A 30.0-kg crate is being dragged along a rough horizontal surface. The coefficient of kinetic friction between the crate and the surface is 0.250. What horizontal force must be exerted on the crate for it to accelerate at 2.50 m/s2?

To solve this problem, we need to find the horizontal force required to make a 30.0-kg crate accelerate at 2.50 m/s² on a rough surface with a coefficient of kinetic friction of 0.250.

### Step 1: Calculate the force due to friction

The force of kinetic friction (\(F_f\)) can be calculated using the formula:

\[
F_f = \mu_k \cdot F_n
\]

Where:
- \(\mu_k = 0.250\) (coefficient of kinetic friction)
- \(F_n\) is the normal force, which, for a horizontal surface, is equal to the weight of the crate. The weight (\(F_g\)) is:

\[
F_g = m \cdot g
\]

Where:
- \(m = 30.0 \, \text{kg}\) (mass of the crate)
- \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity)

So:

\[
F_g = 30.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 294.3 \, \text{N}
\]

Thus, the normal force \(F_n\) is 294.3 N. Now, the frictional force \(F_f\) is:

\[
F_f = 0.250 \times 294.3 \, \text{N} = 73.575 \, \text{N}
\]

### Step 2: Use Newton's second law

To accelerate the crate, the net force \(F_{\text{net}}\) must overcome both the frictional force and provide the necessary force to accelerate the crate. Newton's second law is:

\[
F_{\text{net}} = m \cdot a
\]

Where:
- \(a = 2.50 \, \text{m/s}^2\) (given acceleration)

So, the net force is:

\[
F_{\text{net}} = 30.0 \, \text{kg} \times 2.50 \, \text{m/s}^2 = 75.0 \, \text{N}
\]

### Step 3: Calculate the total horizontal force

The total horizontal force \(F_{\text{total}}\) that must be exerted is the sum of the force required to overcome friction and the force required to accelerate the crate:

\[
F_{\text{total}} = F_f + F_{\text{net}}
\]

Substituting the values:

\[
F_{\text{total}} = 73.575 \, \text{N} + 75.0 \, \text{N} = 148.575 \, \text{N}
\]

### Final Answer:

The horizontal force that must be exerted on the crate is approximately **148.6 N**.

---

Would you like any further details or have any additional questions?

Here are some related questions to explore:

1. How would the required force change if the coefficient of kinetic friction increased?
2. What is the role of normal force in determining frictional force in this scenario?
3. How would the situation differ if the crate was on an inclined surface?
4. What happens to the frictional force if the crate’s mass is doubled?
5. How does the acceleration change if the applied force is less than the required horizontal force?

**Tip:** When working with friction, remember that the normal force is usually equal to the weight of the object if the surface is horizontal and flat.

Learn how to calculate the horizontal force required to move a 30.0-kg crate on a rough surface with a kinetic friction coefficient of 0.250, including step-by-step solutions using Newton's Second Law and friction formulas. Suitable for students in Grades 9-12.

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