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Question

A liquid drop of 1.00 g falls from a height of a cliff 1.00 km. It hits the ground with a speed of 50 ms1.what is the work done by the unknown resistive force.

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Solution

Given that,

Mass of liquid m=1.00g=.001kg

Height h=1km=1000m

Speed v=50m/s

We know that,

Work done by gravitational force is loss of potential energy is

P.E=mgh

P.E=0.001×10×1000

P.E=10J

Now, gain in kinetic energy

K.E=12mv2

K.E=12×0.001×50×50

K.E=1.25J

Now,

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

W=101.25

W=8.75J

Hence, the work done by resistive force is 8.75 J


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