Answer to Question #85077 in Quantum Mechanics for najmul

Question

A minimum force 5N is required to make a body of mass 2kg move on a horizontal floor.
But a force 4N is required to maintain its motion with a uniform velocity. Calculate
coefficient of static friction and coefficient of kinetic friction.

Answer 1

From the definition of static friction force

F_sf=μ_s N

where F_sf is static friction force; μ_s is coefficient of static friction; N is normal force. According to the condition of the problem F_sf=5N(a minimum force is required to make a body move is equal to static friction force), N=mg(the body lies on a horizontal floor), m=2kg is mass of the body, g=9.8 m/s^2 is acceleration of gravity. Therefore

μ_s=F_sf/N=5/(2∙9.8)≈0.255

From the definition of kinetic friction force

F_kf=μ_k N

where F_kf is kinetic friction force; μ_k is coefficient of kinetic friction; N is normal force. According to the condition of the problem F_kf=4N(force is required to maintain its motion with a uniform velocity is equal to static friction force), N=mg(the body move on a horizontal floor), m=2kg is mass of the body, g=9.8 m/s^2 is acceleration of gravity. Therefore

μ_k=F_kf/N=4/(2∙9.8)≈0.204