Question

Question 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?

Chapter Work, Energy And Power Page 136

Answer 1

Hi

s=500/2=250m
because we take half jorney of drop.

so we get
w=.082j

when drop at top it is rest .
so only potential energy is exit
hence
E=mgh
here h is completely distance
i.e
h=2s
energy of drop when it is top
E=2w
E=2*.082=
0.164j

when drop fall from the top it velocity is 10m/s

here is potential energy is zero only kinetic energy is exist
E=1/2mv^2
E=1/2*10^3*4/3*3.14(2x10^-3)^3 x(10)^2
so energy when drop falling
E=1.6x10^-3=.0016j

so
Resistive force = energy at ground - energy at top
=.0016-.164
=-.162j
hope it help u

Answer 2

radius of drop (r) = 2mm = 2× 10^-3 m
we know, density of water = 10³ Kg/m³

mass of rain drop = volume of rain drop × density of water
= 4/3πr³ × 10³ Kg
= 4/3 × 3.14 × (2× 10^-3)³ × 10³ Kg
= 3.35 × 10^-5 Kg

rain drop falls from the height of 500m
distance travelled in 1st half journey = 500/2 = 250 m

wrokdone by Gravitational force on the drop in 1st half journey of it = Gravitational force on the rain drop × distance travelled in 1st half journey
= mg × 250
= 3.35 × 10^-5× 9.8 × 250 Joule
= 820 × 10^-4 Joule
= 0.082 Joule

Gravitational force acting in 2nd half journey will be same becoz distance travelled by rain drop in both journeys are same.

hence, work done by Gravitational force in 2nd half journey = 0.082 Joule

_____________________________________
we know,
free fall body has initial velocity = 0
so, initial velocity of rain drop = 0
final velocity of rain drop = 10 m/s

hence, change in kinetic energy of rain = Final kinetic energy - initial kinetic energy
= 1/2mv² - 1/2mu²
= 1/2× 3.35 × 10^-5 × (10)² - 0
= 1.675 × 10^-3 Joule

now,
Total work done by Gravitational force on rain drop = 0.082 + 0.082 = 0.164 J

According to Work - energy theorem,

Workdone by the resistive force in the entire journey + total work done by Gravitational force on rain drop = change in kinetic energy

work done by the resistive force in the entire journey = 1.675 × 10^-3 -0.164
= -0.162 joule