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12 July, 01:14

A 10.0 kg mass is placed on a frictionless, horizontal surface. The mass is connected to the end of a horizontal compressed spring which has a spring constant 339 N/m. When the spring is released, the mass has an initial, positive acceleration of 10.2 m/s2. What was the displacement of the spring, as measured from equilibrium, before the block was released? Watch the sign of your answer.

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  1. 12 July, 03:59
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    The displacement of the spring, as measured from equilibrium, is 0.301 m

    Explanation:

    Hi there!

    Using Hooke's law, we can calculate the displacement of the spring:

    F = - kx

    Where:

    F = restoring force exerted by the spring.

    k = spring constant.

    x = displacement of the spring.

    The force exerted by the spring can also be calculated using the Newton law:

    F = m · a

    Where:

    F = force.

    m = mass of the object.

    a = acceleration.

    Then, combining both laws:

    F = - k · x = m · a

    -k · x = m · a

    -339 N/m · x = 10.0 kg · 10.2 m/s²

    x = 10.0 kg · 10.2 m/s² / - 339 N/m

    x = - 0.301 m

    The displacement of the spring, as measured from equilibrium, is 0.301 m
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