91Ó°ÊÓ

Mass$m=3.5\text{ â¶Ä‰k²µ}$

Distance,$s=0.90\text{″¾}$

Force,$F=25\text{ N}$

Initial speed,$v_0=0\text{ }\frac{\text{m}}{\text{s}}$

Final speed,$v_f=1.60\frac{\text{m}}{\text{s}}$

This problem is based on the kinematic equations of moti0on that describe the motion of an object at constant acceleration. Also, this deals with Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. By using these equations, the acceleration of the book can be found. Further, by using Newton's law of motion, the coefficient of kinetic friction between the book and floor can be computed.

Formula:

The velocity in kinematic equation can be written as,

${v_f}^2={v_0}^2+2as$

By using Kinematic equation of motion, acceleration of the book is,

${v_f}^2={v_0}^2+2as$

Substitute the value and solve as:

$\begin{array}{l}a=\frac{{v_f}^2−{v_0}^2}{2S}\\ =\frac{{1.60}^2−0^2}{2\left(0.90\right)}\\ =1.42\text{ }\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

According to the Newton’s 2nd law of motion:

$$\begin{gathered} \underset{}{\overset{}{∑}}F=ma \\ F-f_s=ma \\ F-{\mathrm{μ}}_k\left(mg\right)=ma \end{gathered}$$

Substitute the values and solve as:

$\begin{array}{rcl}{\mathrm{μ}}_k& =& \frac{F-ma}{mg}\\ & =& \frac{25-\left(3.5\right)\left(1.42\right)}{\left(3.5\right)\left(9.81\right)}\\ & =& 0.58\\ & & \end{array}$

Hence, the coefficient of kinetic friction between the book and floor is$0.58$