(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?

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, .

(a)

The force due to static friction is given as,

Here, is the coefficient of static friction.

Solving as,

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,

Thus, the value of coefficient of kinetic friction is 0.061.

Learn more about the kinetic friction here:

link

when earth is in rotational motion around the sun,it displaces from its place.

when a ball is placed on an inclined surface it does both the rotational and circular motion.

explanation:

i guess it’s a and b