A 4.8-kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t=0 s, the block has a dis

Question

A 4.8-kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t=0 s, the block has a displacement of -0.50m, a velocity of -0.80m/s and an acceleration of +8.3m/s2 The force constant of the spring is closest to:______.
A) 62 N/m
B) 67 N/m
C) 56 N/m
D) 73 N/m
E) 80 N/m

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Khánh Gia 5 years 2021-08-06T01:58:50+00:00 1 Answers 92 views 0

Answers ( )

  1. sori
    0
    2021-08-06T02:00:39+00:00
    Reply

    Answer:

    E) 80 N/m

    Explanation:

    Given;

    mass of the block, m = 4.8 kg

    displacement of the block, x = -0.5 m

    velocity of the block, v = -0.8 m/s

    acceleration of the block, a = 8.3 m/s²

    From Newton’s second law of motion;

    F = ma

    Also, from Hook’s law;

    F = -Kx

    where;

    k is the force constant

    Thus, ma = -kx

    k = -ma/x

    k = -(4.8 x 8.3) / (-0.5)

    k = 79.7 N/m

    k ≅ 80 N/m

    Therefore, the force constant of the spring is closest to 80 N/m

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