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.
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
Divide each side by (mass):
Acceleration = (force) / (mass)
= (100 N) / (50 kg)
= 2 m/s²
Answer:
Explanation:
A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

F is the force applied and x is the elongation of the spring
k is the spring constant.
negative sign indicates the change in direction from equilibrium position.
In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

Inserting this into Hooke's Law

computing it for x,

This is the model which will tell the length of the spring against change in the mass located in the pan.