Question

A 0.16-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 140 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 = -1.2×10−2 m, find the acceleration of the block.

Answer 1

Answer:

The magnitude of the  acceleration  is  $a = 10.5 m/s^2$

Explanation:

From the question we are told that

   The mass of the block is $m = 0.16 \ kg$

   The spring constant is $k = 140\ N/m$

    At first displacement is  $x = + 0.080 m$

 At  second displacement is  $x = -1.2 *10^{-2} \ m$

The acceleration at second displacement is mathematically represented as

                $a = \frac{k}{m} * x$

                $a = - \frac{140}{0.16} * 1.2 *10^{-2}$

                 $a = - 10.5 m/s^2$

Therefore the magnitude of the acceleration is

                 $a = 10.5 m/s^2$

And the direction is  in the negative x-axis