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 __________.

Answer 1

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.