A warehouse worker is pushing a 90.0 kg crate with a horizontal force at 279 N a

Question

A warehouse worker is pushing a 90.0 kg crate with a horizontal force at 279 N at a speed of v = 0.850 m/s across the warehouse flow. Me encounters a rough horizontal section of the floor that is 0.75 m long and where the coefficient of kinetic friction between the crate and floor is 0.35/. Determine the magnitude and direction of the net force acting on the crate white it is pushed over the rough section of the floor Determine the net work done on the crate while it is pushed over the rough section of the floor. Find the speed of the create when it, reaches the end of the rough surface.

Explanation / Answer

a) Magnitude of frictional force = 0.357*9.81*90 = 315.195N
Magnitude of net force = 315.195-279 = 36.195N in the opposite direction
b)
Net work done = (Net force)*(Total displacement) = 36.195*0.75 = 27.146J
c)0.5*90*vf2 = 0.5*90*0.852 - 27.146 (Net work done = change in kinetic energy of the crate)
Therefore, vf = 0.345 m/s

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