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katrin2010 [14]

2 years ago

7

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:________.

1 answer:

SVETLANKA909090 [29]2 years ago

7 0

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

$v = -A w sin (w* t )$

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

=> $v = -0.04 \ m/s$

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