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A factory worker pushes a 35.0-kg crate a distance of 5.0 m along a level floor at constant speed by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done my the normal force? By gravity?

Ok so because the crate is being pushed at constant speed, [tex] \Sigma F_{x} = 0 [/tex] and [tex] \Sigma F_{y} = 0 [/tex]. For part (a) wouldnt the force the worker has to apply be equal to [tex] f_{k} = \mu_{k} n = (0.25)(343 N) = 85.75 N [/tex]?

(b) Since [tex] W = f\bullet s [/tex], [tex] W = 85.75\times\cos 0\times(5.0 m) = 428.75 N [/tex]. In part (c) would this just be the same answer, but opposite sign?

(d) The work would be 0, because the forces are perpendicular to the path of the box?

Thanks

Ok so because the crate is being pushed at constant speed, [tex] \Sigma F_{x} = 0 [/tex] and [tex] \Sigma F_{y} = 0 [/tex]. For part (a) wouldnt the force the worker has to apply be equal to [tex] f_{k} = \mu_{k} n = (0.25)(343 N) = 85.75 N [/tex]?

(b) Since [tex] W = f\bullet s [/tex], [tex] W = 85.75\times\cos 0\times(5.0 m) = 428.75 N [/tex]. In part (c) would this just be the same answer, but opposite sign?

(d) The work would be 0, because the forces are perpendicular to the path of the box?

Thanks

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