The gravitational force component along the incline is given by:
\[ F_{\text{gravity}} = mg \sin \theta \]
where:
- \( m = 100 \, \text{kg} \) (mass of the log)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( \theta = 15^\circ \) (angle of the incline)
\[ F_{\text{gravity}} = 100 \times 9.81 \times \sin(15^\circ) \]
\[ F_{\text{gravity}} = 100 \times 9.81 \times 0.2588 \]
\[ F_{\text{gravity}} = 253.8 \, \text{N} \]
The normal force is given by:
\[ F_{\text{normal}} = mg \cos \theta \]
\[ F_{\text{normal}} = 100 \times 9.81 \times \cos(15^\circ) \]
\[ F_{\text{normal}} = 100 \times 9.81 \times 0.9659 \]
\[ F_{\text{normal}} = 947.0 \, \text{N} \]
The frictional force is given by:
\[ F_{\text{friction}} = \mu F_{\text{normal}} \]
For the first 100 meters (\( \mu_1 = 0.33 \)):
\[ F_{\text{friction1}} = 0.33 \times 947.0 \]
\[ F_{\text{friction1}} = 312.5 \, \text{N} \]
For the next 100 meters (\( \mu_2 = 0.67 \)):
\[ F_{\text{friction2}} = 0.67 \times 947.0 \]
\[ F_{\text{friction2}} = 634.5 \, \text{N} \]
The total force required to pull the log is the sum of the gravitational force component and the frictional force for each segment.
For the first 100 meters:
\[ F_{\text{total1}} = F_{\text{gravity}} + F_{\text{friction1}} \]
\[ F_{\text{total1}} = 253.8 + 312.5 \]
\[ F_{\text{total1}} = 566.3 \, \text{N} \]
For the next 100 meters:
\[ F_{\text{total2}} = F_{\text{gravity}} + F_{\text{friction2}} \]
\[ F_{\text{total2}} = 253.8 + 634.5 \]
\[ F_{\text{total2}} = 888.3 \, \text{N} \]
The work done is given by:
\[ W = F \times d \]
For the first 100 meters:
\[ W_1 = F_{\text{total1}} \times 100 \]
\[ W_1 = 566.3 \times 100 \]
\[ W_1 = 56630 \, \text{J} \]
For the next 100 meters:
\[ W_2 = F_{\text{total2}} \times 100 \]
\[ W_2 = 888.3 \times 100 \]
\[ W_2 = 88830 \, \text{J} \]
The total work done is the sum of the work done for each segment:
\[ W_{\text{total}} = W_1 + W_2 \]
\[ W_{\text{total}} = 56630 + 88830 \]
\[ W_{\text{total}} = 145460 \, \text{J} \]
\[
\boxed{W_{\text{total}} = 145460 \, \text{J}}
\]
Answered
