A 29 kg chair initially at rest on a horizontalfloor requires a 373 N horizontal force to setit in motion. Once the chair is in motion, a342 N horizontal force keeps it moving at aconstant velocity.The acceleration of gravity is 9.81 m/s2.a) What is the coefficient of static frictionbetween the chair and the floor?008 (part 2 of 2) 10.0 pointsb) What is the coefficient of kinetic frictionbetween the chair and the floor?
When the chair is at rest, the force applied has to overcome the force of static friction before the chair can move. The max force applied before the chair moves = F(s) Then F(s) = F(static friction) F(s) = μs x N .....(μs=coefficient of static friction) F(s) = μs x mg μs = F(s)/mg = 373 / 29(9.81) = 1.31 Once the chair is in motion with constant velocity, then the applied force overomes the force of kinetic friction Then F(k) = F(kinetic friction) F(k) = μk x N .....(μk=coefficient of kinetic friction) F(k) = μk x mg μk = F(k)/mg = 342 / 29(9.81) = 1.20
Force due to gravity = 29 kg * 9.81 m/s^2 coefficient of static friction = 373 N / 29 kg * 9.81 m/s^2 coefficient of kinetic friction = 342 N / 29 kg * 9.81 m/s^2
