A balloon filled with helium rises against gravity increasing its pote

1.	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 is constant.
Answer» As dragging viscous force of air on balloon is neglected so there is Net Buoyant Force = Vρg Let m, V, ρ_HE, denote respectively the mass, volume and density of helium balloon and ρ_air density of air Volume V of balloon displaces volume V of air. So, V(ρ_air − ρ_HE)g = ma, or, V(ρ_air − ρ_HE)g = \(m\frac{dv}{dt}........(1)\) Integrating equation (1) with respect to t, we have V(ρ_air − ρ_HE)gt = mv \(\frac{1}{2}mv^2=\frac{1}{2}m\frac{v^2}{m^2}\)(ρ_air − ρ_HE)²g²t² \(=\frac{1}{2m}V^2\)(ρ_air − ρ_HE)²g²t² ......(2) If the balloon rises to a height h from h = ut + \(\frac{1}{2}\)at² We get, \(h=\frac{1}{2}at^2=\frac{1}{2}\frac{V(ρair − ρHE)}{m}gt^2\) (∵ u = 0).........(3) From Eqs. (3) and (2) \(\frac{1}{2}mv^2=[V(ρ_{air} − ρ_{HE})g][\frac{1}{2m}V(ρ_{air} − ρ_{HE})gt^2]\) = V(ρ_air − ρ_HE)gh Rearranging the terms. ⇒ \(\frac{1}{2}mv^2+Vρ_{HE}gh=Vρ_{air}gh\) Or KE_balloon + PE_balloon = changing in PE of air. So, as the balloon goes up, and equal volume of air comes down, increase in PE and KE of the balloon is at cost of PE of air [which comes down].

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 is constant.

Answer»

As dragging viscous force of air on balloon is neglected so there is Net Buoyant Force = Vρg

Let m, V, ρ_HE, denote respectively the mass, volume and density of helium balloon and ρ_air density of air

Volume V of balloon displaces volume V of air.

So,

V(ρ_air − ρ_HE)g = ma, or,

V(ρ_air − ρ_HE)g = \(m\frac{dv}{dt}........(1)\)

Integrating equation (1) with respect to t, we have

V(ρ_air − ρ_HE)gt = mv

\(\frac{1}{2}mv^2=\frac{1}{2}m\frac{v^2}{m^2}\)(ρ_air − ρ_HE)²g²t²

\(=\frac{1}{2m}V^2\)(ρ_air − ρ_HE)²g²t² ......(2)

If the balloon rises to a height h from h = ut + \(\frac{1}{2}\)at²

We get,

\(h=\frac{1}{2}at^2=\frac{1}{2}\frac{V(ρair − ρHE)}{m}gt^2\)

(∵ u = 0).........(3)

From Eqs. (3) and (2)

\(\frac{1}{2}mv^2=[V(ρ_{air} − ρ_{HE})g][\frac{1}{2m}V(ρ_{air} − ρ_{HE})gt^2]\)

= V(ρ_air − ρ_HE)gh

Rearranging the terms.

⇒ \(\frac{1}{2}mv^2+Vρ_{HE}gh=Vρ_{air}gh\)

Or KE_balloon + PE_balloon = changing in PE of air.

So, as the balloon goes up, and equal volume of air comes down, increase in PE and KE of the balloon is at cost of PE of air [which comes down].