A bar is supported by three identical springs A, B, and C. Springs A and B are identical to each other and support the bar symmetrically. A downward force of 28.0 N is applied to the bar. If a force o

The problem states that $6 \text{ N}$ causes an extension of $2 \text{ cm}$.
First, convert the extension to meters: $2 \text{ cm} = 0.02 \text{ m}$.
Using Hooke's Law, $F = ke$, where $k$ is the spring constant and $e$ is the extension.

$$k = \frac{F}{e} = \frac{6 \text{ N}}{0.02 \text{ m}} = 300 \text{ N/m}$$

For spring C:

$$e_C = \frac{F_C}{k} = \frac{25.5 \text{ N}}{300 \text{ N/m}} = 0.085 \text{ m}$$

Convert to centimeters: $0.085 \text{ m} \times 100 \text{ cm/m} = 8.5 \text{ cm}$.

For spring A:

$$e_A = \frac{F_A}{k} = \frac{14.0 \text{ N}}{300 \text{ N/m}} = 0.04666... \text{ m}$$

Convert to centimeters: $0.04666... \text{ m} \times 100 \text{ cm/m} \approx 4.67 \text{ cm}$.

For spring B:

$$e_B = \frac{F_B}{k} = \frac{14.0 \text{ N}}{300 \text{ N/m}} = 0.04666... \text{ m}$$

Convert to centimeters: $0.04666... \text{ m} \times 100 \text{ cm/m} \approx 4.67 \text{ cm}$.

The extension on each spring is:
Spring A: $\boxed{\text{4.67 cm}}$
Spring B: $\boxed{\text{4.67 cm}}$
Spring C: $\boxed{\text{8.5 cm}}$