Let's solve the problem step by step by analyzing the forces acting on the two blocks and applying Newton's second law of motion. We'll handle both scenarios: (1) $m_1$ moving up the incline, and (2) $m_1$ moving down the incline.

Known Data:

• $m_1 = 7.5 \, \text{kg}$
• $m_2 = 24.5 \, \text{kg}$
• Angle of incline, $\theta = 52.5^\circ$
• Coefficient of kinetic friction, $\mu_k = 0.24$
• Gravitational acceleration, $g = 9.8 \, \text{m/s}^2$

Forces on $m_1$ (on the incline):

• Gravitational force component along the incline: $F_{g_1} = m_1 g \sin\theta$
• Normal force: $N_1 = m_1 g \cos\theta$
• Frictional force: $F_{\text{friction}} = \mu_k N_1 = \mu_k m_1 g \cos\theta$
• Tension in the string: $T$

Forces on $m_2$ (hanging vertically):

• Gravitational force: $F_{g_2} = m_2 g$
• Tension in the string: $T$

Scenario 1: $m_1$ Moving Up the Incline

In this case, the tension in the string must overcome both the component of $m_1$'s weight along the incline and the frictional force.

Equations for $m_1$ (moving up):

$T - F_{\text{friction}} - m_1 g \sin\theta = m_1 a$ Substitute for the friction force and component of weight: $T - \mu_k m_1 g \cos\theta - m_1 g \sin\theta = m_1 a$ $T - (0.24)(7.5)(9.8)\cos(52.5^\circ) - (7.5)(9.8)\sin(52.5^\circ) = 7.5a$

Equation for $m_2$:

Since $m_2$ is moving down: $m_2 g - T = m_2 a$ Substitute $g = 9.8 \, \text{m/s}^2$: $(24.5)(9.8) - T = 24.5a$ Solve for $T$: $T = 240.1 - 24.5a$

Solving for $a$:

Now, substitute the expression for $T$ into the equation for $m_1$:

$240.1 - 24.5a - (0.24)(7.5)(9.8)\cos(52.5^\circ) - (7.5)(9.8)\sin(52.5^\circ) = 7.5a$ Simplify and solve for $a$.

Scenario 2: $m_1$ Moving Down the Incline

In this case, the forces acting on $m_1$ are helping it move down, while friction and the tension in the string resist this motion.

Equations for $m_1$ (moving down):

$m_1 g \sin\theta - F_{\text{friction}} - T = m_1 a$ Substitute for friction and the component of weight: $(7.5)(9.8)\sin(52.5^\circ) - (0.24)(7.5)(9.8)\cos(52.5^\circ) - T = 7.5a$

Equation for $m_2$:

$T - m_2 g = -m_2 a$ Substitute $g = 9.8 \, \text{m/s}^2$: $T - (24.5)(9.8) = -24.5a$ Solve for $T$: $T = 240.1 - 24.5a$

Solving for $a$:

Substitute $T$ into the equation for $m_1$ and solve for $a$.

Let me calculate the two accelerations for both cases.The magnitude of the acceleration in each case is as follows:

1. When $m_1$ is moving up the incline, the acceleration of the system is approximately $5.35 \, \text{m/s}^2$.

2. When $m_1$ is moving down the incline, the acceleration of the system is approximately $6.02 \, \text{m/s}^2$ (in the downward direction).

Would you like further details or have any questions?

Here are 5 related questions to extend your understanding:

1. How does the angle of incline affect the acceleration of the system?
2. What would happen if there were no friction on the incline?
3. How does increasing $m_2$ impact the acceleration of the system?
4. What is the effect of reducing the coefficient of kinetic friction on the system's motion?
5. How would the system behave if the masses of $m_1$ and $m_2$ were equal?

Tip: When dealing with inclines, always break forces into components along the incline (parallel) and perpendicular to it for easy analysis!