How much work is done to slide a crate for a distance of 25 meters along a loading dock by pulling it with a force of 180 N, where the dock is at an angle of \(45^\circ\) from the horizontal?
Step-by-step Solution:
The work done \(W\) when a force \(F\) is applied to move an object over a distance \(d\) along a direction making an angle \(\theta\) with the horizontal is given by the formula: \[ W = F \cdot d \cdot \cos(\theta) \] Where: - \(F = 180 \, \text{N}\) (the force applied), - \(d = 25 \, \text{m}\) (the distance the crate is moved), - \(\theta = 45^\circ\) (the angle between the force and the horizontal). Now, substitute the given values into the formula: \[ W = 180 \cdot 25 \cdot \cos(45^\circ) \] We know that \(\cos(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.707\). So: \[ W = 180 \cdot 25 \cdot 0.707 \] \[ W = 180 \cdot 17.675 \] \[ W = 3181.98 \, \text{J} \] Thus, the work done is approximately \(3.18198 \times 10^3\) J. Therefore, the correct answer is A: \(3.18198 \times 10^3\) J.