To analyze the work done by Trevor when pulling the wagon both to the corner and back, we can apply the formula for work:
\[ W = F \cdot d \cdot \cos(\theta) \]
where \(W\) is the work done, \(F\) is the magnitude of the force applied, \(d\) is the distance moved in the direction of the force, and \(\theta\) is the angle between the force and the direction of motion.
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Going to the corner:
- Distance \(d = 50 , \text{m}\)
- Force \(F = 50 , \text{N}\) (let's assume the force is consistent)
- Angle \(\theta = 0^\circ\) (force is parallel to the ground)
The work done going to the corner is: \[ W_{\text{to corner}} = F \cdot d \cdot \cos(0) = 50 , \text{N} \cdot 50 , \text{m} \cdot 1 = 2500 , \text{J} \]
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Coming back:
- Distance \(d = 50 , \text{m}\)
- Force \(F = 50 , \text{N}\) (same force)
- Angle \(\theta = 30^\circ\)
The work done coming back is: \[ W_{\text{back}} = F \cdot d \cdot \cos(30^\circ) = 50 , \text{N} \cdot 50 , \text{m} \cdot \cos(30^\circ) = 50 , \text{N} \cdot 50 , \text{m} \cdot \left(\frac{\sqrt{3}}{2}\right) \approx 2500 , \text{J} \cdot 0.866 \approx 2165 , \text{J} \]
Now, comparing the work done in both scenarios:
- Work going to the corner: 2500 J
- Work coming back: approximately 2165 J
Thus, coming back, Trevor does less work because he is pulling at an angle which reduces the effective component of the force in the direction of motion.
The correct choice is: He does less work coming back because the force moving the wagon is only \(50 \cos(30^\circ)\).
3 answers