The given question is incomplete. The complete question is as follows.
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.
Explanation:
The given data is as follows.
= 20 N,
= 25 N, a = -0.9
W = 83 N
m = 
= 8.46
Now, we will balance the forces along the y-component as follows.
N = W +
= 83 + 25 = 108 N
Now, balancing the forces along the x component as follows.
= ma
= 7.614 N
Also, we know that relation between force and coefficient of friction is as follows.

= 
= 0.0705
Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.
Answer:
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m = mass of the person = 82 kg
g = acceleration due to gravity acting on the person = 9.8 m/s²
F = normal force by the surface on the person
f = kinetic frictional force acting on the person by the surface
μ = Coefficient of kinetic friction = 0.45
The normal force by the surface in upward direction balances the weight of the person in down direction , hence
F = mg eq-1
kinetic frictional force on the person acting is given as
f = μ F
using eq-1
f = μ mg
inserting the values
f = (0.45) (82) (9.8)
f = 361.6 N