It is well known that a raindrop or a small pebble falls under the influence of the downward gravitational force and the opposing resistive force. The resistive force is known to be proportional to the speed of the drop. Consider a drop or small pebble of 1 g falling (from rest) from a diff of height 1.00 km. It hits the ground with a speed of 50.0 m $s^{- 1}$ . What is the work done by the unknown resistive force?

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Solution

Given,

$m a s s = 1 g, h e i g h t = 1 k m, s p e e d = 50 m / s$

So the work done by the opposing force is,

the drag force is $k v$ where k is the constant and v is the velocity.

we know that

$k v = m g$

$k = \frac{m g}{v}$

$\frac{\frac{1}{1000} \times 10}{50} = 2 \times 10^{- 4}$

Now, by the work-energy theorem,

Change in KE is equal to the sum of work done by the gravitation and the work done by the drag,

$\frac{1}{2} (10^{- 3}) (50)^{2} = 10 +$ work done by the drag.

Since the gravitational force act as $\frac{1}{1000} \times 10 \times 1000 = 10 J$

$1.2 = 10 +$ work done by the drag

Work was done by the drag $= - 8.75 J$

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