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17 January, 05:22

A 25.0 [cm] long spring stands vertically on the ground, with its lower end secured in a base. A brick with mass, 1.5 [kg] is held 40 [cm] directly above the spring and dropped onto the spring. The spring compresses to a length of 17.0 [cm] before starting to launch the brick back upward. What is the spring's spring constant?

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Answers (2)
  1. 17 January, 05:34
    0
    1837.5 N/m

    Explanation:

    From the law of conservation of energy,

    The potential energy of the brick = energy stored in the spring

    mgh = 1/2ke² ... Equation 1

    Where m = mass of the brick, h = height, g = acceleration due to gravity, k = spring constant, e = extension

    Make k the subject of the equation

    k = 2mgh/e² ... Equation 2

    Given: m = 1.5 kg, e = 25-17 = 8 cm = 0.08 m, h = 40 cm = 0.4 m, g = 9.8 m/s²

    Substitute into equation 2

    k = 2 (1.5) (9.8) (0.4) / (0.08²)

    k = 1837.5 N/m

    Hence the spring constant of the spring = 1837.5 N/m
  2. 17 January, 05:34
    0
    compression in spring = 25-17 = 8

    change in potential energy of brick when it joins the spring = mg * (40 + 8)

    which is equal to the stored potential energy in sprint = 1/2*k * (8^2)

    so k = 3/2mg = 3/2*1.5*10 = 2250 N/m
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