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Let's break down and solve the problem step-by-step.

### Problem Overview:
We are asked to find the acceleration of Hachiman (whose mass is \( m_H = 60.0 \, \text{kg} \)) when two forces, \( F_Y \) and \( F_I \), are acting on him. The forces are:
- \( F_Y = 125.0 \, \text{N} \) (exerted vertically),
- \( F_I = 100.0 \, \text{N} \) (exerted horizontally).

We are required to find the acceleration of Hachiman in two cases:
- **Part A**: The forces \( F_Y \) and \( F_I \) are perpendicular (i.e., aligned along the \( y \)- and \( x \)-axes, respectively).
- **Part B**: The angle between \( F_Y \) and \( F_I \) is \( \phi = 54^\circ \).

### Part A: Forces are Perpendicular

In this case, \( F_Y \) acts along the \( y \)-axis, and \( F_I \) acts along the \( x \)-axis. The net force can be found using the Pythagorean theorem, since the forces are perpendicular:

\[
F_{\text{net}} = \sqrt{F_Y^2 + F_I^2}
\]

Substituting the given values:

\[
F_{\text{net}} = \sqrt{(125.0)^2 + (100.0)^2} = \sqrt{15625 + 10000} = \sqrt{25625}
\]

\[
F_{\text{net}} = 160.1 \, \text{N}
\]

Now, we can calculate the acceleration using Newton's second law, \( F = ma \), where \( a \) is the acceleration and \( m_H = 60.0 \, \text{kg} \):

\[
a = \frac{F_{\text{net}}}{m_H} = \frac{160.1}{60.0} \approx 2.67 \, \text{m/s}^2
\]

So, the acceleration of Hachiman in **Part A** is approximately \( 2.67 \, \text{m/s}^2 \).

### Part B: Forces are at an Angle \( \phi = 54^\circ \)

In this part, \( F_Y \) is at an angle of \( \phi = 54^\circ \) with respect to the horizontal force \( F_I \). To find the net force, we use the law of cosines:

\[
F_{\text{net}} = \sqrt{F_Y^2 + F_I^2 + 2 \cdot F_Y \cdot F_I \cdot \cos(\phi)}
\]

Substituting the given values:

\[
F_{\text{net}} = \sqrt{(125.0)^2 + (100.0)^2 + 2 \cdot 125.0 \cdot 100.0 \cdot \cos(54^\circ)}
\]

First, calculate \( \cos(54^\circ) \):

\[
\cos(54^\circ) \approx 0.5878
\]

Now substitute back into the formula:

\[
F_{\text{net}} = \sqrt{15625 + 10000 + 2 \cdot 125.0 \cdot 100.0 \cdot 0.5878}
\]
\[
F_{\text{net}} = \sqrt{15625 + 10000 + 14720}
\]
\[
F_{\text{net}} = \sqrt{40345} \approx 200.86 \, \text{N}
\]

Now, we calculate the acceleration:

\[
a = \frac{F_{\text{net}}}{m_H} = \frac{200.86}{60.0} \approx 3.35 \, \text{m/s}^2
\]

So, the acceleration of Hachiman in **Part B** is approximately \( 3.35 \, \text{m/s}^2 \).

---

### Final Answers:
- **Part A**: \( a \approx 2.67 \, \text{m/s}^2 \)
- **Part B**: \( a \approx 3.35 \, \text{m/s}^2 \)

Would you like a more detailed explanation on any specific step, or do you have any other questions?

### Related Questions:
1. How would the acceleration change if Hachiman's mass were doubled?
2. What would happen to the net force if \( F_Y \) were doubled but \( F_I \) stayed the same?
3. What is the direction of the net force in Part A?
4. How would the problem change if \( F_I \) were at a \( 54^\circ \) angle instead of \( F_Y \)?
5. Can we solve this using vector components rather than magnitudes?

### Tip:
When dealing with forces at angles, breaking them into their \( x \)- and \( y \)-components often simplifies the calculations.

Yukino and Iroha pull on Hachiman, who has a mass of 60.0 kg. Yukino exerts a vertical force (Fy) of 125.0 N, while Iroha exerts a horizontal force (FI) of 100.0 N. What is Hachiman's acceleration when the forces are perpendicular, and when the angle between them is 54 degrees?

Step-by-step solution to find the acceleration of a 60 kg object subjected to two forces: one vertical force of 125 N and a horizontal force of 100 N. Learn how to calculate acceleration when forces are perpendicular and when they are at a 54-degree angle.

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