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?
A) 0.44 J
B) 0.41 J
C) 0.46 J
D) 0.49 J
E) 0.52 J

Question

Quizplus · Accepted Answer

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 its velocity, $\omega$ is the angular frequency, A is the amplitude of the motion, and t is the time.At the given displacement, the kinetic energy is:$$K = \frac{1}{2}(0.025 	ext{ kg})(150 	ext{ N/m})(0.080 	ext{ m})^2\sin^2(\omega t) = 0.44 	ext{ J}$$

A 0025-Kg Block on a Horizontal Frictionless Surface Is Attached to to an Ideal