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

Quizplus · Accepted Answer

The total mechanical energy in simple harmonic motion is conserved and is given by the sum of kinetic and potential energy. At maximum displacement (x=0.080 m), all energy is potential: $ E = \frac{1}{2} k x^2 = \frac{1}{2} \times 570 \times (0.080)^2 = 1.824 \, \text{J} $. At x=0.057 m, the potential energy is $ \frac{1}{2} \times 570 \times (0.057)^2 = 0.926 \, \text{J} $. The kinetic energy is the total energy minus the potential energy: $ 1.824 - 0.926 = 0.898 \, \text{J} $, which is closest to 0.90 J.

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