A factory worker pushes a 28.4 kg crate a distance of 4.9 m along a level floor

Question

A factory worker pushes a 28.4 kg crate a distance of 4.9 m along a level floor at constant velocity by pushing downward at an angle of 32degrees below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.26.

How much work is done on the crate by friction during this displacement?
Express your answer using two significant figures.

How much work is done on the crate by this force when the crate is pushed a distance of 4.9 ?
Express your answer using two significant figures.

Explanation / Answer

f=friction MAN= force by man m=mass a=acceleration N=normal force u=mu S=displacement 1) Sum up all the forces in the x direction and you get MAN-f=ma Since it's going at a constant velocity, the acceleration is 0 so it leaves your equation with: MAN=f (added f to the other side) Remember f=un Now sum up the forces in the Y direction to find your normal force and you get: N-mg=ma -----> N =mg Now plug N into the first equation MAN=umg MAN=.26(28.4kg)(9.8m/s^2) 2)Work=Force*Displacement Work=MAN*S =72.36n*4.9m =354.57

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