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28 August, 10:54

A 68 kgkg driver gets into an empty taptap to start the day's work. The springs compress 2.1*10-2 mm. What is the effective spring constant of the spring system in the taptap?

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  1. 28 August, 12:23
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    K = 3.1 8 * 10⁴ Nm

    Explanation:

    Given:

    m = 68 Kg

    x = 2.1 * 10 ⁻² m (in (2.1*10-2 mm) i think m is repeated just like kgkg in the start of question)

    g=9.81 m/s²

    K=?

    According to Hooke's Law

    F = K x

    also F=mg

    ⇒ mg = Kx

    ⇒ K = mg/x

    K = 68 Kg * 9.81 m/s² / 2.1 * 10 ⁻² m

    K = 3.1 8 * 10⁴ Nm
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