Answer:
a) 8.80 N
b) 4.78 N
Explanation:
The weight of the book is
[tex]W = mg = 2.03\times9.81 = 19.9143 \text{ N}[/tex]
By Newton's third law, the reaction from the table, R, is equal to its weight.
R = 19.9143 N
From the law of solid friction,
[tex]F = \mu R[/tex]
F is the maximum frictional force and [tex]\mu[/tex] is the coefficient of friction.
a) The force needed to begin moving the book is that needed to overcome static friction
[tex]F = 0.442\times19.9143 = 8.80\text{ N}[/tex]
b) The force needed to move the book at constant speed once it begins moving is that needed to overcome kinetic friction
[tex]F = 0.240\times19.9143 = 4.78\text{ N}[/tex]
The force is needed to begin moving the book is of 8.80 N.
The force is needed to keep the book moving at constant speed once it begins moving is of 4.78 N.
Given data:
The mass of book is, m = 2.03 kg.
The coefficient of static friction between the book and the desk is, [tex]\mu = 0.442[/tex].
The coefficient of kinetic friction is, [tex]\mu_{k} = 0.240[/tex].
In this problem, concept of frictional force is applied, which says that the force opposing the motion is known as frictional force. And it is given as,
[tex]F = \mu \times N[/tex]
Here, N is the normal force, which is due to weight of book (W).
N = W = mg
Solving as,
[tex]F = \mu \times W\\\\F = 0.442 \times (mg)\\\\F = 0.442 \times (2.03 \times 9.8)\\\\F = 8.80 \;\rm N[/tex]
Thus, we can conclude that the force is needed to begin moving the book is of 8.80 N.
And now force needed to move the book at constant speed once it begins moving is that needed to overcome kinetic friction. Then,
[tex]F' = \mu_{k} \times W \\\\F' = 0.240 \times (mg)\\\\F' = 0.240 \times (2.03 \times 9.8)\\\\F' = 4.78 \;\rm N[/tex]
Thus, the force is needed to keep the book moving at constant speed once it begins moving is of 4.78 N.
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