Laws of Motion

$$\begin{gathered} F_1=k{x}_1 \\ 5=k{x}_1 \\ x_1=\frac{5}{k} \\ 7=k{x}_2 \\ x_2=\frac{7}{k} \end{gathered}$$

$$\begin{gathered} x=5x_1-2x_2 \\ x=\frac{25}{k}-\frac{14}{k}\Rightarrow x=\frac{11}{k} \\ F=kx \\ F=k\times \frac{11}{k} \\ F=11N \end{gathered}$$

Let:

• $M$ be the initial mass of the balloon.
• $T$ be the tension (upward force) acting on the balloon.

Using Newton’s second law for the initial state:

$$T - Mg = Ma$$

$$T = M(g + a)$$

Let $m$ be the mass that is dropped. The new mass of the balloon becomes $(M - m)$. The equation of motion after dropping mass:

$$T - (M - m)g = (M - m) \cdot 3a$$

Substituting $T = M(g + a)$:

$$M(g + a) - (M - m)g = (M - m) \cdot 3a$$

Expanding:

$$Mg + Ma - Mg + mg = 3a(M - m)$$

$$Ma + mg = 3aM - 3am$$

Rearrange terms:

$$Ma - 3aM = - mg - 3am$$

$$Ma - 3aM = -m(g + 3a)$$

$$m = \frac{2aM}{g + 3a}$$​

The mass that should be dropped is $\frac{2aM}{g+3a}$.