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June 2016 p42 q7
3908
A particle A of mass 1.6 kg rests on a horizontal table and is attached to one end of a light inextensible string. The string passes over a small smooth pulley P fixed at the edge of the table. The other end of the string is attached to a particle B of mass 2.4 kg which hangs freely below the pulley. The system is released from rest with the string taut and with B at a height of 0.5 m above the ground, as shown in the diagram. In the subsequent motion A does not reach P before B reaches the ground.
(i) Given that the table is smooth, find the time taken by B to reach the ground.
(ii) Given instead that the table is rough and that the coefficient of friction between A and the table is \(\frac{3}{8}\), find the total distance travelled by A. You may assume that A does not reach the pulley.
Solution
(i) For the smooth table, apply Newton's second law to the system:
\(2.4g - T = 2.4a\)
\(T = 1.6a\)
Combine to solve for \(a\):
\(2.4g = (1.6 + 2.4)a\)
\(a = 6 \text{ m/s}^2\)
Using \(s = \frac{1}{2} a t^2\):
\(0.5 = \frac{1}{2} \times 6 \times t^2\)
\(t = 0.408 \text{ s}\)
(ii) For the rough table, the frictional force \(F = \frac{3}{8} \times 1.6g = 6 \text{ N}\).