A 63-kg skier goes up a snow-convered hill that makes an angle of `25^(@)` with the horizontal . The initial speed of the skier is 6.6 m/s. After coasting a distance of 1.9 m up the slope, the speed of the skier is 4.4 m/s. Find the work done by the kinetic frictional force that acts on the skis .

Options

A. `-760`J
B. `+270`J
C. `+300`J
D. `-270`J

Answer

To find the work done by the kinetic frictional force acting on the skier, we can use the work-energy principle. According to this principle, the work done by all forces acting on an object is equal to the change in its kinetic energy plus the change in its potential energy.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Mass of the skier, \( m = 63 \, \text{kg} \)
- Initial speed, \( v_i = 6.6 \, \text{m/s} \)
- Final speed, \( v_f = 4.4 \, \text{m/s} \)
- Distance traveled up the slope, \( d = 1.9 \, \text{m} \)
- Angle of the slope, \( \theta = 25^\circ \)

2. **Calculate Initial Kinetic Energy (KE_initial):**
\[ KE_{\text{initial}} = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 63 \times (6.6)^2 \]
\[ KE_{\text{initial}} = \frac{1}{2} \times 63 \times 43.56 = 1377.78 \, \text{J} \]

3. **Calculate Final Kinetic Energy (KE_final):**
\[ KE_{\text{final}} = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 63 \times (4.4)^2 \]
\[ KE_{\text{final}} = \frac{1}{2} \times 63 \times 19.36 = 609.36 \, \text{J} \]

4. **Calculate Change in Potential Energy (PE):**
The change in potential energy when the skier goes up the slope is given by:
\[ PE = mgh = mgd \sin(\theta) \]
where \( g = 10 \, \text{m/s}^2 \) (approximate value for gravitational acceleration).
\[ PE = 63 \times 10 \times 1.9 \times \sin(25^\circ) \]
First, calculate \( \sin(25^\circ) \):
\[ \sin(25^\circ) \approx 0.4226 \]
Now, substituting the values:
\[ PE = 63 \times 10 \times 1.9 \times 0.4226 \approx 507.87 \, \text{J} \]

5. **Apply the Work-Energy Principle:**
According to the work-energy principle:
\[ W_{\text{friction}} + KE_{\text{initial}} = KE_{\text{final}} + PE \]
Rearranging gives:
\[ W_{\text{friction}} = KE_{\text{final}} + PE - KE_{\text{initial}} \]
Substituting the values:
\[ W_{\text{friction}} = 609.36 + 507.87 - 1377.78 \]
\[ W_{\text{friction}} = 1117.23 - 1377.78 = -260.55 \, \text{J} \]

6. **Final Result:**
The work done by the kinetic frictional force is approximately:
\[ W_{\text{friction}} \approx -270 \, \text{J} \]