Answer:
Total energy \[=3\times
{{10}^{7}}\,J\]
Efficiency
of car engine = 0.5
\ Output or energy used to drive car
\[=\text{efficiency
}\times 3\times {{10}^{7}}J=0.5\times 3\times {{10}^{7}}J\]
\[=1.5\times
{{10}^{7}}J\]
Thus,
energy used to drive car by the combustion of 1 litre petrol \[=1.5\times
{{10}^{7}}J\]
Work
done = energy used
If
F = force of friction acting on the car
Then
\[F\times S=1.5\times {{10}^{7}}\]
\[F=\frac{1.5\,\times
\,{{10}^{7}}}{S}\,\,=\,\frac{1.5\times {{10}^{7}}}{15\,\times {{10}^{3}}}\]
\[(\because
\,\,S=15km)\]
\[={{10}^{3}}N.\]
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