How far will the spring stretch?

A: When a 4.23kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.95cm. If the 4.23kg mass is removed how far will the spring stretch if a 1.58kg is hung on it instead?

B:How much work must an external agent do to stretch the spring 4.01cm from its unstretched position?

1 Answer

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  • Jim
    Lv 7
    4 years ago
    Favorite Answer

    weight of 4.23 kg mass = 41.5 N

    F = kx

    41.5 = k(2.95E-2)

    k = 41.5/2.95 E2 = 14.1E2 = 1410 N/m

    weight of 1.58 kg mass = 15.5 N

    F = kx = 1410x

    15.5 = 1410x

    x = 15.5/1410 = 0.0110 m = 1.10 cm ANS

    SPE = 1/2kx² = (0.5)(1410)(4.01E-2)² = 1.13 J ANS B

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