Final answer:
The work done by the 182 N force, acting at a 24-degree angle above the horizontal, on a 15.4 kg block dragged over a rough surface for a 57.6 m displacement is calculated to be approximately 8,510 J. This accounts for the force component parallel to the direction of displacement.
Step-by-step explanation:
The work done by the constant force of 182 N, acting at an angle of 24 degrees above the horizontal, on the 15.4 kg block dragged over a rough surface is found by considering the component of the force parallel to the direction of displacement. The work is calculated using the formula \(W = F \cdot d \cdot \cos(\theta)\), where \(F\) is the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and displacement vectors. Substituting the known values, the work done is the product of the force component in the direction of displacement and the displacement distance, resulting in the total work done by the force during the 57.6 m displacement of the block on the rough surface.