Answer: The kinetic energy of the ball when it is 0.26 m from its equilibrium position is 0 J.
Explanation:
The ball is present in simple harmonic motion. During this motion, the speed of ball will be maximum at a distance from the equilibrium point.
We are given that the amplitude is 0.3 m and kinetic energy at x = 0.26 m will be calculated as follows.
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
Here, v = 0
So, [tex]K.E = \frac{1}{2}mv^{2}[/tex]
[tex]K.E = \frac{1}{2}m(0)^{2}[/tex]
= 0 J
Thus, we can conclude that the kinetic energy of the ball when it is 0.26 m from its equilibrium position is 0 J.
