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Two workers are sliding 360 kg crate across the floor. One worker pushes forward on the crate with a force of 380 N while the other pulls in the same direction with a force of 280 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

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2 answers
  1. the crate is not accelerating
    ... so the frictional force equals the forces of the workers

    m * g * μ = 380 N + 280 N ... μ = 660 / (360 * 9.8)

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  2. Mg = 360*9.8 = 3528 N. = Wt. of crate = Normal force, Fn.
    380+280 = 660 N. = Force applied.
    u*Fn = u*3528 = Force of kinetic friction.

    660-3528u = M*a
    660-3528u = M*0 = 0
    u = 660/3528.

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