A 0.39-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant)is 570 N/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 m. Its kinetic energy is closest to
A)0.90 J.
B)0.84 J.
C)0.95 J.
D)1.0 J.
E)1.1 J.

Question

A 0.39-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant)is 570 N/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 m. Its kinetic energy is closest to
A)0.90 J.
B)0.84 J.
C)0.95 J.
D)1.0 J.
E)1.1 J.

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 x = 0.057 m, the kinetic energy is:$$K = \frac{1}{2}(0.39 	ext{ kg})(570 	ext{ N/m})(0.080 	ext{ m})^2\sin^2(\omega t)$$$$K = 0.90 	ext{ J}$$Therefore, the correct answer is A.

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