Given data
The weight of the block is W = 48 N
The angle of inclination is theta = 30o
The spring constant is k = 3270 N/m
The length of the plan is s = 24 m
(a)
The expression for the energy at the top of the inclination is given as:
[tex]PE=W\times s\times\sin \theta[/tex]
The expression for the kinetic energy stored in the block due to the block moving down the plane is given as:
[tex]\begin{gathered} KE=\frac{1}{2}mv^2 \\ KE=\frac{1}{2}\frac{W}{g}v^2 \end{gathered}[/tex]
From the energy conservation principle, the expression is written as follows:
[tex]\begin{gathered} KE=PE \\ \frac{1}{2}\frac{w}{g}v^2=W\times s\times\sin 30^o \\ v=\sqrt[]{2gs\sin \theta} \end{gathered}[/tex]
Substitute the value in the above equation.
[tex]\begin{gathered} v=\sqrt[]{2\times9.8m/s^2\times24\text{ m}\times\sin 30\circ} \\ v=15.33\text{ m/s} \end{gathered}[/tex]
The expression for the maximum distance that the spring was compressed from equilibrium is given as:
[tex]\begin{gathered} U_s=KE \\ \frac{1}{2}kx^2=\frac{1}{2}\frac{W}{g}\times v^2 \\ x=\sqrt{\frac{Wv^2}{gk}} \end{gathered}[/tex]
Substitute the value in the above equation.
[tex]\begin{gathered} x=\sqrt[]{\frac{48\text{ N}\times(15.33)^2}{9.8m/s^2\times3270\text{ N/m}}} \\ x=0.593\text{ m} \end{gathered}[/tex]
Thus, the maximum distance that was compressed from equilibrium is 0.593 m.
(b)
Thus, the speed of the block just before it collides with the spring is 15.33 m/s.
