question_answer 68)

                  A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of \[50\,\,m\,{{s}^{-1}}\].                 Calculate                 (a) the loss of P.E. of the drop.                 (b) the gain in K.E. of the drop.                 (c) Is the gain in K.E. equal to loss of P.E.? It not why.                 Take \[g=10\,m{{s}^{-2}}\]                

Answer:

                  (a) P.E. at \[1\,km\,=mgh={{10}^{-3}}\times 10\times {{10}^{3}}=10\,J\]                 P.E at ground = 0                 \ Loss in P.E. = 10 J ? 0 = 10 J                 (b) K.E. of the drop at 1 km = 0                 K.E. of the drop when it just hit the ground                 \[=\,\frac{1}{2}\,m{{\upsilon }^{2}}=\frac{1}{2}\,\times {{10}^{-3}}\,=\,2500\]                 \[=1.25\,J\]                 Gain in K.E. \[=12.5\,J\]                 (c) Gain in K.E. is greater than loss is P.E. This is because, a part of P.E. is used to overcome the force of friction of air.