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24 March, 04:27

A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F (t) is applied to the end of the rope, and the height of the crate above its initial position is given by y (t) = (2.80 m/s) t + (0.61 m/s^3) t^3.

(a) What is the magnitude of the force F when 4.10s?

(b) is the magnitude's unit N but the system doesn't accept it?

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  1. 24 March, 04:32
    0
    122.2

    Explanation:

    From the question, the equation of the force acting on the crate is,

    F-mg = ma

    F = ma+mg ... Equation 1

    Where F = upward force, m = mass of the crate, a = acceleration of the crate, g = acceleration due to gravity of the crate.

    Given: m = 5.0 kg, g = 9.8 m/s²

    And a,

    a = d²y/dt²

    Where,

    y (t) = 2.8t+0.61t³

    d²y/dt² = 3.66t ... equation 2

    when t = 4.0 s.

    substitute into equation 2

    d²y/dt² = a = 3.66 (4)

    a = 14.64 m/s²

    substitute into equation 1,

    F = 5 (14.64) + 5 (9.8)

    F = 73.2+49

    F = 122.2 N.
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