A balloon weighing W descends with an acceleration of a. When weight w is removed from the balloon, the balloon has an upward acceleration of a. In such a situation, the value of w should be

where g is acceleration due to gravity.

Concept:

To solve this problem, we use Newton's second law of motion. When a balloon descends or ascends, it experiences forces due to gravity and buoyancy. The weight removed from the balloon changes its net acceleration.

Calculation:

Let:

• \( W \) = Weight of the balloon (initial)
• \( w \) = Weight removed
• \( a \) = Acceleration (upward and downward)
• \( g \) = Acceleration due to gravity

1. When the balloon descends with acceleration \( a \):

The net downward force is:

\( W - F_b = \frac{W}{g} \cdot a \)

Rearranging, we get:

\( F_b = W - \frac{W \cdot a}{g} \)    ...(1)

2. When weight \( w \) is removed, the balloon accelerates upwards with acceleration \( a \):

The net upward force is:

\( F_b - (W - w) = \frac{W - w}{g} \cdot a \)

Using equation (1) for \( F_b \):

\( W - \frac{W \cdot a}{g} - (W - w) = \frac{(W - w) \cdot a}{g} \)

Simplifying this:

\( w = \frac{W \cdot a}{g} + \frac{(W - w) \cdot a}{g} \)

3. Solving for \( w \):

Combine terms and solve:

\( w(1 + \frac{a}{g}) = \frac{2W \cdot a}{g} \)

\( w = \frac{2W \cdot a}{a + g} \)

The weight \( w \) that should be removed is:

\( w = \frac{2aW}{a + g} \)