A 0.16-kg block on a horizontal frictionless surface is attached to an ideal spring whose force constant (spring constant)is 360 N/m.The block is pulled from its equilibrium position at x = 0.000 m to a position x = +0.080 m and is released from rest.The block then executes simple harmonic motion along the horizontal x-axis.When the position is x = -0.037 m,what is the acceleration of the block? A)83 m/s² B)43 m/s² C)64 m/s² D)270 m/s² E)370 m/s²

Question

Quizplus · Accepted Answer

The acceleration $a$ of the block in simple harmonic motion can be found using the formula $a = -\omega^2 x$, where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency, $k$ is the spring constant, $m$ is the mass of the block, and $x$ is the displacement from the equilibrium position. Given $k = 360 \, \text{N/m}$, $m = 0.16 \, \text{kg}$, and $x = -0.037 \, \text{m}$, we find $\omega = \sqrt{\frac{360}{0.16}}$ and then calculate $a$. The negative sign in the formula indicates the direction of acceleration is opposite to the direction of displacement, which is a characteristic of simple harmonic motion. The exact numerical calculation leads to an acceleration value that matches the one provided in option A.

A 016-Kg Block on a Horizontal Frictionless Surface Is Attached to to an Ideal