A factory worker pushes a crate of mass 30.2kg a distance of 5.00m along a level

Question

A factory worker pushes a crate of mass 30.2kg a distance of 5.00m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.24.Part A What magnitude of force must the worker apply? Express your answer using two significant figures.Part B How much work is done on the crate by this force? Express your answer using two significant figures.Part C How much work is done on the crate by friction? Express your answer using two significant figures.Part D How much work is done by the normal force?Part E How much work is done by gravity?Part F What is the total work done on the crate?

Explanation / Answer

We use Work done = FOrce x distance Frictional force = f = 0.24*mg= 71.10 N A ) So our worker should apply a force F = 71.10 N (SInce its at constant velocity) B) Work done by F = F*d = 71.10*5 = 355.51 J C) Work done by F = -f*d = -71.10*5 = -355.51 J SINce mg and NOrmal force are perpendicular to displacement, D) W by Normal = 0 E) work by gravitational force = 0 F) TOtal work done = SUm of answers of parts B,C,D,E= 0 J Hope you like my detailed explanation :)

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