In moving a 35.0-kg desk from one side of a classroom to the other, a professor finds that a horizontal force of 275 N is necessary to set the desk in motion, and a force of 195 N is necessary to keep it in motion at a constant speed. What are the coefficients of (a) static and (b) kinetic friction between the desk and the floor?

Static friction acts until the body starts moving. As soon as the body starts to move, kinetic friction acts on it. Hence, according to the question, 275N corresponds to static friction and 195N corresponds to kinetic friction.

(a) Coefficient of static friction, μs = ?

We know,

fs = μs X mg

μs = fs/mg

μs = [tex]\frac{275}{35 X 9.8}[/tex]

μs = 0.8

Therefore, coefficient of static friction, μs is 0.8

(b) Coefficient of kinetic friction, μk = ?

We know,

fk = μk X mg

μk = fk/mg

μk = [tex]\frac{195}{35 X 9.8}\\ \\[/tex]

μk = 0.57

Therefore, coefficient of kinetic friction, μk is 0.57

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-11-19T12:34:12+01:00

(a) The coefficient of static friction between the desk and the floor is 0.8.

(b) The coefficient of kinetic friction between the desk and the floor is 0.57.

The given parameters;

mass of the desk, m = 35 kg
horizontal force to set in motion, = 275 N
force to keep in motion, = 195 N

The coefficient of static friction is calculated as follows;

[tex]\mu_ s = \frac{F_s}{F_n} \\\\\mu_ s= \frac{275}{35 \times 9.8} \\\\\mu_ s = 0.8[/tex]

The coefficient of kinetic friction is calculated as follows;

[tex]\mu_ k = \frac{F_k}{F_n} \\\\\mu_ k = \frac{195}{35 \times 9.8} \\\\\mu_k = 0.57[/tex]

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