A factory worker pushes a 30.7 kg crate a distance of 5.0 m along a level floor

Question

A factory worker pushes a 30.7 kg crate a distance of 5.0 m along a level floor at constant velocity by pushing downward at an angle of 32 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26. a- What magnitude of force must the worker apply to move the crate at constant velocity? b- How much work is done on the crate by this force when the crate is pushed a distance of 5.0 m ? c-How much work is done on the crate by friction during this displacement? Express your answers using two significant figures.

Explanation / Answer

let the force applied be F

friction force = k * (mg + F * sin(32))

since the velocity is constant so net force = 0

so,

F * cos(32) = k * (mg + F * sin(32))

F * cos(32) = 0.26 * (30.7 * 9.8 + F * sin(32))

magnitude of the applied force F = 110.132 N

work done = force * distance

work done = 110.132 * cos(32) * 5

work done = 466.986 J

since the net force is 0 then the net work will also be 0 so,

work done by friction force = -466.986 J

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