Q.A baloon filled with helium rises against gravity increasing its potential energy. The speed of the baloon 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.

The key idea is that the balloon is not an isolated system — it exchanges energy with the surrounding air. Mechanical energy of the balloon alone is not conserved because an external buoyant force does work on it.

Reasoning:

1. The balloon experiences an upward buoyant force $F_b = \rho_{\text{air}} V g$, which is greater than its weight $mg$ (since helium is less dense than air). This net upward force does positive work on the balloon.

2. As the balloon rises, its gravitational potential energy increases ($mgh$), and its kinetic energy also increases ($\frac{1}{2}mv^2$). The sum of these two — the balloon's mechanical energy — clearly increases. …

The key confusion here is natural: if the balloon gains both height (potential energy) and speed (kinetic energy), where does that extra energy come from? Mechanical energy conservation only holds when only conservative forces (like gravity) do work. But a rising balloon experiences two forces: gravity downward and buoyancy upward. Buoyancy is not a conservative force in the usual sense — it is a contact force from the surrounding fluid, and it can do net work on the balloon.

Let’s break it down.

1. Identify the forces and the work they do.

   The balloon has weight $mg$ downward and a buoyant force $F_b = \rho_{\text{air}} V g$ upward, where $V$ is the balloon’s volume and $\rho_{\text{air}}$ is constant (given). Since the balloon rises, the buoyant force is larger than the weight. The net upward force is $F_{\text{net}} = (\rho_{\text{air}} V - m)g$, and it does positive work on the balloon as it rises. That work increases the balloon’s total mechanical energy (kinetic + gravitational potential).

2. Where does the buoyant force’s energy come from?

   The buoyant force arises because the surrounding air has a pressure gradient due to gravity. When the balloon rises, it displaces air that falls downward — effectively, the gravitational potential energy of the air–balloon system is being redistributed. The balloon gains energy at the expense of the air’s gravitational potential energy. In a more complete analysis (including the air), total energy is conserved, but the balloon alone does not form a closed system.

3. Apply the work–energy theorem to the balloon.

   The work–energy theorem says: the net work done on an object equals its change in kinetic energy. Here,

Gravity does negative work: $W_{\text{gravity}} = -mg\Delta h$. Buoyancy does positive work: $W_{\text{buoyancy}} = \rho_{\text{air}} V g \Delta h$. So

The change in gravitational potential energy of the balloon is $\Delta U = mg\Delta h$. Adding, the total mechanical energy change of the balloon is

This is positive — mechanical energy of the balloon is not conserved; it increases by the work done by buoyancy.

4. What if we include the air? …

Concept: Mechanical Energy Conservation Requires a Closed (Isolated) System

Step 1: Identify all forces on the balloon

Weight $mg$ (down) and buoyant force $F_b=\rho_{air}Vg$ (up). Since the balloon rises, $F_b>mg$.

Step 2: Recognise buoyancy as an external force that does work

Buoyancy is a contact force exerted by the surrounding air — not an internal conservative force of an isolated balloon–Earth system. It does positive net work on the balloon as it rises through a height $\Delta h$:

Step 3: Apply the work–energy theorem to the balloon alone

Step 4: Add the change in the balloon's gravitational PE

So the balloon's own mechanical energy is NOT conserved — it increases.

Step 5: Extend the system to include the displaced air …