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15 December, 09:05

When a 2.60-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.86 cm. (a) What is the force constant of the spring? N/m (b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it? cm (c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?

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  1. 15 December, 11:01
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    a) k = 891.82 N/m

    b) e = 0.0143 m = 1.43 cm

    c) W = 5.02 J

    Explanation:

    Step 1: Data given

    Mass = 2.60 kg

    the spring stretches 2.86 cm = 0.0286

    Step 2: What is the force constant of the spring?

    Force constant, k = force applied / extension produced

    k = (2.60kg * 9.81N/kg) / 0.0326 m

    k = 891.82 N/m

    b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it

    Extension = F/k = (1.30 kg * 9.81) / 891.82 = 0.0143 m = 1.43 cm

    Half the mass means half the extension

    c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?

    W = average force used * distance

    W = 1/2 * k*e * e = 1/2 k*e²

    W = 1/2 * 891.82 * (0.075) ² = W = 5.02 J
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