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Consider a drop of rain water having mass 1g falling from a height of 1 km. It hits the ground with a speed of 50m/s. Take ‘g’ constant with a value 10 m/s². The work by the (i) gravitational force and the (ii) resistive force of air is

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(i)10J; (ii) -8.75J

(i) -10J; (ii) -8.25J

(i) 1.25J; (ii) -8.25J

(i) 100J; (ii) 8.75J

answer is A.

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Detailed Solution

$W_{g} = mgh = 0.001 \times (1 \times 10^{3}) \times 10 = 10 J$

$W_{f} = ∆ KE - W_{g} = \frac{1}{2} {mv}^{2} - mgh = \frac{1}{2} \times 10^{- 3} \times 50 \times 50 - 10 = - 8.75 J$

$W_{f} = - 8.75 J$
work done by friction in negative.

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