Question

A ball vibrates back and forth from the free end of an ideal spring having a force constant (spring constant) of 20 N/m. If the amplitude of this motion is 0.30 m, what is the kinetic energy of the ball when it is 0.26 m from its equilibrium position

Answer 1

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.

             $K.E = \frac{1}{2}mv^{2}$

Here,   v = 0

So,          $K.E = \frac{1}{2}mv^{2}$

          $K.E = \frac{1}{2}m(0)^{2}$

                       = 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.