6.13 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 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 m s–1 ?

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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    6.13 The radius of the rain drop = 2 mm = 2 *10-3 m

    The height of drop, s = 500 m

    Density of water,  ?  = 103 kg/ m3

    Mass of the rain drop = volume *density = (4/3) ? r3 *?  = 3.35 *10-5 kg

    The gravitational force on the raindrop, F = mg = 3.28 *10-4 N

    Work done by the gravity on the drop is = mgs where s = 250 m

    Work done = 0.082 J

    The work done during the second half will remain same.

    The total energy of the raindrop will be conserved during the motion.

    Total energy at the top

    E1 = mgh where h = 500 m, E1 = 0.164 J

    Due to resistive force, the energy of the drop on reaching the ground

    E2 = (1/2)mv2 where

    ...more

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Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.

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