Consider a drop of rain water having mass \(1\, g\) falling from a height of \(1\, km.\) It hits the ground with a speed of \(50\, m s^{-1}\). Take \(g\) constant with a value \(10 \, m s^{-1}\). The work done by the \((i)\) gravitational force and the \((ii)\) resistive force of air is

\(\begin{gathered}
  \,\,\,Here,\,m = 1\,g = {10^{ - 3}}kg,h = 1\,km \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 100\,m, \hfill \\
  v = 50\,m{s^{ - 1}},g = 10m{s^{ - 2}} \hfill \\
  \left( i \right)\,The\,work\,done\,by\,the\,gravitational\, \hfill \\
  force \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = mgh = {10^{ - 3}} \times 10 \times 1000 = 10\,J \hfill \\ 
\end{gathered} \) \(\begin{gathered}
  \left( {ii} \right)The\,total\,work\,done\,by\,gravitational\, \hfill \\
  force\,and\,the\,resistive\,force\,of\,air\,is\, \hfill \\
  equal\,to\,change\,in\,kinetic\,energy\,of\, \hfill \\
  rain\,drop. \hfill \\
  \therefore \,{W_g} + {W_r} = \frac{1}{2}m{v^2} - 0 \hfill \\
  10 + {W_r} = \frac{1}{2} \times {10^{ - 3}} \times 50 \times 50\,or\, \hfill \\
  {W_r} =  - 8.75\,J \hfill \\ 
\end{gathered} \)