To find the maximum speed of the mass-spring system, we can use the conservation of mechanical energy.
The potential energy stored in the spring when stretched by 7.3 centimeters is given by:
\[ U = \frac{1}{2} k x^2 \]
Where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.
We can find \( k \) using Hooke's Law:
\[ F = -kx \]
Given that the mass is 0.91 kg and the acceleration due to gravity is 9.80 m/s², the force exerted on the spring when the mass is hung from it is \( F = mg \).
So, \( k = \frac{mg}{x} \).
Now, we can find the maximum speed \( v \) using the conservation of energy:
\[ \frac{1}{2} k x^2 = \frac{1}{2} mv^2 \]
\[ v = \sqrt{\frac{k}{m}x^2} \]
Given that \( x = 0.086 \) m (converted from 8.6 cm) and \( g = 9.80 \) m/s², we can calculate \( k \) and then find \( v \).
