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The problem involves two packages sliding down a ramp. Package A has a mass of 5.0 kg, and Package B has a mass of 15.0 kg. The coefficient of friction for Package B is 0.25, and the ramp is inclined at an angle of 20 degrees. You are asked to find the maximum coefficient of friction for Package A that allows both packages to slide down the 2.0-meter ramp.

### Step-by-step solution:

1. **Forces acting on each package:**
   - Both packages are subject to gravitational force, normal force, and frictional force.
   - The component of gravity acting down the incline is \( F_{\text{gravity}} = mg\sin(\theta) \).
   - The frictional force is \( F_{\text{friction}} = \mu mg\cos(\theta) \).

2. **For Package B (mass = 15.0 kg, \( \mu_B = 0.25 \)):**
   - Gravitational force down the incline: 
     \[
     F_{\text{gravity, B}} = m_B g \sin(20^\circ) = 15 \cdot 9.8 \cdot \sin(20^\circ)
     \]
     \[
     F_{\text{gravity, B}} \approx 15 \cdot 9.8 \cdot 0.342 = 50.2 \, \text{N}
     \]
   - Frictional force:
     \[
     F_{\text{friction, B}} = \mu_B m_B g \cos(20^\circ) = 0.25 \cdot 15 \cdot 9.8 \cdot \cos(20^\circ)
     \]
     \[
     F_{\text{friction, B}} \approx 0.25 \cdot 15 \cdot 9.8 \cdot 0.94 = 34.6 \, \text{N}
     \]

3. **Net force on Package B:**
   \[
   F_{\text{net, B}} = F_{\text{gravity, B}} - F_{\text{friction, B}} = 50.2 \, \text{N} - 34.6 \, \text{N} = 15.6 \, \text{N}
   \]

4. **For Package A (mass = 5.0 kg):**
   - The condition is that both packages slide together, so the net force on Package A should equal the net force on Package B (since they move together).

   The friction coefficient \( \mu_A \) is what we need to find.

5. **Net force on Package A:**
   The gravitational force acting down the incline is:
   \[
   F_{\text{gravity, A}} = m_A g \sin(20^\circ) = 5 \cdot 9.8 \cdot \sin(20^\circ)
   \]
   \[
   F_{\text{gravity, A}} \approx 5 \cdot 9.8 \cdot 0.342 = 17.0 \, \text{N}
   \]

   The frictional force is:
   \[
   F_{\text{friction, A}} = \mu_A m_A g \cos(20^\circ) = \mu_A \cdot 5 \cdot 9.8 \cdot 0.94
   \]
   \[
   F_{\text{friction, A}} = 46.06 \cdot \mu_A
   \]

6. **Setting the net force on Package A equal to Package B:**
   \[
   F_{\text{net, A}} = F_{\text{gravity, A}} - F_{\text{friction, A}} = 17.0 - 46.06 \cdot \mu_A
   \]
   This net force should equal the net force on Package B, which is 15.6 N:
   \[
   17.0 - 46.06 \cdot \mu_A = 15.6
   \]
   Solving for \( \mu_A \):
   \[
   17.0 - 15.6 = 46.06 \cdot \mu_A
   \]
   \[
   1.4 = 46.06 \cdot \mu_A
   \]
   \[
   \mu_A = \frac{1.4}{46.06} \approx 0.0304
   \]

Thus, the maximum coefficient of friction for Package A is approximately **0.0304**, which does not match any of the provided answer choices. However, this might imply a mistake or a different interpretation of the problem. Would you like to discuss further details or reconsider other aspects?

---

Here are some related questions for further understanding:

1. How does increasing the incline angle affect the frictional forces?
2. What happens if the coefficient of friction for Package B changes to a different value?
3. How would the result change if Package A had a higher mass than Package B?
4. Can we calculate the time it takes for the packages to slide down the ramp?
5. What if there was no friction at all? How would that change the motion of both packages?

**Tip:** Always check the units of your forces, masses, and accelerations to ensure consistent calculations.

Two packages at UPS are sliding down the 20-degree ramp shown below. Package A has a mass of 5.0 kg. Package B has a mass of 15.0 kg and a coefficient of friction of 0.25. What is the maximum coefficient of friction for package A that allows the packages to reach the bottom of the 2.0-meter slope?

Learn how to calculate the maximum coefficient of friction for a package sliding down a 20-degree incline with detailed steps. This physics problem covers key concepts in Newton's laws of motion, frictional forces, and force equilibrium on inclined planes. Suitable for high school students in grades 10-12.

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