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A 15.6 kg block is dragged over a rough, horizontal surface by a constant force of 172 N acting at an angle of 29.5° above the horizontal. The block is displaced 34.4 m and the coefficient of kinetic friction is 0.247. Find the work done by the 172 N force. The acceleration of gravity is 9.8 m/s². Answer in units of J.

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Final answer:

The work done by a 172 N force at a 29.5° angle above the horizontal on a block displaced 34.4 m across a rough surface is calculated using the component of the force in the direction of displacement, multiplying by the displacement and the cosine of the angle.

Step-by-step explanation:

The student's question involves calculating the work done by a force applied at an angle to a block being dragged across a rough horizontal surface. When calculating the work done by the force, we need to consider the force's horizontal component, as only that component does work in the direction of displacement. Thus, the work done by the applied force is given by the formula:

Work = Force × Displacement × cos(θ),

where θ is the angle of the force above the horizontal. For the given 172 N force at an angle of 29.5° acting over a displacement of 34.4 m, the horizontal component of the force is 172 N × cos(29.5°). The work done by this force is:

Work = (172 N × cos(29.5°)) × 34.4 m,

After calculating the cosine value and multiplying it by the other values, the work done by the applied force can be found, which is expressed in joules (J).

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