Question

a drop of rainwater having mass 1g falling from a height of 1km it hits the ground with a speed of 50m/s take g constant with a value 10m/s^2 the work done by the resistive force of air

Answer 1

Answer:

10J, -8.75J

Explanation:

As we know work energy equation,

Wg-Wa=△KE

Wg= work done due to gravity and

Wa= Workdone by resistive force of air

△KE= Change in kinetic energy, m=10^{-3}kg,h= 10^{3}m

Here v2=50,v1=0

So mgh+Wa=mv2^{2}/2-mv1^{2}/2

10^{-3}×10×10^{3}+Wa=10^{-3}×50^{2}/2-0

10+Wa=2.5/2 10=Wa=1.25 Wa=1.25-10 Wa=-8.75J Resistive force of airNow ,Gravitational force workdone =mgh= 10^{-3}×10×10^{3}=10J `

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