A drop with initial mass M falls under the action of the force of gravity and evaporates uniformly losing a mass m each second.What is the work performed by the force of gravity during the time interval from the start to the complete evaporation of the drop?(air resistances may be neglected).
The answer is g2 (M3/6m2)....
plz solve
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1 Answers
2305Let the drop fall through a height ∂x in time ∂t
m = ∂M∂t
∂M = m∂t
or,M = ∫m∂t =mt.............(1)
∂x∂t= ∂gt2/2∂t=gt
∂x = gt ∂t...........(2)
∂F = g∂M = mg ∂t..............(3)
We know,
W = F*x, since F and x changes with time we have to integrate twice
∂2W = ∂F*∂x = mg ∂t *gt ∂t = mg2t ∂t2
Therefore W = ∫∫mg2t ∂t2
Integrating it once we get,
W = ∫mg2t22 ∂t = mg2t36
In the integrals the limits are from 0 to t.
From (1) we get that,
W = M3g26m2 which is the desired result.
