Question

When you push a 1.87-kg book resting on a tabletop, it takes 2.02 n to start the book sliding. once it is sliding, however, it takes only 1.47 n to keep the book moving with constant speed. what is the coefficient of static friction between the book and the tabletop?

Answer 1

`F_(friction) = \mu N = \mu mg \\ \\ \mu = (F_(friction))/(mg)`

F static friction
m mass
g gravitational acceleration
μ static friction coefficient


`\mu = (2.02N)/(1.87kg(9.81 (m)/(s^2) )) = 0.11`