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
A) 0.90 J.
B) 0.84 J.
C) 0.95 J.
D) 1.0 J.
E) 1.1 J.

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, and its value is equal to the total energy of the system, which is given by:$$E = \frac{1}{2}kA^2 = \frac{1}{2}(570 \mathrm {~N} / \mathrm { m })(0.080 \mathrm {~m})^2 = 1.83 \mathrm {~J}$$Therefore, the kinetic energy of the block when it is at x = 0.057 m is:$$K = E - U = 1.83 \mathrm {~J} - \frac{1}{2}kA^2 = 1.83 \mathrm {~J} - \frac{1}{2}(570 \mathrm {~N} / \mathrm { m })(0.057 \mathrm {~m})^2 = 0.90 \mathrm {~J}$$

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