A 30.0-kg packing crate in a warehouse is pushed to the loading dock by a worker

Question

A 30.0-kg packing crate in a warehouse is pushed to the loading dock by a worker who applies a horizontal force. The coefficient of kinetic friction between the crate and the floor is 0.20. The loading dock is 15.0 m southwest of the initial position of the crate.

A) If the crate is pushed 10.6 m south and then 10.6 m west, what is the total work done on the crate by friction?

B) If the crate is pushed along a straight-line path to the dock, so that it travels 15.0 m southwest, what is the work done on the crate by friction?

I found the right answers online, but I don't understand them. All the right answers that I found did not include the vertical and horizontal components of mg or the force applied and I want to understand why. I thought that since the force being applied is horizontal, then I would have to find its components and same thing with the weight mg. Please give me a detailed explanation.

Explanation / Answer

friction work = µ*m*g*d

so ,

a) work = 0.20 * 30.0kg * 9.8m/s² * (10.6 + 10.6)m = 1250 J

b) work = 0.20 * 30.0kg * 9.8m/s² * 15.0m = 882 J

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