Question 2 (2 points,
In the figure, a force F is applied to a FedEx box of mass m2 = 1.0 kg. 1. The force is directed up a plane tilted by 0 = 42° The box is connected by cord to a UPS box of mass F n 3.0 kg on the floor. The floor and plane are frictionless. The pulley is frictionless, and the masses of the pulley and cord are negligible. What is the acceleration of the system if the applied force is 18.5 N?

This problem involves a system with two masses connected by a pulley. One mass (\(m_1 = 3.0 \, \text{kg}\)) is on a horizontal frictionless surface, and the other mass (\(m_2 = 1.0 \, \text{kg}\)) is on an inclined plane tilted at an angle \(\theta = 42^\circ\). The goal is to find the acceleration of the system when a force \(F = 18.5 \, \text{N}\) is applied to the box on the incline. Here's the step-by-step approach:

### Step 1: Free-body diagrams

- **For the UPS box on the floor (\(m_1 = 3.0 \, \text{kg}\)):**  
  This mass is on a horizontal frictionless surface. The only force acting on it in the direction of motion is the tension \(T\) in the cord connecting it to the other box.
  
  Newton's Second Law for this mass is:
  \[
  m_1 a = T
  \]
  where \(a\) is the acceleration of the system.

- **For the FedEx box on the incline (\(m_2 = 1.0 \, \text{kg}\)):**  
  The forces acting on this box are:
  1. The applied force \(F = 18.5 \, \text{N}\) directed up the incline.
  2. The gravitational force \(m_2 g\) (which can be split into components parallel and perpendicular to the incline). The component of gravity acting down the incline is \(m_2 g \sin \theta\).
  3. The tension \(T\) in the cord pulling the box down the incline.
  
  Newton's Second Law for this mass in the direction along the incline is:
  \[
  m_2 a = F - T - m_2 g \sin \theta
  \]

### Step 2: Set up the system of equations
We now have two equations:
1. \( m_1 a = T \)
2. \( m_2 a = F - T - m_2 g \sin \theta \)

We can substitute \(T\) from the first equation into the second:

\[
m_2 a = F - m_1 a - m_2 g \sin \theta
\]

### Step 3: Solve for \(a\)
Rearranging the equation for \(a\):

\[
a(m_1 + m_2) = F - m_2 g \sin \theta
\]

\[
a = \frac{F - m_2 g \sin \theta}{m_1 + m_2}
\]

### Step 4: Plug in the values
Given:
- \(m_1 = 3.0 \, \text{kg}\)
- \(m_2 = 1.0 \, \text{kg}\)
- \(\theta = 42^\circ\)
- \(F = 18.5 \, \text{N}\)
- \(g = 9.8 \, \text{m/s}^2\)

We calculate the acceleration \(a\):

\[
a = \frac{18.5 - (1.0)(9.8) \sin 42^\circ}{3.0 + 1.0}
\]

Let's calculate the final answer.The acceleration of the system is approximately \(2.99 \, \text{m/s}^2\).

Would you like further details on how this was derived, or any clarification? Here are five related questions you can explore:

1. How would the problem change if there was friction on the incline?
2. What if the angle of the incline was different, say 30° or 60°?
3. How does the tension in the cord compare to the force applied?
4. What would happen if the applied force was less than the force needed to overcome gravity on the incline?
5. How would the system behave if both masses were placed on the horizontal surface?

**Tip:** In problems involving inclined planes, always break forces into parallel and perpendicular components to simplify the analysis.

In the figure, a force F is applied to a FedEx box of mass m2 = 1.0 kg. The force is directed up a plane tilted by 0 = 42°. The box is connected by cord to a UPS box of mass m1 = 3.0 kg on the floor. The floor and plane are frictionless. The pulley is frictionless, and the masses of the pulley and cord are negligible. What is the acceleration of the system if the applied force is 18.5 N?

Learn how to calculate the acceleration of a system involving two masses connected by a pulley, one on a frictionless inclined plane and the other on a horizontal surface. Apply Newton's Second Law, use free-body diagrams, and solve using trigonometric functions. Suitable for high school students in Grades 10-12.

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