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 eq

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:________.

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Kim Cúc 7 months 2021-08-05T21:42:48+00:00 1 Answers 2 views 0

Answers ( )

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    2021-08-05T21:43:55+00:00

    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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