A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distance d. When t

Question

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distance d. When the block is released, it slides across a surface that is frictionless except for a rough patch of width 5.0 cm that has a coefficient of kinetic friction = 0.44. Find d, such that the block’s speed after crossing the rough patch is 2.3 m/s.

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Philomena 5 years 2021-08-29T23:55:49+00:00 1 Answers 80 views 0

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  1. Farah
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    2021-08-29T23:56:51+00:00
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    Answer:

    Compression distance: d \approx 0.102\,m

    Explanation:

    According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy (K), in joules, at the expense of elastic potential energy (U), in joules, and overcoming work losses due to friction (W_{l}), in joules:

    K + W_{l} = U (1)

    By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

    \frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2} (2)

    Where:

    m – Mass of the block, in kilograms.

    v – Final velocity of the block, in meters per second.

    \mu – KInetic coefficient of friction, no unit.

    g – Gravitational acceleration, in meters per square second.

    s – Width of the rough patch, in meters.

    k – Spring constant, in newtons per meter.

    d – Compression distance, in meters.

    If we know that m = 1.2\,kg, v = 2.3\,\frac{m}{s}, \mu = 0.44, g = 9.807\,\frac{m}{s^{2}}, s = 0.05\,m and k = 730\,\frac{N}{m}, then the compression distance of the spring is:

    \frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}

    m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}

    d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }

    d \approx 0.102\,m

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