On a level road stretch of s = 3 km a car of mass m = 1.5 t…

On a level road stretch of $s = 3\ \mathrm{km}$ a car of mass $m = 1.5\ \mathrm{t}$ speeds up from $v_1 = 36\ \mathrm{km/h}$ to $v_2 = 72\ \mathrm{km/h}$; the rolling friction coefficient is $\mu = 0.02$. Find (a) the engine's work on this stretch, (b) its average power (assuming uniform acceleration).

Explanation of Method

Engine work and power while speeding up.

Step 1: (a) Work

\[ \Delta E_k = \tfrac12 \cdot 1500\,(20^2 - 10^2) = 225\ \mathrm{kJ}, \]

\[ W_f = \mu mg\,s = 0.02 \cdot 14\,715 \cdot 3000 = 882.9\ \mathrm{kJ}. \]

\[W = \Delta E_k + W_f \approx 1108\ \mathrm{kJ} = 1.11\ \mathrm{MJ}\]

Step 2: (b) Average power

\[ t = \frac{s}{(v_1 + v_2)/2} = \frac{3000}{15} = 200\ \mathrm{s}, \]

\[ \bar P = \frac{W}{t}. \]

\[\bar P \approx 5.54\ \mathrm{kW}\]

$\Delta E_k = 225$ $\text{friction } 882.9$ $\text{engine work}$ $1107.9\ \mathrm{kJ}$ $t = \frac{s}{\bar v} = \frac{3000}{15} = 200\ \mathrm{s}$ $\bar P = \frac{W}{t} \approx 5.54\ \mathrm{kW}$

Where the engine's work goes on the 3 km stretch (kJ). Four fifths are spent against friction; only a fifth increases the car's speed.

Result

\[\boxed{\,W = \Delta E_k + W_f \approx 1108\ \mathrm{kJ} = 1.11\ \mathrm{MJ}\,}\]

\[\boxed{\,\bar P \approx 5.54\ \mathrm{kW}\,}\]

Remark

Check. The original results are correct.