Question

(i) a force of 35.0 n is required to start a 6.0-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 35.0-n force continues, the box accelerates at 0.60 m/s2. what is the coefficient of kinetic friction?

Answer 1

a. The force applied would be equal to the frictional force.

F = us Fn

where, F = applied force = 35 N, us = coeff of static friction, Fn = normal force = weight

 

35 N = us * (6 kg * 9.81 m/s^2)

us = 0.595

 

b. The force applied would now be the sum of the frictional force and force due to acceleration

F = uk Fn + m a

where, uk = coeff of kinetic friction

 

35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)

uk = 0.533