A 0.39-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant)is $$570 \mathrm {~N} / \mathrm { m } .$$ The block is pulled from its equilibrium position at x = 0.000 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 position of the block is $x = 0.057 \mathrm {~m}$ its kinetic energy is closest to

Question

Quizplus · Accepted Answer

The correct answer is A because the kinetic energy of the block is maximum when it passes through its equilibrium position. The kinetic energy of the block is given by:$$K = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2A^2\sin^2(\omega t)$$where m is the mass of the block, v is the velocity of the block, $\omega$ is the angular frequency of the motion, A is the amplitude of the motion, and t is the time.At x = 0.057 m, the block is at a displacement of A/2 from its equilibrium position. Therefore, the kinetic energy of the block is:$$K = \frac{1}{2}m\omega^2A^2\sin^2(\omega t) = \frac{1}{2}(0.39 \mathrm{~kg})(570 \mathrm{~N} / \mathrm{m})(0.080 \mathrm{~m})^2\sin^2(\omega t) = 0.90 \mathrm{~J}$$

A 039-Kg Block on a Horizontal Frictionless Surface Is Attached to to an Ideal