91Ó°ÊÓ

The given data can be listed below,

• The mass of the box is,$m=6.0\text{ k²µ}$.
• The velocity of the box is,$v=3.0\text{″¾/²õ}$.
• The force constant of spring is,$k=75\text{ N/³¦³¾}$ .

The work-energy equation, which shows a direct correlation between a particle's velocity and the net work performed on it.

The total work done on the box is given by,

$\begin{array}{c}{W}_{tot}=K_f−K_i\\ =\frac{1}{2}m{v}_f^2−\frac{1}{2}m{v}_i^2\end{array}$

…(¾±)

Here, m is the mass of the box and$v_f$ is the final velocity of the box whose value is zero as it comes to rest in last, $v_1$is the initial velocity of the box.

Substitute all the values in the above,

$\begin{array}{c}{W}_{tot}=\frac{1}{2}m{\left(0\right)}^2−\frac{1}{2}m{\left(3.0\text{″¾/²õ}\right)}^2\\ =−\frac{1}{2}\left(6.0\text{ k²µ}\right){\left(3.0\text{″¾/²õ}\right)}^2\\ =−27\text{ J}\end{array}$

According to the conservation law, the work done due to spring force and the total work done are equal. The work done due to spring force is given by,

$W=−\frac{1}{2}k{x}^2$ …(¾±¾±)

Here, k is the force constant of the spring, $x$is the compressed length of the spring.

Substitute all the values in the above,

$\begin{array}{c}−27\text{ J}=−\frac{1}{2}\left(75\text{ N/³¦³¾}\right)x^2\\ x=\sqrt{\frac{2×27\text{ J}}{75\text{ N/³¦³¾}}}\\ =8.5\text{ c³¾}\end{array}$

Thus, the maximum compression in the spring is $8.5\text{ c³¾}$.