1. If the coefficient of kinetic friction between a 39.0 kg crate and the floor

Question

1. If the coefficient of kinetic friction between a 39.0 kg crate and the floor is 0.31, what horizontal force is required to move the crate at a steady speed across the floor? A force of 20.0 N is required to start a 3.0 kg box moving across a horizontal concrete floor. (a) What is the coefficient of static friction between the box and the floor? (b) If the 20.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?

Explanation / Answer

1)Force required = u*m*g = 0.31*39*9.81 = 118.6029 N 2) a)Force = u*m*g = 20 =>u = 20/mg = 0.67959 b)Force-umg = ma =>20-u*3*9.81 = 3*0.5 =>u = 0.6286

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