Question

A raindrop of mass 1 g falling from a height of 1 km hits the ground with a speed of $50 \mathrm {~ms} ^ { - 1 }$. If the resistive force is proportional to the speed of the drop, then the work done by the resistive force is :

(Take )

• A. 10 J
• B. -10 J
• C. 8.75 J
• D. -8.75 J

Answer 1

Answer: (D)

-8.75 J

Answer 2

Here, $\mathrm { m } = 1$ $\mathrm { g } = 10 ^ { - 3 }$ kg

H=1km = 1000 m = $10 ^ { 3 } \mathrm {~m}$

The change in kinetic energy of the drop is

$\Delta \mathrm { K } = \frac { 1 } { 2 } \mathrm { mv } ^ { 2 } - 0$

$= \frac { 1 } { 2 } \times 10 ^ { - 3 } \times 50 \times 50 = 1.25 \mathrm {~J}$

The work done by the gravitational force is

$\mathrm { W } _ { \mathrm { g } } = \mathrm { mgh } = 10 ^ { - 3 } \times 10 \times 10 ^ { 3 } 10 \mathrm {~J}$

According to work-energy theorem

$\Delta \mathrm { K } = \mathrm { w } _ { \mathrm { g } } + \mathrm { W } _ { \mathrm { r } }$

Where $W _ { r }$ is the work done by the resistive force on the raindrop. Thus