Answer:
the maximum speed of the ball is 12.65 m/s
Explanation:
Given;
mass of the ball, m = 40 g = 0.04 kg
spring constant, k = 25 N/m
Apply the principle of conservation of energy;
The Elastic potential energy of the spring will be converted into Kinetic of the ball;
[tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2\\\\ kx^2 = mv^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m}} \\\\v = \sqrt{\frac{(25)(0.506)^2}{0.04}} \\\\v = \sqrt{160.0225} \\\\v = 12.65 \ m/s[/tex]
Therefore, the maximum speed of the ball is 12.65 m/s
