A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction bctwecn thc box and the floor is 0.20. The box moves n 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 (A) 12 kg-m/s (B) 48 kg-m/s (C 60 kg-m/s (D) 96 kg-m/s

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AmeerAbdullah AmeerAbdullah · Answer 1 · 2020-03-16T01:37:19+01:00

Answer:

(D) 96 kg-m/s

Explanation:

Let's start off by first calculating the normal force between the box and the floor.

This will be:

Normal Force = 12 * 9.81 = 117.72 N

We can now use the friction equation to find the frictional force on the box when it is moving:

Frictional force = Coefficient of friction * Normal Force

Frictional force = 0.4 * 117.72 = 47.09 N

Finally, since we have the force on the box, we can find the acceleration:

F = Mass * Acceleration

47.09 = 12 * Acceleration

Acceleration = 3.92 m/s^2

Final speed after 2 seconds:

[tex]V=U+a*t[/tex]

[tex]V = 4 +(-3.92)*(2)[/tex]

V= -3.84 m/s

Since we know the initial and final speeds, we can calculate the change in momentum:

Change in momentum = Final Momentum - Initial Momentum

Change in momentum = [tex]3.84*12-(-4)*12[/tex]

Change in momentum = 94.08 kg*m/s

Thus we can see that option (D) is the closest answer.