Looking for classes? Ksquare Career Institute, Bengaluru →
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 value 10 m/$s^2$. The work done by the (i) gravitational force and the (ii) resistive force of air is
Explanation
$W_g = mgh = 10$ J; $W_a = \Delta K - W_g$.
Detailed Solution
$W_g + W_a = K_f - K_i$
$mgh + W_a = \frac{1}{2}mv^2 - 0$
$10^{-3}\times10\times10^{3} + W_a = \frac{1}{2}\times10^{-3}\times(50)^2$
$W_a = -8.75$ J, i.e. work done by air resistance is –8.75 J and work done by gravity is 10 J.