A 4.0 kg bundle starts up a `30^(@)` incline with 150 J of kinetic energy. How far will it slide up the incline if the coefficient of kinetic friction between bundle and incline is 0.36 ?

Answer

To solve the problem, we will use the work-energy principle, which states that the work done by all forces acting on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the bundle, \( m = 4.0 \, \text{kg} \)
- Initial kinetic energy, \( KE_i = 150 \, \text{J} \)
- Angle of incline, \( \theta = 30^\circ \)
- Coefficient of kinetic friction, \( \mu = 0.36 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (approximate value)

2. **Calculate the Forces Acting on the Bundle:**
- The gravitational force acting down the incline:
\[ F_{\text{gravity}} = mg \sin(\theta) = 4.0 \times 10 \times \sin(30^\circ) = 4.0 \times 10 \times \frac{1}{2} = 20 \, \text{N} \]
- The normal force acting on the bundle:
\[ F_{\text{normal}} = mg \cos(\theta) = 4.0 \times 10 \times \cos(30^\circ) = 4.0 \times 10 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \, \text{N} \]
- The frictional force acting down the incline:
\[ F_{\text{friction}} = \mu F_{\text{normal}} = 0.36 \times 20\sqrt{3} \approx 12.47 \, \text{N} \]

3. **Calculate the Total Work Done Against Gravity and Friction:**
- The total force acting down the incline is the sum of the gravitational force and the frictional force:
\[ F_{\text{total}} = F_{\text{gravity}} + F_{\text{friction}} = 20 + 12.47 = 32.47 \, \text{N} \]
- The work done against these forces as the bundle moves a distance \( s \) up the incline is given by:
\[ W = F_{\text{total}} \cdot s \]

4. **Apply the Work-Energy Principle:**
- The work done is equal to the change in kinetic energy:
\[ W = KE_f - KE_i \]
- Since the final kinetic energy \( KE_f = 0 \) (the bundle comes to rest), we have:
\[ -W = KE_i \implies -F_{\text{total}} \cdot s = -150 \]
- Thus,
\[ F_{\text{total}} \cdot s = 150 \]

5. **Solve for Distance \( s \):**
- Rearranging the equation gives:
\[ s = \frac{150}{F_{\text{total}}} = \frac{150}{32.47} \approx 4.63 \, \text{m} \]

### Final Answer:
The bundle will slide up the incline approximately **4.63 meters** before coming to rest.