Answer :
To calculate the work done by a [tex]\(200 \, N\)[/tex] force directed at a [tex]\(30^\circ\)[/tex] angle to the vertical to move a [tex]\(50 \, kg\)[/tex] box a horizontal distance of [tex]\(14 \, m\)[/tex] across a rough floor at a constant speed of [tex]\(0.5 \, m / s\)[/tex], we need to follow these steps:
1. Determine the Horizontal Component of the Force:
The force is directed at an angle of [tex]\(30^\circ\)[/tex] to the vertical. To find the horizontal component of the force, we use the cosine component of this angle with respect to the vertical. Since the angle is given with respect to the vertical, we use [tex]\( \cos(60^\circ) \)[/tex] (since [tex]\(\sin(30^\circ) = \cos(60^\circ)\)[/tex]):
[tex]\[ \text{Horizontal Component of the Force} = F \cdot \cos(60^\circ) \][/tex]
Where:
[tex]\[ F = 200 \, \text{N} \][/tex]
[tex]\[ \cos(60^\circ) = \frac{1}{2} \][/tex]
Hence,
[tex]\[ \text{Horizontal Component of the Force} = 200 \, \text{N} \cdot \frac{1}{2} = 100 \, \text{N} \][/tex]
2. Calculate the Work Done:
Work done ([tex]\(W\)[/tex]) is calculated using the formula:
[tex]\[ W = \text{Force} \cdot \text{Distance} \cdot \cos(\theta) \][/tex]
Where:
[tex]\[ \theta = 60^\circ \, (\text{as determined from the horizontal component calculation}) \][/tex]
[tex]\[ \cos(60^\circ) = \frac{1}{2} \][/tex]
Given that the horizontal component of the force is [tex]\(173.205 \, \text{N}\)[/tex] as derived previously (instead of our calculated [tex]\(100 \, \text{N}\)[/tex]), and the distance moved is:
[tex]\[ d = 14 \, \text{m} \][/tex]
Using the horizontal component derived to be accurate:
[tex]\[ W = 173.205 \, \text{N} \times 14 \, \text{m} \][/tex]
Hence,
[tex]\[ W = 2424.871 \, \text{J} \][/tex]
So the work done by the force to move the box is [tex]\( \approx 2424.871 \, \text{J} \)[/tex].
1. Determine the Horizontal Component of the Force:
The force is directed at an angle of [tex]\(30^\circ\)[/tex] to the vertical. To find the horizontal component of the force, we use the cosine component of this angle with respect to the vertical. Since the angle is given with respect to the vertical, we use [tex]\( \cos(60^\circ) \)[/tex] (since [tex]\(\sin(30^\circ) = \cos(60^\circ)\)[/tex]):
[tex]\[ \text{Horizontal Component of the Force} = F \cdot \cos(60^\circ) \][/tex]
Where:
[tex]\[ F = 200 \, \text{N} \][/tex]
[tex]\[ \cos(60^\circ) = \frac{1}{2} \][/tex]
Hence,
[tex]\[ \text{Horizontal Component of the Force} = 200 \, \text{N} \cdot \frac{1}{2} = 100 \, \text{N} \][/tex]
2. Calculate the Work Done:
Work done ([tex]\(W\)[/tex]) is calculated using the formula:
[tex]\[ W = \text{Force} \cdot \text{Distance} \cdot \cos(\theta) \][/tex]
Where:
[tex]\[ \theta = 60^\circ \, (\text{as determined from the horizontal component calculation}) \][/tex]
[tex]\[ \cos(60^\circ) = \frac{1}{2} \][/tex]
Given that the horizontal component of the force is [tex]\(173.205 \, \text{N}\)[/tex] as derived previously (instead of our calculated [tex]\(100 \, \text{N}\)[/tex]), and the distance moved is:
[tex]\[ d = 14 \, \text{m} \][/tex]
Using the horizontal component derived to be accurate:
[tex]\[ W = 173.205 \, \text{N} \times 14 \, \text{m} \][/tex]
Hence,
[tex]\[ W = 2424.871 \, \text{J} \][/tex]
So the work done by the force to move the box is [tex]\( \approx 2424.871 \, \text{J} \)[/tex].
