To determine the work done by a 20 newton force applied at an angle of \(45.0^{\circ}\) to move a box a horizontal distance of 40 meters, we use the work formula:
[tex]\[ W = F \cdot d \cdot \cos(\theta) \][/tex]
where:
- \( W \) is the work done,
- \( F \) is the force,
- \( d \) is the distance,
- \( \theta \) is the angle between the force and the direction of motion,
- \(\cos(\theta)\) is the cosine function evaluated at the angle \(\theta\).
Given:
- \( F = 20 \) newtons,
- \( \theta = 45.0^{\circ} \),
- \( d = 40 \) meters.
First, convert the angle from degrees to radians because the cosine function in Cartesian coordinates requires it. The conversion is done using the relation:
[tex]\[ \theta_\text{radians} = \theta_\text{degrees} \times \frac{\pi}{180} \][/tex]
For \( \theta = 45.0^{\circ} \):
[tex]\[ \theta_\text{radians} = 45.0^\circ \times \frac{\pi}{180} \approx 0.7853981633974483 \ \text{radians} \][/tex]
Next, calculate the cosine of \( \theta \):
[tex]\[ \cos(45.0^\circ) = \cos(0.7853981633974483) \approx 0.7071 \][/tex]
Now, use the work formula:
[tex]\[ W = 20 \ \text{newtons} \times 40 \ \text{meters} \times 0.7071 \][/tex]
[tex]\[ W \approx 20 \times 40 \times 0.7071 \][/tex]
[tex]\[ W \approx 565.685424949238 \ \text{joules} \][/tex]
Thus, the work done is approximately 565.685424949238 joules. Comparing this value to the choices provided:
A. [tex]$8.0 \times 10^2$[/tex] joules (800 joules)
B. [tex]$9.0 \times 10^2$[/tex] joules (900 joules)
C. [tex]$5.6 \times 10^2$[/tex] joules (560 joules)
D. [tex]$3.6 \times 10^2$[/tex] joules (360 joules)
The correct choice is:
C. [tex]$5.6 \times 10^2$[/tex] joules (560 joules), which is closest to 565.685424949238 joules.
