Consider a drop of rainwater having a mass of $1 ext{ g}$ falling from a height of $1 ext{ km}$. It hits the

Question

Consider a drop of rainwater having a mass of $1	ext{ g}$ falling from a height of $1	ext{ km}$. It hits the ground with a speed of $50	ext{ m/s}$. Take $g$ as constant with a value $10	ext{ m/s}^2$. The work done by the (i) gravitational force and the (ii) resistive force of air is:

Accepted Answer

1. **Calculate Work Done by Gravity ($W_g$):** The work done by the gravitational force is $W_g = mgh$. 
   Given: $m = 1	ext{ g} = 10^{-3}	ext{ kg}$, $h = 1	ext{ km} = 1000	ext{ m}$, $g = 10	ext{ m/s}^2$.
   $$ W_g = 10^{-3} 	imes 10 	imes 1000 = 10	ext{ J} $$ [Class 11 Physics, Ch 6, Sec 5.2, Example 5.2]
2. **Calculate Change in Kinetic Energy ($\Delta K$):** The drop starts from rest ($v_i=0$) and reaches a final speed $v_f=50	ext{ m/s}$.
   $$ \Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - 0 = \frac{1}{2} 	imes 10^{-3} 	imes (50)^2 = \frac{2500}{2000} = 1.25	ext{ J} $$
3. **Apply Work-Energy Theorem:** The change in kinetic energy is equal to the total work done by all forces (gravity + resistive force $W_r$).
   $$ \Delta K = W_g + W_r $$
   $$ 1.25	ext{ J} = 10	ext{ J} + W_r $$
   $$ W_r = 1.25 - 10 = -8.75	ext{ J} $$
   Thus, the work done by gravity is $10	ext{ J}$ and by the resistive force is $-8.75	ext{ J}$.

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