A0.025-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?

K=0.4368J

Explanation:

The mechanical energy will be the sum of the elastic energy and the kinetic energy of the block: . Their formulas are and for any point in the movement (although E must remain constant since the movement is frictionless), where k is the spring constant, the displacement, m the mass of the block and v its velocity.

When the block is released from rest, at that point its mechanical energy is equal to the elastic energy, which is at its maximum () since the displacement is at its maximum ().

Since mechanical energy is conserved, the mechanical energy at that point must be equal also to the mechanical energy at the point asked, which we will call P (when the displacement is ), so we will have:

So the kinetic energy at point P is

And substituting values we have:

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2