A horizontal force is applied to a system of two blocks (m1 = 300g, m2 = 500g) resting on a horizontal surface. The coefficient of friction is 0.4. If the blocks accelerate at 200 cm/s2, what is the m

$$m_1 = 300 \, \text{g} = 0.3 \, \text{kg}$$ $$m_2 = 500 \, \text{g} = 0.5 \, \text{kg}$$ $$a = 200 \, \text{cm/s}^2 = 2 \, \text{m/s}^2$$

The coefficient of friction $\mu = 0.4$ and gravitational acceleration $g = 9.8 \, \text{m/s}^2$ are already in appropriate units.

Since the two blocks are in contact and move together, we can treat them as a single system with a total mass $M$.

$$M = m_1 + m_2 = 0.3 \, \text{kg} + 0.5 \, \text{kg} = 0.8 \, \text{kg}$$

On a horizontal table, the normal force $N$ is equal to the total weight of the system.

$$N = M \cdot g = 0.8 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 7.84 \, \text{N}$$

The friction force $F_f$ is given by the product of the coefficient of friction and the normal force.

$$F_f = \mu \cdot N = 0.4 \times 7.84 \, \text{N} = 3.136 \, \text{N}$$

The net force acting on the system is the applied force $F_{applied}$ minus the friction force $F_f$. This net force causes the acceleration $a$.

$$F_{net} = F_{applied} - F_f = M \cdot a$$

Rearrange the equation to solve for $F_{applied}$:

$$F_{applied} = M \cdot a + F_f$$

Substitute the calculated values:

$$F_{applied} = (0.8 \, \text{kg} \times 2 \, \text{m/s}^2) + 3.136 \, \text{N}$$ $$F_{applied} = 1.6 \, \text{N} + 3.136 \, \text{N}$$ $$F_{applied} = 4.736 \, \text{N}$$

Rounding $4.736 \, \text{N}$ to two decimal places gives $4.74 \, \text{N}$.
This matches option C.

The final answer is $\text{C. 4.74 N}$.