Answer:
The maximum speed of the object is 0.662 m/s.
Explanation:
Given that,
Mass of the object, m = 0.67 kg
Spring constant of the spring, k = 15 N/m
The spring is pulled 14 cm or 0.14 m from the equilibrium position and released.
To find,
The maximum speed of the object.
Solution,
The maximum speed of the object is given by :
[tex]v=A\omega[/tex]........(1)
Where
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega=\sqrt{\dfrac{15}{0.67}}[/tex]
[tex]\omega=4.73\ rad/s[/tex]
So,
[tex]v=0.14\times 4.73[/tex]
v = 0.662 m/s
So, the maximum speed of the object is 0.662 m/s.
