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
The coefficient of static friction between the box and the floor is 0.60
The value of coefficient of kinetic friction is 0.061
Given data:
The magnitude of force to start the box is, F = 35.0 N.
The mass of box is, m = 6.0 kg.
The magnitude of acceleration is, [tex]a = 0.60 \;\rm m/s^{2}[/tex].
(a)
The force due to static friction is given as,
[tex]f = \mu mg[/tex]
Here, [tex]\mu[/tex] is the coefficient of static friction.
Solving as,
[tex]35.0= \mu \times 6 \times 9.8\\\mu \approx 0.60[/tex]
Thus, the coefficient of static friction between the box and the floor is 0.60.
(b)
If box accelerates, the frictional force comes to play to oppose the motion. Therefore,
[tex]f = F\\\mu_{k} \times m\times g = ma\\\mu_{k}\times g = a\\\mu_{k}\times 9.8 = 0.60\\\mu_{k} = 0.061[/tex]
Thus, the value of coefficient of kinetic friction is 0.061.
Learn more about the kinetic friction here:
https://brainly.com/app/ask?q=kinetic+friction
