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The problem asks to calculate the force \( F \) required to accelerate a mass \( m = 2357 \, \text{kg} \) from rest to a velocity \( v = 7.6 \, \text{m/s} \) over a distance of \( s = 13.4 \, \text{m} \), given the formula:

\[
F = \frac{mv^2}{2s}
\]

### Step-by-Step Solution:
1. **Given values:**
   - Mass, \( m = 2357 \, \text{kg} \)
   - Velocity, \( v = 7.6 \, \text{m/s} \)
   - Distance, \( s = 13.4 \, \text{m} \)

2. **Substitute these values into the equation:**

\[
F = \frac{2357 \times (7.6)^2}{2 \times 13.4}
\]

3. **Perform the calculations:**
   - \( v^2 = 7.6^2 = 57.76 \)
   - \( m \times v^2 = 2357 \times 57.76 = 136113.92 \)
   - \( 2 \times s = 2 \times 13.4 = 26.8 \)
   - Now divide:
   \[
   F = \frac{136113.92}{26.8} \approx 5081.78 \, \text{N}
   \]

4. **Round the result to 3 decimal places:**

\[
F \approx 5081.781 \, \text{N}
\]

Thus, the constant force required is approximately **5081.781 N**.

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Do you have any questions or need more details on any step?

### Related Questions:
1. What is the formula for calculating work done when applying force over a distance?
2. How does the distance over which force is applied affect the required force?
3. What are the effects of friction when calculating force in real-life scenarios?
4. Can this formula be applied to non-constant acceleration cases?
5. How would doubling the velocity affect the required force?

### Tip:
Always ensure that units are consistent (e.g., mass in kg, velocity in m/s, distance in meters) when working with formulas in physics.

Calculate the constant force F, required to increase the velocity v, of a 2,357 kg mass, from rest to 7.6 m/s, over a distance of 13.4 m, given that F = (mv^2) / (2s).

Learn how to calculate the constant force required to increase the velocity of a 2,357 kg mass from rest to 7.6 m/s over a distance of 13.4 m using the formula F = (mv^2) / (2s). This problem covers key physics and kinematics concepts, including Newton's Second Law. Suitable for students in Grades 9-12.

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