Question

A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is what is the kinetic energy of the block?

Answer 1

Complete Question

A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 150 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 0.024 m, what is the kinetic energy of the block?

Answer:

The kinetic energy is  $KE = 0.4368\ J$

Explanation:

From the question we are told that

   The mass of the block is $m= 0.025\ kg$

   The spring constant is $k = 150 N/m$

   The length of first  displacement  is $x_1 = 0.80 \ m$

     The length of first  displacement  is $x_2 = 0.024 \ m$

At the $x_2$ the kinetic energy is mathematically evaluated as

     $KE = \Delta E$

Where $\Delta E$ is the change in energy stored on the spring which is mathematically represented as

            $\Delta E = \frac{1}{2} k (x_1 ^2 - x_2^2)$

=>        $KE = \frac{1}{2} k (x_1 ^2 - x_2^2)$

Substituting value

          $KE = \frac{1}{2} * 150 * (0.08^2 - 0.024^2)$

          $KE = 0.4368\ J$