A car of mass $$\(1600 \mathrm{~kg}\)$$ travels at constant speed $$\(20 \mathrm{~m} \mathrm{~s}^{-1}\)$$ up a straight road inclined at an angle of $$\(\sin ^{-1} 0.12\)$$ to the horizontal. Calculate, in $$\(\mathrm{kW}\)$$, the power developed by the engine of the car. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_w21_qp_41 Year：2021 Question No：5(c)
Answer:
\(P=\left(\frac{4040}{3}+1600 \times g \times 0.12\right) \times 20\)
\({\left[=\frac{196000}{3}\right] }\)
\(P=65.3 \mathrm{~kW}\)
Alternative method for question
\(P=\frac{1960000}{30}\)
\(P=65.3 \mathrm{~kW}\)
Alternative method for question
\(P=\frac{9800}{3} \times 20\)
\(P=65.3 \mathrm{~kW}\)
\({\left[=\frac{196000}{3}\right] }\)
\(P=65.3 \mathrm{~kW}\)
Alternative method for question
\(P=\frac{1960000}{30}\)
\(P=65.3 \mathrm{~kW}\)
Alternative method for question
\(P=\frac{9800}{3} \times 20\)
\(P=65.3 \mathrm{~kW}\)
Knowledge points:
4.5.4 use the definition of power as the rate at which a force does work, and use the relationship between power, force and velocity for a force acting in the direction of motion Including calculation of (average) power;P = Fv
Solution:
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