Let's break down the given problem step by step.

Given:

• Mass of the cargo box, $m = 90 \, \text{kg}$
• Applied force, $F_{\text{applied}} = 500 \, \text{N}$
• Coefficient of friction, $\mu = 0.52$
• Gravitational acceleration, $g = 9.8 \, \text{m/s}^2$

Step 1: Calculate the normal force

The normal force $F_{\text{normal}}$ is equal to the weight of the box since there is no vertical acceleration. So, we can find it as: $F_{\text{normal}} = m \cdot g = 90 \cdot 9.8 = 882 \, \text{N}$

Step 2: Calculate the force of friction

The force of friction $F_{\text{friction}}$ can be calculated using the coefficient of friction: $F_{\text{friction}} = \mu \cdot F_{\text{normal}} = 0.52 \cdot 882 = 458.64 \, \text{N}$

Step 3: Calculate the net force

The net force $F_{\text{net}}$ acting on the box is the difference between the applied force and the force of friction: $F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 500 - 458.64 = 41.36 \, \text{N}$

Step 4: Calculate the acceleration

Now, using Newton's second law $F = m \cdot a$, we can find the acceleration $a$ as: $a = \frac{F_{\text{net}}}{m} = \frac{41.36}{90} \approx 0.459 \, \text{m/s}^2$

Final Answer:

The acceleration of the cargo box is approximately $0.459 \, \text{m/s}^2$.

Would you like further details or have any questions on the process?

Related Questions:

1. How would the acceleration change if the mass of the box were increased?
2. What happens to the acceleration if the coefficient of friction is reduced to 0.3?
3. How can you calculate the stopping distance of the box given an initial velocity?
4. What if the box is being pulled at an angle? How would that affect the normal force and friction?
5. How would the solution differ if the surface were inclined?

Tip:

Always check whether the force you're applying is sufficient to overcome friction before calculating acceleration!