To solve this problem, we need to determine the coefficient of kinetic friction ($\mu_k$) between the object $m_1$ and the surface it is moving on. Here's the approach we will take:

Step 1: Draw Free Body Diagrams

We have two masses, $m_1$ and $m_2$, connected by a light string passing over a frictionless pulley. The given angles and accelerations will help us set up the forces on both objects.

• For mass $m_1$ (on the surface), the forces acting on it are:

  1. Gravitational force, $F_{g1} = m_1 g$, acting downward.
  2. Normal force, $N$, acting vertically upward.
  3. Frictional force, $f_k = \mu_k N$, acting opposite to the direction of motion.
  4. Tension in the string, $T$, pulling horizontally toward the pulley.

• For mass $m_2$ (hanging vertically), the forces are:

  1. Gravitational force, $F_{g2} = m_2 g$, acting downward.
  2. Tension in the string, $T$, pulling upward.

Step 2: Equations of Motion

For $m_1$, using Newton's second law: $\Sigma F_x = m_1 a \quad \text{(horizontal direction)}$ $T - f_k = m_1 a$ $T - \mu_k N = m_1 a$ Where $N = m_1 g \cos \phi$, because the normal force balances the vertical component of the gravitational force.

For $m_2$, using Newton's second law: $\Sigma F_y = m_2 a \quad \text{(vertical direction)}$ $m_2 g - T = m_2 a$

Step 3: Solve for $\mu_k$

From the equation for $m_2$, solve for the tension $T$: $T = m_2 g - m_2 a$

Now, substitute this expression for $T$ into the equation for $m_1$: $m_2 g - m_2 a - \mu_k m_1 g \cos \phi = m_1 a$ Rearranging to solve for $\mu_k$: $\mu_k = \frac{m_2 g - m_2 a - m_1 a}{m_1 g \cos \phi}$

Step 4: Substitute Given Values

Given:

• $m_1 = 6.57 \, \text{kg}$
• $m_2 = 7.30 \, \text{kg}$
• $a = 0.878 \, \text{m/s}^2$
• $g = 9.81 \, \text{m/s}^2$
• $\phi = 61^\circ$

Substitute these into the formula for $\mu_k$:

$\mu_k = \frac{(7.30)(9.81) - (7.30)(0.878) - (6.57)(0.878)}{(6.57)(9.81) \cos(61^\circ)}$

Now, calculating this expression:

$\mu_k = \frac{71.493 - 6.405 - 5.775}{6.57 \times 9.81 \times \cos(61^\circ)}$ $\mu_k = \frac{59.313}{6.57 \times 9.81 \times 0.4848}$ $\mu_k = \frac{59.313}{31.378}$ $\mu_k \approx 1.89$

Thus, the coefficient of kinetic friction required is approximately 1.89.

Do you need further explanation or details on any part of the process?

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Here are 5 questions you might find interesting to explore next:

1. How does the coefficient of friction affect the acceleration of the system?
2. What would happen if the pulley had friction?
3. How would the problem change if the surface were inclined instead of flat?
4. How does the angle $\phi$ influence the normal force and friction?
5. How can we calculate the work done by the frictional force in this system?

Tip: Always make sure to break down forces into components, especially when dealing with inclined planes or angles in problems like this. It simplifies the analysis significantly!