One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched by a length

Question

One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched by a length of 4.3 cm. How long is the spring when a 3.3 kg mass is suspended from it in cm?

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Xavia 2 months 2021-07-31T19:39:25+00:00 1 Answers 6 views 0

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    2021-07-31T19:40:36+00:00

    Answer:

    2.63 cm

    Explanation:

    Hooke’s law gives that the force F is equal to cy where c is spring constant and x is extension

    Making c the subject of the formula then

    c=\frac {F}{y}

    Since F is gm but taking the given mass to be F

    c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930

    By substitution now considering F to be 3.3 kg

    y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm

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