A liquid drop of $1.00 g$ falls from a height of a cliff $1.00 k m$ . It hits the ground with a speed of $50 m s^{- 1}$ .what is the work done by the unknown resistive force.

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Solution

Given that,

Mass of liquid $m = 1.00 g = .001 k g$

Height $h = 1 k m = 1000 m$

Speed $v = 50 m / s$

We know that,

Work done by gravitational force is loss of potential energy is

$P . E = m g h$

$P . E = 0.001 \times 10 \times 1000$

$P . E = 10 J$

Now, gain in kinetic energy

$K . E = \frac{1}{2} m v^{2}$

$K . E = \frac{1}{2} \times 0.001 \times 50 \times 50$

$K . E = 1.25 J$

Now,

Loss of potential energy = gain in kinetic energy +work done by resistive force

$W = 10 - 1.25$

$W = 8.75 J$

Hence, the work done by resistive force is $8.75 J$

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Q. A raindrop of mass

1 g

falling from a height of

1 k m

hits the ground with a speed of

50 m s^{- 1}

. If the resistive force is proportional to the speed of the drop, then the work done by the resistive force is (Take

g = 10 m s^{- 2})

It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass $1.00$ g falling from a 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?