A vertical spring with a spring constant of 450 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30-kg

Question

A vertical spring with a spring constant of 450 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30-kg ball is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.5 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the uncompressed spring was the block dropped

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Farah 3 years 2021-08-05T23:59:43+00:00 1 Answers 46 views 0

Answers ( )

  1. Latifah
    0
    2021-08-06T00:01:01+00:00
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    Answer:

    h = 1.91 m

    Explanation:

    From the law of conservation of energy, we know that:

    Potential\ Energy\ of\ Ball = Elastic\ Petential\ Energy\ of\ Spring\\mgh = \frac{1}{2}kx^2

    where,

    m = mass of ball = 0.3 kg

    g = acceleration due to gravity = 9.81 m/s²

    h = height of ball = ?

    k = spring constant = 450 N/m

    x = compressed length = 2.5 cm = 0.025 m

    Therefore,

    (0.3\ kg)(9.81\ m/s^2)h = \frac{1}{2} (450\ N/m)(0.025\ m)^2\\\\h = \frac{(450\ N/m)(0.025\ m)^}{(2)(0.3\ kg)(9.81\ m/s^2)}

    h = 1.91 m

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