Answer:
The speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.
Explanation:
Mass attached to the spring, m = 10 kg
maximum displacement of the spring, A = 0.44 m
The spring constant is calculated from Hook's law;
F = kx
mg = kx
k = (mg) / x
k = (10 x 9.8) / 0.44
k = 222.73 N/m
The angular speed of the spring is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega =\sqrt{\frac{222.73}{10} } \\\\\omega = 4.72 \ rad/s[/tex]
The speed when the box reaches a point 10 cm from the equilibrium position is calculated as;
[tex]v = \omega \sqrt{A^2-x^2} \\\\v = 4.72\sqrt{0.44^2-0.1^2}\\\\v = 2.02 \ m/s[/tex]
Therefore, the speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.
