91Ó°ÊÓ

$$\begin{gathered} \text{Weightofthebox}=10kg \\ \text{Thedistanceoftheslide}=4.0m \\ \text{Springconstant},k=250N/m \end{gathered}$$

By using the conservation of energy equation.

$K_i+U_{si}+U_{gi}=K_f+U_{sf}+U_{gf}$

Kinetic energy at the initial and final positions is zero.

Therefore,

localid="1649400284183" $$\begin{gathered} U_{si}+U_{gi}=U_{sf}+U_{gf} \\ \frac{1}{2}k(s_i{)}^2+mg{y}_i=\frac{1}{2}m{v}_1^2+\frac{1}{2}k(s_f{)}^2+mg{y}_f \\ {y}_i=s_i\sin \left(30°\right) \\ {y}_f=s_f\sin \left(30°\right) \\ \frac{1}{2}k(0{)}^2+mg{s}_i\sin \left(30°\right)=mg{s}_f\sin \left(30°\right)+\frac{1}{2}k(s_f{)}^2 \\ mg{s}_i\sin \left(30°\right)=mg{s}_f\sin \left(30°\right)+\frac{1}{2}k(s_f{)}^2 \\ \left(10kg\right)\left(9.8m/s^2\right)\left(4m\right)\left(\frac{1}{2}\right)=mg{s}_f\sin \left(30°\right)+\frac{1}{2}k(s_f{)}^2 \\ 196J=mg{s}_f\sin \left(30°\right)+\frac{1}{2}k\left(s_f{)}^2 \\ \frac{1}{2}\left(250N/m\right)\right(s_f{)}^2+\left(10kg\right)\left(9.8m/s^2\right)\left(\frac{1}{2}\right)s_f=196J \\ \left(125N/m\right)(s_f{)}^2+\left(49kgm/s^2\right)s_f=196J⋯⋯\left(1\right) \\ \text{Bysolvingtheequation(1)weget} \\ {\text{s}}_f\text{=1.46}m\text{and}-1.07m \end{gathered}$$

$\mathrm{The}\mathrm{maximum}\mathrm{compression}\mathrm{of}\mathrm{the}\mathrm{spring}\text{is}1.46\mathrm{m}$

Weight of the box$=10kg$

The distance of the slide$=4.0m$.

Spring constant, $k=250N/m$

The net force at the initial contact s,

$F_s=-mg\sin \left(30°\right)$

The net force at the maximum displacement

$F_s=-ks-mg\sin \left(30°\right)$

The maximum speed at the compression of the spring at $F=0$is

localid="1649399872122" $$\begin{gathered} 0=-ks-mg\sin \left(30°\right) \\ s=\frac{-mg\sin \left(30°\right)}{k} \\ s=\frac{-\left(10kg\right)\left(9.8m/s^2\right)\left({\displaystyle \frac{1}{2}}\right)}{250N/m} \\ s=\frac{-49kg.m/s^2}{250N/m} \\ s=-0.196m \end{gathered}$$

At $19.6cm$the box has its maximum speed.