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A book slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m. Calculate the coefficient of kinetic friction ????k between the book and the carpet. Assume the only forces acting on the book are friction, weight, and the normal force.

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dsdrajlin

A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.

A book slides across a level, carpeted floor with an initial speed (u) of 3.25 m/s. It comes to rest (final speed = v = 0m/s) after 3.25 m (s).

Since this is a uniformly decelerated rectilinear motion, we can calculate the acceleration (a) using the following kinematic expression.

[tex]v^{2} = u^{2} + 2as\\\\(0m/s)^{2} = (3.25m/s)^{2} + 2a(3.25m)\\\\a = -1.63 m/s^{2}[/tex]

The negative sign only explains that the friction opposes the movement. We can calculate the force exerted by the friction (F) using Newton's second law of motion.

[tex]F = m \times a[/tex]    [1]

where,

  • m: mass of the book

We can also calculate the force of friction using the following expression.

[tex]F = k \times N = k \times m \times g[/tex]    [2]

where,

  • k: coefficient of kinetic friction
  • N: normal force
  • g: gravity (9.81 m/s²)

Given [1] = [2], we get,

[tex]m \times a = k \times m \times g\\\\k = \frac{1.63 m/s^{2} }{9.81 m/s^{2}} = 0.166[/tex]

A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.

Learn more: https://brainly.com/question/13754413

hamzaahmeds

This question involves the concepts of the equations of motion, Newton's Second Law, and Frictional Force.

The coefficient of kinetic friction is found to be "0.15".

First, we will use the third equation of motion to find out the acceleration of the book:

[tex]2as = v_f^2-v_i^2\\[/tex]

where,

a = acceleration = ?

s = distance covered = 3.25 m

vf = final speed  = 0 m/s

vi = initial speed = 3.25 m/s

Therefore,

[tex]2a(3.25\ m) = (0\ m/s)^2-(3.25\ m/s)^2\\\\a=\frac{-10.56\ m^2/s^2}{7\ m/s}[/tex]

a = - 1.51 m/s² (negative sign indicates deceleration)

Now the force can be calculated using Newton's Second Law of motion:

F = ma

This force is also equal to the frictional force:

F = μmg

comparing both forces, we get:

ma = μmg

a = μg

[tex]\mu = \frac{a}{g} = \frac{1.51\ m/s^2}{9.81\ m/s^2}[/tex]

μ = 0.15

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion.

Ver imagen hamzaahmeds