A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points of 15.0 N. If the coefficient of kinetic friction

Question

A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
of 15.0 N. If the coefficient of kinetic friction between the book and the
table is 0.35, find its acceleration. *​

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Vân Khánh 3 years 2021-07-14T02:13:20+00:00 1 Answers 68 views 0

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  1. Vodka
    0
    2021-07-14T02:14:45+00:00
    Reply

    Answer:

    Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

    Explanation:

    Weight of the book:

    W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

    If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

    As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

    When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

    • \mu_{\rm k}, the coefficient of kinetic friction, times
    • F(\text{normal force}), the normal force that’s acting on it.

    That is:

    \begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

    Friction acts in the opposite direction of the object’s motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

    \begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

    Apply Newton’s Second Law to find the acceleration of this book:

    \displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

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