 x s/a)
- t = square root( 2 x 1.4/4.36)
- t = square root(0.6422) = 0.8sec.
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