Question

When you push a 1.90-kg book resting on a tabletop, you have to exert a force of 2.10 N to start the book sliding. Once it is sliding, however, you can use a force of only 1.06 N to keep the book moving with constant speed. Part A What is the coefficient of static friction between the book and the tabletop?

Answer 1

Static frictional force = ƒs = (Cs) • (Fɴ)
              2.26 = (Cs) • m • g
              2.26 = (Cs) • (1.85) • (9.8)
           Cs = 0.125 
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
              1.49 = (Cκ) • m • g
              1.49 = (Cκ) • (1.85) • (9.8)
           Cκ = 0.0822