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Thank you for making answers clear so I can rate accordingly, as soon as I can.

A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 285 N at a speed of v = 0.845 m/s across the warehouse floor. He 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 0.355.

(a) Determine the magnitude and direction of the net force acting on the crate while it is pushed over the rough section of the floor.

(b) Determine the net work done on the crate while it is pushed over the rough section of the floor.
J

(c) Find the speed of the crate when it reaches the end of the rough surface.
m/s

magnitude     N direction     ---Select--- up down in the same direction as the motion of the crate in the opposite direction as the motion of the crate

Explanation / Answer

friction force = mu x N

N = mg = 90 x 9.8 = 882N

friction force = 0.355 x 882 = 313.11 N

net force on the block = Force applied - friction ofrce = 285 N - 313.11 N

now the friction force is greater than the applied force hence there will be no motion

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