To determine the average external force that acted on the mass during the given time interval, we can apply Newton's second law of motion, which is:

$F_{\text{avg}} = \frac{\Delta p}{\Delta t}$

where:

• $F_{\text{avg}}$ is the average external force,
• $\Delta p$ is the change in momentum,
• $\Delta t$ is the change in time.

Step 1: Calculate the initial and final momentum

Momentum ($p$) is given by the equation:

$p = m \cdot v$

where:

• $m$ is the mass,
• $v$ is the velocity.

Initial momentum:

The initial velocity is $5 \, \text{m/s}$ due north. With a mass of $4.00 \, \text{kg}$:

$p_{\text{initial}} = 4.00 \, \text{kg} \times 5 \, \text{m/s} = 20 \, \text{kg} \cdot \text{m/s} \, \text{(north)}$

Final momentum:

The final velocity is $2 \, \text{m/s}$ due south, so:

$p_{\text{final}} = 4.00 \, \text{kg} \times (-2) \, \text{m/s} = -8 \, \text{kg} \cdot \text{m/s} \, \text{(south)}$

Step 2: Calculate the change in momentum

$\Delta p = p_{\text{final}} - p_{\text{initial}} = (-8) - (20) = -28 \, \text{kg} \cdot \text{m/s}$

Step 3: Calculate the average force

The time interval is $\Delta t = 6.00 \, \text{seconds}$. Using the formula for average force:

$F_{\text{avg}} = \frac{-28 \, \text{kg} \cdot \text{m/s}}{6.00 \, \text{seconds}} = -4.67 \, \text{N}$

The negative sign indicates the force is directed towards the south, opposite to the initial direction of motion.

Thus, the magnitude of the average external force is:

$F_{\text{avg}} = 4.67 \, \text{N} \, \text{(south)}$

Final Answer:

The average external force that acted on the mass is $4.67 \, \text{N}$ towards the south.

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Would you like further details or explanations? Here are 5 related questions for practice:

1. What would the average force be if the time interval was halved?
2. How would the result change if there was friction involved?
3. What is the total displacement of the mass during the 6.00-second interval?
4. How would the problem change if the mass had a varying speed over time?
5. What is the significance of the direction of force in momentum change?

Tip: Always remember that momentum is a vector quantity, meaning both magnitude and direction are important in calculating changes.