A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The obje

Question

A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N/m. The object is displaced 3.54 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released

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Hải Đăng 3 years 2021-08-02T07:54:17+00:00 1 Answers 28 views 0

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  1. Dulcie
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    2021-08-02T07:56:08+00:00
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    Answer:

    17.54N in -x direction.

    Explanation:

    Amplitude (A) = 3.54m

    Force constant (k) = 5N/m

    Mass (m) = 2.13kg

    Angular frequency ω = √(k/m)

    ω = √(5/2.13)

    ω = 1.53 rad/s

    The force acting on the object F(t) = ?

    F(t) = -mAω²cos(ωt)

    F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

    F(t) = -17.65 * cos (5.355)

    F(t) = -17.57N

    The force is 17.57 in -x direction

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