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Work, Power and Energy

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Question : 13 of 30
Marks: +1, -0
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speedthereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey?
What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s−110\ \mathrm{m}\ \mathrm{s}^{-1}?
Solution:  
Radius of raindrop, r=2 mm=2×10−3 mr = 2\ \mathrm{mm} = 2 \times 10^{-3}\ \mathrm{m}
Distance covered by drop in each half of the journey
h=5002=250 mh = \frac{500}{2} = 250\ \mathrm{m}
Mass of raindrop = Volume of drop × Density
=43πr3ρ(ρ=103 kg m3=density of water)= \frac{4}{3} \pi r^{3} \rho (\rho = 10^{3}\ \mathrm{kg}\ \mathrm{m}^{3} = \text{density of water})
=3.35×10−5 kg= 3.35 \times 10^{-5}\ \mathrm{kg}
Work done by gravitational force during each half
=mgh=3.35×10−5×9.8×250=0.082 J= mgh = 3.35 \times 10^{-5} \times 9.8 \times 250 = 0.082\ \mathrm{J}
Whether the rain drop falls with decreasing acceleration or with uniform speed, the work done by the gravitational force on the drop remains same.
If there were no resistive force, energy of drop on reaching the ground E1=mghE_1 = mgh =3.35×10−5×9.8×500=0.164 J= 3.35 \times 10^{-5} \times 9.8 \times 500 = 0.164\ \mathrm{J}
Actual energy, E2=12mv2E_{2} = \frac{1}{2} m v^{2} =12×3.35×10−5×(10)2= \frac{1}{2} \times 3.35 \times 10^{-5} \times (10)^{2} =1.675×10−3 J= 1.675 \times 10^{-3}\ \mathrm{J}
Work done by the resistive force
W=E2−E1W = E_{2} - E_{1} =1.675×10−3−0.164= 1.675 \times 10^{-3} - 0.164 =−0.162 J= -0.162\ \mathrm{J}
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