To solve the problem, let’s break it into parts:

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Given Information:

• Mass of box 1 ($m_1$) = $4.00 \, \text{kg}$
• Mass of box 2 ($m_2$) = $6.00 \, \text{kg}$
• Acceleration ($a$) = $5.00 \, \text{m/s}^2$ (same for both boxes since they are connected)
• Surface is frictionless.

a) Acceleration of the 4.00 kg box

Since the boxes are connected, they share the same acceleration:

$a = 5.00 \, \text{m/s}^2$

Thus, the acceleration of the 4.00 kg box is also:

$\boxed{5.00 \, \text{m/s}^2}$

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b) Free-body diagram for the 4.00 kg box and tension $T$:

The forces acting on the 4.00 kg box are:

1. Tension $T$: Acts to the right, pulling the box.
2. Net Force $F_{\text{net}}$: Causes acceleration.

From Newton’s second law:

$F_{\text{net}} = m_1 a$

Substitute values:

$F_{\text{net}} = (4.00 \, \text{kg})(5.00 \, \text{m/s}^2) = 20.0 \, \text{N}$

The net force on the 4.00 kg box is equal to the tension $T$, so:

$T = 20.0 \, \text{N}$

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c) Free-body diagram for the 6.00 kg box:

The forces acting on the 6.00 kg box are:

1. Tension $T$: Acts to the left, pulling the box.
2. Applied force $F$: Acts to the right.
3. Net Force $F_{\text{net}}$: Causes acceleration.

Direction of the net force: To the right (same as $F$).

Comparison of forces $T$ and $F$:

• $F$ is greater than $T$ because $F$ is responsible for accelerating the combined system, while $T$ only accelerates the 4.00 kg box.

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d) Calculate the magnitude of $F$:

Using Newton’s second law for the 6.00 kg box:

$F_{\text{net}} = F - T$

$m_2 a = F - T$

Substitute values:

$(6.00 \, \text{kg})(5.00 \, \text{m/s}^2) = F - 20.0 \, \text{N}$

$30.0 \, \text{N} = F - 20.0 \, \text{N}$

Solve for $F$:

$F = 50.0 \, \text{N}$

Thus, the applied force $F$ is:

$\boxed{50.0 \, \text{N}}$

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Summary of Results:

1. Acceleration of the 4.00 kg box: $5.00 \, \text{m/s}^2$
2. Tension in the rope ($T$): $20.0 \, \text{N}$
3. Net force direction for the 6.00 kg box: Right
4. Magnitude of applied force ($F$): $50.0 \, \text{N}$

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Would you like further clarification or additional diagrams? 😊

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Follow-Up Questions:

1. How does the acceleration change if the rope’s mass is considered?
2. What would happen to the tension if friction were present?
3. How would the forces change if the applied force $F$ were halved?
4. What if the masses of the boxes were swapped?
5. How does the free-body diagram illustrate the system’s motion?

Tip:

Always start with a clear free-body diagram to visualize all forces acting on an object. This simplifies applying Newton’s laws!