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In a room that is 2.69 m high, a spring (unstrained length = 0.28 m) hangs from the ceiling. A board whose length is 2.22 m is attached to the free end of the spring. The board hangs straight down, so that its 2.22-m length is perpendicular to the floor. The weight of the board (145 N) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?

  • Physics -

    Elongation of the spring is
    x=2.69 –(2.22+0.28) = 0.19 m
    mg= kx
    k=mg/x- 145/0.19=763.16 N/m

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