A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitcase slides 2.3 m before stopping. Use work and energy to find the suitcase’s coefficient of kinetic friction on the floor.

To solve this problem, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. Since the suitcase stops, its final kinetic energy will be zero, and the initial kinetic energy will be converted to work done against friction.

### Step-by-Step Solution

1. **Identify the Initial Kinetic Energy (KE) of the Suitcase:**

   The initial kinetic energy of the suitcase is given by:
   \[
   KE = \frac{1}{2} m v^2
   \]
   where:
   - \( m = 15 \, \text{kg} \)
   - \( v = 1.2 \, \text{m/s} \)

   Substituting values:
   \[
   KE = \frac{1}{2} \times 15 \, \text{kg} \times (1.2 \, \text{m/s})^2
   \]
   \[
   KE = \frac{1}{2} \times 15 \times 1.44
   \]
   \[
   KE = 10.8 \, \text{J}
   \]

2. **Calculate the Work Done by Friction (W):**

   The work done by friction is equal to the force of friction times the distance over which it acts:
   \[
   W = f_k \cdot d
   \]
   where:
   - \( f_k \) is the kinetic friction force, which can be expressed as \( f_k = \mu_k \cdot N \)
   - \( d = 2.3 \, \text{m} \) (the distance the suitcase slides)

   Since the suitcase is sliding on a horizontal surface, the normal force \( N \) is equal to the gravitational force \( mg \). Thus, \( f_k = \mu_k mg \).

   Therefore, the work done by friction becomes:
   \[
   W = \mu_k mg \cdot d
   \]

3. **Set Up the Work-Energy Equation:**

   According to the work-energy theorem, the work done by friction (which is negative because it opposes motion) is equal to the initial kinetic energy:
   \[
   -\mu_k mg \cdot d = -KE
   \]

   Simplifying:
   \[
   \mu_k mg \cdot d = KE
   \]

   Solving for \( \mu_k \):
   \[
   \mu_k = \frac{KE}{mg \cdot d}
   \]

4. **Plug in the Values:**

   Substituting the values:
   \[
   \mu_k = \frac{10.8 \, \text{J}}{15 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 2.3 \, \text{m}}
   \]
   \[
   \mu_k = \frac{10.8}{338.1}
   \]
   \[
   \mu_k \approx 0.032
   \]

### Final Answer
The coefficient of kinetic friction \( \mu_k \) is approximately \( 0.032 \).

Would you like further details or explanations?

---

1. How does the work-energy theorem apply to frictional forces?
2. What other methods could we use to find the coefficient of friction if the work-energy method was not an option?
3. Why is it important to consider the direction of frictional work in these calculations?
4. How would the coefficient of friction change if the initial velocity of the suitcase were different?
5. How does this approach change if the suitcase is moving up or down an inclined plane?

**Tip:** Always confirm that forces like gravity and friction are accurately represented in your energy equations to avoid miscalculations in work or energy.

Find the coefficient of kinetic friction for a 15 kg suitcase sliding 2.3 m with an initial speed of 1.2 m/s using the work-energy principle. This solution covers physics fundamentals like kinetic energy, frictional force, and work-energy application, ideal for Grades 11-12.

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