A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic fricti

Question

A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200. What average power is produced by friction as the rock stops?

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Yến Oanh 1 year 2021-08-31T21:50:00+00:00 1 Answers 94 views 0

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    2021-08-31T21:51:56+00:00

    Answer:

    Therefore,

    The average power produced by friction as the rock stops is 156.86 Watts.

    Explanation:

    Given:

    mass = m =20.0 kg

    initial velocity,

    [tex]v_{i}=8.00\ m/s[/tex]

    coefficient of kinetic friction,

    [tex]\mu_{k}=0.200\ m/s[/tex]

    To Find:

    average power, P =?

    Solution:

    According to Newton’s Second Law,

    [tex]Force=mass\times Acceleration[/tex]

    Also Friction force,

    [tex]Friction\ Force=-\mu_{k}mg[/tex]

    Therefore,

    [tex]Force=Friction\ Force[/tex]

    Substituting the values we get

    [tex]m\times a=-\mu_{k}mg\\\\a=-0.2\times 9.8=-1.96\ m/s^{2}[/tex]

    Since Rock is getting stopped therefore acceleration is negative.

    Final velocity will also be ZERO.

    [tex]v_{f}=0\ m/s[/tex]

    Now by Kinematic Equation

    [tex]v_{f}=v_{i}+at[/tex]

    Substituting the values we will get time ‘t’,

    [tex]t=-\dfrac{8}{-1.96}=4.08\ s[/tex]

    Now  according to work theorem ,

    [tex]W=Change\ in\ Kinetic\ Energy[/tex]

    which is equal to

    [tex]W=\dfrac{1}{2}m((v_{f})^{2}-(v_{i})^{2})[/tex]

    Substituting the values we get

    [tex]W=-\dfrac{1}{2}\times 20\times 8^{2}=-640\ Joules[/tex]

    Now ,

    [tex]Power=|\dfrac{Work}{time}|[/tex]

    Substituting the values we get

    [tex]Power=|\dfrac{-640}{4.08}|=156.86\ Watt[/tex]

    Therefore,

    The average power produced by friction as the rock stops is 156.86 Watts.

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