Physics

Physics

Problem

A 10 kg10\,\text{kg} crate is dragged along a horizontal floor for 4 m4\,\text{m} by a constant 50 N50\,\text{N} force that makes an angle θ\theta above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.30.3 and g=10 m s−2g = 10\,\text{m}\,\text{s}^{-2}. For what value of θ\theta is the net work done on the crate zero? a) 38∘38^{\circ} b) 53∘53^{\circ} c) 72∘72^{\circ} d) 18∘18^{\circ}
Intuitive explanation

The key insight is that net work depends on the net force, which changes as you vary the angle θ\theta—specifically, the vertical component of the applied force reduces the normal force and thus friction. You need to find the angle where the work done by the applied force exactly cancels the work done against friction over the 4 m4\,\text{m} distance. Set up the force balance in the vertical direction to find the normal force as a function of θ\theta, then use that to calculate friction, and finally equate the magnitudes of work.

Intuitive explanation

The key insight is that net work depends on the net force, which changes as you vary the angle θ\theta—specifically, the vertical component of the applied force reduces the normal force and thus friction. You need to find the angle where the work done by the applied force exactly cancels the work done against friction over the 4 m4\,\text{m} distance. Set up the force balance in the vertical direction to find the normal force as a function of θ\theta, then use that to calculate friction, and finally equate the magnitudes of work.

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