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To calculate the work done, we use the formula:

\[
W = F \cdot d \cdot \cos(\theta)
\]

where:
- \(W\) is the work done,
- \(F\) is the applied force (500 N),
- \(d\) is the displacement (4.00 m),
- \(\theta\) is the angle between the force and the direction of motion (45.0°),
- \(\cos(\theta)\) accounts for the fact that the force is applied at an angle.

Let's calculate it:

\[
W = 500 \, \text{N} \times 4.00 \, \text{m} \times \cos(45.0^\circ)
\]

Since \(\cos(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707\), we can plug in the values:

\[
W = 500 \times 4 \times 0.707
\]

Let me compute that.The work done is approximately \( 1414.2 \, \text{J} \) (Joules).

Would you like more details or have any questions? Here are some related questions:

1. How is the formula for work derived from basic physics principles?
2. What is the difference between work done at an angle and along a straight path?
3. How would the work change if the angle was 90°?
4. What are some real-world examples where work is done at an angle?
5. Can work be negative, and in what cases would that happen?

**Tip:** When calculating work, always ensure the force is in the direction of motion, or use the cosine of the angle between the force and the displacement to adjust.

A force of 500 N is applied at a 45.0° angle and moves a crate 4.00 m. How much work is done?

Learn how to calculate the work done by a 500 N force applied at a 45.0° angle moving a crate over 4.00 m. This problem uses the formula W = F * d * cos(θ) and covers core physics concepts of work and energy. Ideal for Grades 10-12.

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