A 13.3 kg box sliding across the ground decelerates at 2.42 m/s2. What is the
coefficient of kinetic friction?
(No unit)

0.25

Explanation:

According to newtons law of motion

\sum F_x = ma

F_f =  ma

nR = ma

nmg = ma

ng = a

n = a/g

g is the acceleration due to gravity

Given

a = 2.42m/s²

g = 9.8m/s²

Substitute into the formula;

n = 2.42/9.8

n = 0.25

Hence the coefficient of kinetic friction is 0.25

reserve is the answer