Raindrop Falling Problem Solution Using Work-Energy Theorem
-Dr.Sanjaykumar Pawar
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A raindrop falling under gravity with air resistance opposing motion, explained using work-energy theorem. |
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Example 5.2: Work Done by Gravity and Resistive Force
📘 Question
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined.
Consider a drop of mass 1.00 g falling from a height of 1.00 km. It hits the ground with a speed of 50.0 m/s.
(a) What is the work done by the gravitational force?
(b) What is the work done by the unknown resistive force?
🧠Given Data
- Mass, m = 1.00 g = 0.001 kg
- Height, h = 1.00 km = 1000 m
- Initial speed, u = 0 m/s
- Final speed, v = 50.0 m/s
- Acceleration due to gravity, g = 9.8 m/s²
✳️ Solution
(a) Work done by gravitational force
Work done by gravity is given by:
Wg = mgh
Wg = 0.001 × 9.8 × 1000 = 9.8 J
✔ Final Answer: Work done by gravity = 9.8 J
(b) Work done by resistive force
Using Work-Energy Theorem:
Net Work = Change in Kinetic Energy
Step 1: Kinetic Energy
KE = ½ mv² = ½ × 0.001 × (50)²
KE = 0.0005 × 2500 = 1.25 J
Step 2: Net work
Wnet = 1.25 J
Step 3: Work balance
Wnet = Wg + Wresistive
1.25 = 9.8 + Wr
Step 4: Solve
Wr = 1.25 − 9.8 = -8.55 J
✔ Final Answer: Work done by resistive force = -8.55 J
📌 Final Answers
- Work done by gravity = 9.8 J
- Work done by resistive force = -8.55 J
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