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  • A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy?

A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.

Answers (1)

The net buoyant force in this case = vpg

= vol of air displaced x net density in upward direction x g

= V (p(air) - p(He)) g

 

If we assume a to be acceleration in upward direction,

Ma = V (p(air) - p(He)) g --------- (1)

{p(air) - density of air

P(He) - density of helium}

m dv/dt = V (p(air) - p(He)) g

m dv = V (p(air) - p(He)) g. dt

when we integrate both sides,

mv = V (p(air) - p(He)) g. t

v = V/m (p(air) - p(He)) g.t

kinetic energy of balloon,

\frac{1}{2}mv^{2} = \frac{1}{2} \frac{V^{2}}{m} (p(air) - p(He))^{2} g^{2}t^{2}-------------(2)

Let the balloon rise to height h,

a = V/m (p(air) - p(He)) g

h = ut + \frac{1}{2} at^{2}

h= V/2m (p(air) - p(He)) gt2 ---------------- (3)

on rearranging the terms from the three equations we get,

\frac{1}{2} mv^{2} = {\frac{V}{2m} (p(air) - p(He)) gt^{2}} { V (p(air) - p(He)) g }

\frac{1}{2} mv^{2} = V p(air) gh - V p(He)gh

\frac{1}{2}mv^{2} + p(He)Vgh = p(air)Vgh

KE of balloon + PE of balloon = the change in potential energy in air

So, we can conclude that when the balloon goes up, an equal volume of air is displaced downwards. The PE and KE of the balloon keep increasing as the PE of air is changing.

Posted by

infoexpert24

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