Question

A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.10 s is closest to:________.

Answer 1

Answer:

The  value is  $v = -0.04 \ m/s$

Explanation:

From the question we are told that

   The  mass  of the block is  $m = 2.0 \ kg$

   The  force constant  of the spring is  $k = 590 \ N/m$

   The amplitude  is  $A = + 0.080$

   The  time consider is  $t = 0.10 \ s$

Generally the angular velocity of this  block is mathematically represented as

      $w = \sqrt{\frac{k}{m} }$

=>   $w = \sqrt{\frac{590}{2} }$

=>   $w = 17.18 \ rad/s$

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as  

         $v = -A w sin (w* t )$

=>       $v = -0.080 * 17.18 sin (17.18* 0.10 )$

=>       $v = -0.04 \ m/s$

Answer 2

The velocity of the block at the given time is -0.04 m/s.

Angular speed of the block

The angular speed of the block is determined by using the following wave equation;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{590}{2} } \\\\\omega = 17.176 \ rad/s$

Velocity of the block

The velocity of the block at the given time is calculated as follows;'

$v = - A sin(\omega t)\\\\v =- 0.08 \times sin(17.176 \times 0.1)\\\\v = -0.04 \ m/s$

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