A raindrop 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 speed thereafter. 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 ms-1?
Radius of the rain drop r=2mm=2×10−3m Density of water, ρ=103kgm−3 The mass contained in a rain drop, m=ρV ⇒34×3.141×(2×10−3)3×103kg Gravitational force experienced by the rain drop, F=mg ⇒34×3.141×((2×10−3)3×103×9.8)N The work done by the gravity on the drop is given by: W1=Fs =34×3.14×(2×103)3×103×9.8×250=0.082J The work done on the drop in the second half of the journey will be same as that in the first half. ∴W2=0.082 J The total energy of the drop remains conserved during its motion. Total energy at the top: ET=mgh+0 =4/3×3.141×(10−3)3×103×500×10−5 =0.164J Due to the presence of a resistive force, the drop hits the ground with a velocity of 10m/s. ∴Total energy at the ground: EG=21mv2+0=1/2×4/3×3.141×(2×10−3)3×103×9.8×102=1.675×10−3J So, the work done by the resistive force = EG−ET=−0.162J
