Question 178: [Kinematics > Linear motion] (Past Exam Paper – June 2012 Paper 11 Q12)
Diagram shows barrel suspended from frictionless pulley on building. Rope supporting the barrel goes over pulley and is secured to a stake at bottom of the building.
A man stands close to stake. Bottom of the barrel is 18 m above man’s head. The mass of barrel is 120 kg and mass of the man is 80 kg.
Man keeps hold of rope after untying it from the stake and is lifted upwards as barrel falls.
What is man’s upward speed when his head is level with the bottom of barrel? (Use g = 10 m s–2.)
A 6 m s–1 B 8 m s–1 C 13 m s–1 D 19 m s–1
Solution 178:
Answer: A.
There is a downward force on each sides of the rope due to the weights of the barrel and the man. First, the resultant force in the system should be calculated.
(Downward) force (due to weight) of barrel = mg = 120 x 10 = 1200N
(Downward) force (due to weight) of man = mg = 80 x 10 = 800N
Net (resultant) force in the system = 1200 – 800 = 400N
This force acts downwards at the side of the rope where the barrel is located. So, the barrel moves downwards while the man moves upwards.
The force causes a resultant acceleration in the system. Since it is the resultant force in the SYSTEM, the masses of both the man and the barrel should be taken into account.
Resultant force = ma = 400
(120 + 80) a = 400
Acceleration, a = 400 / 200 = 2ms-2
The same acceleration acts on both the man and the barrel.
For the man’s head to be level with the bottom of the barrel, both of them should travel a distance of (18 / 2 =) 9m.
Consider the equation of uniformly accelerated motion:
v2 = u2 + 2as
Initial velocity, u = 0
Acceleration, a = 2ms-2
Distance travelled, s = 9m
v2 = 02 + 2(2)(9)
Final speed = √(2x2x9) = 6ms-1
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