To find the work done by a 20 newton force applied at an angle of [tex]\( 45.0^\circ \)[/tex] to move a box a horizontal distance of 40 meters, we use the formula for work done:
[tex]\[ W = F \cdot d \cdot \cos(\theta) \][/tex]
where:
- [tex]\( W \)[/tex] is the work done,
- [tex]\( F \)[/tex] is the force applied (20 newtons in this case),
- [tex]\( d \)[/tex] is the distance moved (40 meters in this case),
- [tex]\( \theta \)[/tex] is the angle between the force and the direction of movement ([tex]\( 45.0^\circ \)[/tex] in this case),
- [tex]\(\cos\)[/tex] is the cosine function.
Let's walk through the steps:
1. Convert the angle from degrees to radians, because trigonometric functions in physical equations typically use radians. The conversion from degrees to radians is given by:
[tex]\[
\theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180}
\][/tex]
For [tex]\( \theta = 45.0^\circ \)[/tex]:
[tex]\[
\theta_{\text{radians}} = 45.0 \times \frac{\pi}{180} = 0.7853981633974483 \text{ radians}
\][/tex]
2. Calculate the cosine of the angle:
[tex]\[
\cos(0.7853981633974483) = 0.7071067811865476
\][/tex]
3. Substitute the values into the work formula:
[tex]\[
W = 20 \, \text{N} \times 40 \, \text{m} \times 0.7071067811865476
\][/tex]
4. Compute the total work done:
[tex]\[
W = 20 \times 40 \times 0.7071067811865476 = 565.685424949238 \text{ joules}
\][/tex]
Rounding this to one decimal place, we get approximately [tex]\( 565.7 \)[/tex] joules. Now, converting this to scientific notation, we have:
[tex]\[
W = 5.7 \times 10^2 \text{ joules}
\][/tex]
Given the available options in the multiple-choice question:
A. [tex]\( 8.0 \times 10^2 \)[/tex] joules
B. [tex]\( 9.0 \times 10^2 \)[/tex] joules
C. [tex]\( 5.6 \times 10^2 \)[/tex] joules
D. [tex]\( 3.6 \times 10^2 \)[/tex] joules
The closest and correct answer to our calculation is [tex]\( 5.6 \times 10^2 \)[/tex] joules.
Therefore, the correct answer is:
C. [tex]\( 5.6 \times 10^2\)[/tex] joules.
