Given data
*The given mass of the box is m = 4 kg
*The given frictionless table of height is h = 0.2 m
*The spring compressed at a distance is x = 0.06 m
*The table landing a distance is d = 1.2 m
(b)
The formula for the time taken by the box is given by the equation of motion as
[tex]\begin{gathered} h=\frac{1}{2}gt^2 \\ t=\sqrt[]{\frac{2h}{g}} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times0.2}{9.8}} \\ =0.202\text{ s} \end{gathered}[/tex]The formula for the box's speed at the instant it leaves the table is given as
[tex]v=\frac{d}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v=\frac{1.2}{0.202} \\ =5.94\text{ m/s} \end{gathered}[/tex]Hence, the box's speed at the instant it leaves the table is v = 5.94 m/s
(d)
The formula for the spring constant is given by the conservation of energy as
[tex]\begin{gathered} U_s=U_k \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ k=\frac{mv^2}{x^2} \end{gathered}[/tex]*Here U_s is the spring energy
*Here U_k is the kinetic energy
Substitute the known values in the above expression as
[tex]\begin{gathered} k=\frac{(4)(5.94)^2}{(0.06)^2} \\ =3.92\times10^4\text{ N/m} \end{gathered}[/tex]Hence, the spring constant is k = 3.92 × 10^4 N/m
(c)
The formula for the energy initially stored in the spring is given as
[tex]U_s=\frac{1}{2}kx^2[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} U_s=\frac{1}{2}(3.92\times10^4)(0.06)^2 \\ =70.56\text{ J} \end{gathered}[/tex]Hence, the energy initially stored in the spring is U_s = 70.56 J
