Work, Power and EnergyMedium

Question

The rain drop of mass 1 g , starts with zero velocity from a height of 1 km . It hits the ground with a speed of $5\text{ }m/s$. The work done by the unknown resistive force is $\_\_\_\_$ J . (take $g = 10\text{ }m/s^{2}$ )

Options

A.-8.75
B.-8.35
C.-9.55
D.-9.98

Solution

Sol. [Key Concept: Work-energy theorem: $mgh + W_{res} = \frac{1}{2}\ {mv}^{2}$, so $W_{res} = \frac{1}{2}\ {mv}^{2} - mgh$.]

$${w_{g} + w_{\text{res}\text{~}} = \Delta k }{mgh + w_{\text{res}\text{~}} = \frac{1}{2}{mv}^{2} }{\therefore w_{res} = \frac{1}{2}{mv}^{2} - mgh }{= 10^{- 3}\left( \frac{25}{2} - 10^{4} \right) }{= - 9.98\text{ }J }$$

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