91Ó°ÊÓ

The force constant of the spring is k = 1600 N/m

The potential energy for spring compression is U = 3.20 J

The mass of the book is m = 1.20 m

The height of the book is h = 0.800 m

The energy stored due to the variation in the position with some reference position is the potential energy. The potential energy is used to calculate the compression of the spring.

The compression of the spring is calculated as:

$\mathrm{U}=\frac{1}{2}{\mathrm{kx}}^2$

Here, U is the potential energy of spring, x is the compression, and k is the force constant of spring.

Substitute all the values in the above equation and we get,

$$\begin{gathered} 3.20\mathrm{J}=\frac{1}{2}\left(1600\mathrm{N}/\mathrm{m}\right){\mathrm{x}}^2 \\ \mathrm{x}=0.063\mathrm{m} \end{gathered}$$

Therefore, the compression of spring is 0.063 m .

The maximum compressed distance is calculated as:

$\mathrm{mgh}=\frac{1}{2}{\mathrm{ky}}^2$

Here, m is the mass of the book, g is the gravitational acceleration, h the height dropped by the book, k is the force constant, and y is compression of spring.

Substitute all the values in the above equation, and we get,

$$\begin{gathered} \left(1.20\mathrm{kg}\right)\left(9.8\right)\left(0.800\mathrm{m}\right)=\frac{1}{2}\left(1600\mathrm{N}/\mathrm{m}\right){\mathrm{y}}^2 \\ \mathrm{y}=0.11\mathrm{m} \end{gathered}$$

Therefore, the maximum compressed distance for spring due to the book is 0.11 m .