Actually Welcome to the Concept of the Kinematics.
so here we get as,
V^2 = U^2 + 2as
so here, a = -0.2 m/s^2
(0.1)^2 = (0.3)^2 + (-0.2)(s)
=> 0.01 = 0.09 - 0.2s
=> 0.2s = 0.08
=> s = 0.08/0.2
=> s = 0.4 m
Question 14 of 30
2 answers:
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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.