A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction between the box and the floor is 0.20

Question

A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction between the box and the floor is 0.20. The box moves a distance of 4.0 m in 2.0 s. The magnitude of the change in momentum of the box during this time is most nearly __________.

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RuslanHeatt 5 years 2021-08-30T13:19:20+00:00 1 Answers 784 views 0

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  1. Sapo1
    0
    2021-08-30T13:20:52+00:00
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    Explanation:

    The given data is as follows.

           initial speed (u) = 4.0 m/s,     mass (m) = 12 kg

          Distance (s) = 4.0 m,    time (t) = 2.0 sec

    First, we will calculate the acceleration as follows.

               s = ut + \frac{1}{2}at^{2}

             4 = 4.0 \times 2 + \frac{1}{2} \times a \times (2)^{2}

             a = -2 m/s^{2}

    Now, the final speed will be calculated as follows.

                v = u + at

                   = 4.0 + (-2) \times 2

                  = 0

    Therefore, change in momentum will be calculated as follows.

                \Delta p = m(v – u)

                               = 12 \times (0 - 4)

                               = -48 kg m/s

    The negative sign indicates the change in momentum.

    Thus, we can conclude that the change in momentum of the box during this time is most nearly 48 kg m/s.

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