Name: Conservation of Mechanical Energy
Uploaded: 2019-07-19
Duration: 5 min 12 s
Description: When a body falls from a height towards the ground, its potential energy decreases but not the total mechanical energy. Let us understand this using data.

Question

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 cliff 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?

Accepted Answer

Given,
Mass of pebble is, $m = 10^{- 3} kg$
Height of fall is, $h = 1 km = 1000 m$
Hitting velocity of drops, $v = 50 m s^{- 1}$
Frictional force is given by, $f = k v$

From work energy theorem, we know
$W_{ext} + W_{cons .} + W_{non . cons .} = ∆ K$

In the given case, work done by external force is zero, so $W_{ext .} = 0$ , also work done by frictional force is $W_{non cons .} = W_{drag force}$ and there is only conservative force acting on it is due to gravitational force, so $W_{cons .} = - ∆ U$ . Using all these cases in above equation, we have,
$\Rightarrow 0 + (- ∆ U) + W_{Drag force} = ∆ K$
$\Rightarrow W_{Drag force} = ∆ K + ∆ U$
$\Rightarrow W_{Drag force} = (K_{f} + U_{f}) - (K_{i} + U_{i}) \dots (1)$

Given case is shown in below figure,

Considering $(1) and (2)$ as initial and final condition respectively, equation $(1)$ becomes,
$\Rightarrow W_{Drag force} = (K_{2} + U_{2}) - (K_{1} + U_{1})$
$\Rightarrow W_{Drag force} = (\frac{1}{2} m v^{2} + 0) - (0 + m g h)$
Putting all the corresponding values, we have,
$\Rightarrow W_{Drag force} = \frac{1}{2} \times 10^{- 3} \times 50^{2} - (10^{- 3} \times 10 \times 1000)$
$\Rightarrow W_{Drag force} = 1.25 - 10 = - 8.75 J$

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