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` Take `g` constant with a volume `10 m//s^(2)`. The work done by the
(i) gravitational force and the
(ii) resistive force of air is :

Options

A. (i) -10 J (ii) -8.25 J
B. (i) 1.25 J (ii) -8.25 J
C. (i) 100 J (ii) 8.75 J
D. (i) 10 J (ii) - 8.75 J

Answer

D

Explanation

`w_g = mgh = 10^(-3) xx 10 xx10^3 = 10J`
`w_("net") = DeltaKE = 1/2 xx10^(-3) xx 50 xx 50 = (2.5)/2 J = 1.25 J`
By work-energy theorem,
`w_g + w_("res") = w_("net")`
`10 +w_("res") = 1.25`
`:. W_("res") = -8.75 J`