To find the magnitude of the acceleration on the box, we need to analyze the forces acting on it and apply Newton's Second Law of Motion, $F = ma$, where:

• $F$ is the net force acting on the box.
• $m$ is the mass of the box.
• $a$ is the acceleration of the box.

Step 1: Breakdown of forces

• The weight of the box $W = mg$, where $m = 75 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$. $W = 75 \times 9.8 = 735 \, \text{N}$
• One friend applies a vertical force of $500 \, \text{N}$.
• The other friend applies a force of $515 \, \text{N}$ at a 45-degree angle.

Step 2: Resolve the angled force into components

The second friend’s force can be broken into vertical and horizontal components:

• Vertical component: $515 \sin 45^\circ = 515 \times \frac{\sqrt{2}}{2} \approx 364 \, \text{N}$
• Horizontal component: $515 \cos 45^\circ = 515 \times \frac{\sqrt{2}}{2} \approx 364 \, \text{N}$

Step 3: Calculate the net vertical force

The net vertical force $F_{\text{net}}^{\text{vertical}}$ acting on the box is the sum of the two vertical forces minus the weight of the box: $F_{\text{net}}^{\text{vertical}} = 500 \, \text{N} + 364 \, \text{N} - 735 \, \text{N} = 129 \, \text{N}$ This is the force that will cause the box to accelerate vertically.

Step 4: Calculate the vertical acceleration

Now, apply Newton’s Second Law to find the vertical acceleration $a_{\text{vertical}}$: $a_{\text{vertical}} = \frac{F_{\text{net}}^{\text{vertical}}}{m} = \frac{129}{75} \approx 1.72 \, \text{m/s}^2$

Step 5: Calculate the horizontal acceleration

The horizontal component of the second friend's force causes a horizontal acceleration. Since no horizontal force is resisting the movement, the horizontal acceleration $a_{\text{horizontal}}$ is: $a_{\text{horizontal}} = \frac{F_{\text{horizontal}}}{m} = \frac{364}{75} \approx 4.85 \, \text{m/s}^2$

Step 6: Find the magnitude of the total acceleration

To find the total acceleration, we use the Pythagorean theorem: $a_{\text{total}} = \sqrt{a_{\text{vertical}}^2 + a_{\text{horizontal}}^2} = \sqrt{(1.72)^2 + (4.85)^2} \approx \sqrt{2.96 + 23.52} \approx \sqrt{26.48} \approx 5.15 \, \text{m/s}^2$

Final Answer:

The magnitude of the total acceleration on the box is approximately $5.15 \, \text{m/s}^2$.

---

Do you need more details or have any questions about the solution?

Here are 5 related questions to consider:

1. What would happen to the acceleration if the mass of the box increased to 100 kg?
2. How would the result change if the angle of the second friend's lift were 30 degrees instead of 45 degrees?
3. What is the vertical force exerted by each friend on the box if both lifted it straight up?
4. How does the horizontal force affect the overall acceleration of the box?
5. How would the forces change if gravity were different, such as on the Moon?

Tip: When resolving forces into components, always remember to break angled forces into perpendicular parts to simplify