Q#9 (Past Exam Paper – November 2017 Paper 11 Q8)
The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the upthrust force.
The hot-air balloon descends vertically at constant speed. The force of air resistance on the balloon is F.
Which weight of material must be released from the balloon so that it ascends vertically at the same constant speed?
A F B 2F C 3F D 4F
Solution:
Answer: B.
This question provides good practice for the exams. It is both mathematical and requires proper understanding of the theories relevant (such as resultant force, upthrust, air resistance and weight).
The weight W is always downwards. It has a constant value.
The upthrust force U acts upwards. This always has the same value as it depends on the difference in height between the top and bottom surface of the air-balloon.
The direction of the force due to air resistance depends on the motion. Air resistance always opposite motion. If the air-balloon is ascending, the air resistance is downwards and if the air-balloon is descending, the air resistance is upwards.
Also, air resistance depends on speed. The greater the speed, the greater the air resistance.
When descending at constant speed, air resistance = F (upwards).
A constant speed means the acceleration = 0. In other words, the resultant force = 0 [since F = ma]. That is, the sum of upward forces = sum of downward forces.
W = F + U (1)
When ascending,
If speed is the same constant speed, air resistance = F (downwards since it is ascending). Some weight of the material needs to be released for the resultant force to be zero. Let the weight to be released be R.
(W–R) + F = U (2)
From (1),
Upthrust U = W – F (3)
The upthrust force has the same value throughout the motion.
Replace U from (3) in (1).
W – R + F = W – F
– R = – F – F
R = 2F
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