A 41.0-kg crate, starting from rest, is pulled across level floor with a constant horizontal force of 135 N. For the first 15.0 m the floor is essentially frictionless, whereas for the next 12.0 m the coefficient of kinetic friction is 0.320. (a) Calculate the work done by all the forces acting on the crate, during the entire 27.0 m path. (b) Calculate the total work done by all the forces. (c) Calculate the final speed of the crate after being pulled these 27.0 m.

Explanation:

From the information given;

mass of the crate m = 41 kg

constant horizontal force = 135 N

where;

coefficient of kinetic friction = 0.28

a)

To start with the work done by the applied force

Work done by friction:

Work done  by gravity:

Work done by normal force;

b)

total work by all forces:

W = 2100.5  J

c) By applying the work-energy theorem;

total work done = ΔK.E

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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