Two objects are connected by a light string that passes over a frictionless pulley as shown in the figure below. m1 = 9.24 kg, m2 = 7.70 kg, and 𝜑 = 54°. When released from rest, m1 accelerates downward at 1.391 m/s2. For this to happen, the coefficient of kinetic friction must be 
0.134
 
Incorrect: Your answer is incorrect.
 , and to even begin sliding in the first place the coefficient of static friction must be less 
Correct: Your answer is correct.
 than 
1.134
 
Incorrect: Your answer is incorrect.
 . Use g=9.8 m/s2.


This problem involves two masses connected by a string passing over a frictionless pulley, with one mass on an incline. We need to find the coefficient of kinetic friction \(\mu_k\) and verify the coefficient of static friction \(\mu_s\).

Given:
- \(m_1 = 9.24 \, \text{kg}\)
- \(m_2 = 7.70 \, \text{kg}\)
- Incline angle \(\phi = 54^\circ\)
- Acceleration \(a = 1.391 \, \text{m/s}^2\)
- Gravitational acceleration \(g = 9.8 \, \text{m/s}^2\)

### Step 1: Determine the forces on each mass

#### For \(m_1\) (hanging mass):
The forces acting on \(m_1\) are:
- Weight: \(m_1 g\)
- Tension in the string: \(T\)

The equation of motion for \(m_1\), accelerating downwards:
\[
m_1 g - T = m_1 a
\]

#### For \(m_2\) (on the incline):
The forces acting on \(m_2\) are:
- Weight: \(m_2 g\)
- Normal force: \(N\)
- Tension in the string: \(T\)
- Kinetic friction force: \(f_k = \mu_k N\)

The components of the forces:
- The gravitational force acting parallel to the incline: \(m_2 g \sin \phi\)
- The gravitational force acting perpendicular to the incline: \(m_2 g \cos \phi\)

The equation of motion for \(m_2\), accelerating upwards (along the incline):
\[
T - m_2 g \sin \phi - f_k = m_2 a
\]
Substitute \(f_k = \mu_k N\), where \(N = m_2 g \cos \phi\):
\[
T - m_2 g \sin \phi - \mu_k m_2 g \cos \phi = m_2 a
\]

### Step 2: Solve for \(T\)
From the equation for \(m_1\):
\[
T = m_1 g - m_1 a
\]
Substitute the given values:
\[
T = 9.24 \times 9.8 - 9.24 \times 1.391 = 90.552 - 12.843 \approx 77.709 \, \text{N}
\]

### Step 3: Solve for \(\mu_k\)
Substitute \(T\) into the equation for \(m_2\):
\[
77.709 - 7.70 \times 9.8 \times \sin 54^\circ - \mu_k \times 7.70 \times 9.8 \times \cos 54^\circ = 7.70 \times 1.391
\]
First, calculate the trigonometric and other values:
- \(\sin 54^\circ \approx 0.809\)
- \(\cos 54^\circ \approx 0.587\)
- \(7.70 \times 9.8 \times \sin 54^\circ \approx 61.071\)
- \(7.70 \times 9.8 \times \cos 54^\circ \approx 44.413\)
- \(7.70 \times 1.391 \approx 10.711\)

Now substitute these into the equation:
\[
77.709 - 61.071 - \mu_k \times 44.413 = 10.711
\]
Simplify:
\[
16.638 - \mu_k \times 44.413 = 10.711
\]
\[
\mu_k \times 44.413 = 16.638 - 10.711 = 5.927
\]
\[
\mu_k = \frac{5.927}{44.413} \approx 0.134
\]

### Step 4: Solve for the coefficient of static friction (\(\mu_s\))

To prevent \(m_2\) from sliding initially, the static friction must balance the forces along the incline. The maximum static friction force is given by:
\[
f_s = \mu_s N = \mu_s m_2 g \cos \phi
\]
Setting the maximum static friction equal to the parallel component of gravity:
\[
\mu_s m_2 g \cos \phi = m_2 g \sin \phi
\]
\[
\mu_s = \frac{\sin \phi}{\cos \phi} = \tan \phi
\]
\[
\mu_s = \tan 54^\circ \approx 1.376
\]

Thus, the coefficient of static friction must be **less than 1.376** to allow sliding to begin.

### Final Answers:
- The coefficient of kinetic friction \(\mu_k\) is approximately **0.134**.
- The coefficient of static friction \(\mu_s\) is less than **1.376**.

Would you like more details or have any questions?

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Here are 5 related questions:
1. How would increasing the angle of the incline affect the required kinetic friction coefficient?
2. What happens if the frictionless pulley had friction?
3. How would the system behave if \(m_1\) was heavier than \(m_2\)?
4. How is the normal force on \(m_2\) related to the incline angle?
5. How would the motion change if the pulley were not frictionless?

**Tip:** Always break down forces into components parallel and perpendicular to the incline when solving such problems.

Two objects are connected by a light string that passes over a frictionless pulley as shown in the figure below. m1 = 9.24 kg, m2 = 7.70 kg, and 𝜑 = 54°. When released from rest, m1 accelerates downward at 1.391 m/s^2. For this to happen, the coefficient of kinetic friction must be 0.134, and to even begin sliding in the first place the coefficient of static friction must be less than 1.134. Use g=9.8 m/s^2.

Analyze a two-mass system where m1 = 9.24 kg and m2 = 7.70 kg, connected by a string over a pulley, and solve for the kinetic and static friction coefficients. This problem involves using Newton's laws, frictional forces, and trigonometry to find the required coefficients for m1's acceleration of 1.391 m/s^2 on a 54° incline. Perfect for advanced high school physics students.

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