Once the crate in sample problem 4C is in motion, a horizontal force of 53 N keeps the crate moving with a constant velocity. Find ตk, the coefficient of kinetic friction, between the crate and the floor.

Answer: k = 5.4kg/m

where m is the mass of the crate.

Explanation:

The force of kinetic friction is described by the equation:

Ff = k*N

in the opposite direction in wich the object moves, where k is the coefficient of kinetic friction and N is the normal force, that is equal to the weight of the crate. N = m*g where m is the mass of the crate and g is the gravitational acceleration.

If the crate keeps moving with constant velocity, this means that the crate is not accelerating so there is no net force applied on the crate. Then the friction force should be equal in magnitude to the horizontal force of 53N (but with different sign)

then we have:

k*m*g = 53N

k*m = 53N/9.8m/s^2 = 5.4kg

k = 5.4kg/m

In the question we do not have the mass of the crate, so you must put the value in that equation to get the value of k.

Given:

Force required at constant velocity, Fk = 53 N

Force required at rest, Fs = 53 N

Mass, M = 24 kg

Total force, Ft = 0

Fs - Fn = 0

Fk - Fn = 0

Where,

Fn = normal force

= Mass * g * µ

Fs = Mass * g * µs

Fk = Mass * g * µk

Where,

µs = coefficient of static friction

µk = coefficient of kinetic friction

Fk = Mass * g * µk

µk = 53 / (24 * 9.8)

= 0.225

= 0.23.