91Ó°ÊÓ

The given data can be listed below,

• The mass of the book is,$m=2.50\mathrm{kg}$
• The force constant of the spring is,$k=250\mathrm{N}/\mathrm{m}$
• The compression of spring is$d=0.250\mathrm{m}$
• The coefficient of kinetic friction is,${\mathrm{μ}}_k=0.30$

A spring's force constant is determined empirically. It is the force applied when a spring is stretched or compressed by a unit length.

The force on the book and friction forces are acting in opposite directions. So, the work done by both forces are equal and opposite, which is given as,

$\left|W_{spring}\right|=\left|W_{friction}\right|$

the work done due to spring is given by,

$W_{spring}=\frac{1}{2}k{x}_1^2$

Here, k is the force constant and x is the compressed distance.

Substitute the given values in the above equation,

$$\begin{gathered} W_{spring}=\frac{1}{2}\left(250\mathrm{N}/\mathrm{m}\right){\left(-0.25\mathrm{m}\right)}^2 \\ =-7.81\mathrm{Nm}\left(\frac{1\mathrm{J}}{1\mathrm{Nm}}\right) \\ =-7.81\mathrm{J} \end{gathered}$$

The work done by friction is,

$W_{friction}=-{\mathrm{μ}}_{k}mg\left(x_2-x_1\right)$

Here,${\mathrm{μ}}_k$ is the coefficient of friction, m is the mass of the textbook, g is the acceleration due to gravity,$x_1$ is the initial position and$x_2$ is the final position of the textbook.

The final position of the textbook is given as,

$x_2=\frac{W_{spring}}{{\mathrm{μ}}_{k}mg}-x_1$

Substitute the given values in the above,

$$\begin{gathered} x_2=\frac{7.81\mathrm{J}}{0.30×2.5\mathrm{kg}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}-0.25\mathrm{m} \\ =0.81\mathrm{m} \end{gathered}$$

So, the total distance covered by the textbook is given by,

$$\begin{gathered} d={\mathrm{x}}_1+{\mathrm{x}}_2 \\ =0.25\mathrm{m}+0.81\mathrm{m} \\ =1.06\mathrm{m} \end{gathered}$$

Thus, the total distance covered by the textbook from the initial position to the final position is$1.06\mathrm{m}$