A 0.28-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 500 N/m. The block is

Question

A 0.28-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 500 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 = -0.052 m, find the acceleration of the block.

in progress 0
Linh Đan 4 years 2021-07-20T13:05:28+00:00 2 Answers 128 views 0

Answers ( )

  1. sori
    0
    2021-07-20T13:06:37+00:00
    Reply

    Answer:

    The block will accelerate at 92.86m/s²

    Explanation:

  2. thuthuy
    0
    2021-07-20T13:07:05+00:00
    Reply

    Answer:

    The block will accelerate at 92.86m/s²

    Explanation:

    The acceleration of a simple harmonic motion of a spring is expressed as

    a= – kx/m

    Where k = spring constant

    x= displacement

    m= mass of block

    Given data

    Spring constant k = 500N/m

    Displacement x= – 0.052m

    Mass of block m= 0.28kg

    Pluging this parameters into the expression for acceleration we have

    a= – 500*(-0.052)/0.28

    a= 26/0.28

    a= 92.86m/s²

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )