To solve this problem, we will analyze the forces acting on both objects and then use Newton's Second Law to find the tension in the cord. Here's how we break it down:

1. Identify the forces acting on each object:

For Object A (on the incline):

• Weight ($W_A$): The weight of object A is $W_A = m_A \cdot g$, where $m_A = 13.1 \, \text{kg}$ and $g = 9.8 \, \text{m/s}^2$.

  • The component of this weight acting parallel to the incline is $W_{A, \parallel} = m_A \cdot g \cdot \sin(\theta)$.
  • The component acting perpendicular to the incline is $W_{A, \perp} = m_A \cdot g \cdot \cos(\theta)$.

• Friction force ($F_f$): This force opposes the motion and is given by $F_f = \mu_k \cdot N$, where $\mu_k = 0.17$ is the coefficient of kinetic friction and $N$ is the normal force. Since the normal force is the perpendicular component of the weight, we have $N = W_{A, \perp}$.

• Tension force ($T$): This is the tension in the cord, acting up the incline.

For Object B (hanging vertically):

• Weight ($W_B$): The weight of object B is $W_B = m_B \cdot g$, where $m_B = 1.6 \, \text{kg}$.

• Tension force ($T$): The tension in the cord, acting upward, opposes the weight of object B.

2. Write the equations of motion:

For Object A (along the incline):

Using Newton's Second Law, the net force acting along the incline on object A is: $T - F_f - W_{A, \parallel} = m_A \cdot a$ Where:

• $F_f = \mu_k \cdot m_A \cdot g \cdot \cos(\theta)$
• $W_{A, \parallel} = m_A \cdot g \cdot \sin(\theta)$

Thus: $T - (\mu_k \cdot m_A \cdot g \cdot \cos(\theta)) - (m_A \cdot g \cdot \sin(\theta)) = m_A \cdot a$

For Object B (vertically):

Similarly, for object B, the equation of motion is: $W_B - T = m_B \cdot a$ Substituting $W_B = m_B \cdot g$: $m_B \cdot g - T = m_B \cdot a$

3. Solve the system of equations:

From the equations for object A and object B, we can solve for $a$ (the acceleration) and $T$ (the tension in the cord). Let's do this calculation now.The tension in the cord is approximately 23.02 N. The negative acceleration indicates that the system is moving in the direction of object B (the hanging mass), meaning object A is sliding up the incline while object B is falling downward.

Would you like further details on how the tension was derived, or have any questions?

Here are five related questions to explore:

1. How would the tension change if the angle of the incline increased?
2. What would happen to the system if there were no friction?
3. How would the system behave if the masses were equal?
4. How does the coefficient of friction affect the acceleration of object A?
5. What would be the tension if object B's mass increased?

Tip: Always break down inclined plane problems into parallel and perpendicular components for easier analysis.