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A 16.1 kg block is dragged over a rough, horizontal surface by a constant force of 73.1 N acting at an angle of 31.4 degrees above the horizontal. The block is displaced 73.3 m, and the coefficient of kinetic friction is unknown. Find the work done by the 73.1 N force. The acceleration of gravity is 9.8 m/s². Express your answer in units of joules (J).

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Final answer:

The work done by the 73.1 N force on the 16.1 kg block moving over a rough horizontal surface with a displacement of 73.3 m is approximately 4581.4 joules (J).

Step-by-step explanation:

The student is asking how to calculate the work done by a constant force that is applied at an angle to a block moving horizontally over a rough surface. To calculate work done by a force, we use the formula Work (W) = Force (F) × Displacement (d) × cos(θ), where θ is the angle the force makes with the direction of displacement. In this case, the force is 73.1 N, the displacement is 73.3 m, and the angle is 31.4 degrees above the horizontal. Therefore, the work done by the force is calculated as follows:

W = 73.1 N × 73.3 m × cos(31.4°)

First, we calculate the cosine of the angle:

cos(31.4°) ≈ 0.8572

Next, we multiply this value by the force and the displacement to find the work done:

W = 73.1 N × 73.3 m × 0.8572 ≈ 4581.4 J

The work done by the 73.1 N force is approximately 4581.4 joules (J).

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