A car of mass m = 2 t drives up a road that rises h = 4 m…

A car of mass $m = 2\ \mathrm{t}$ drives up a road that rises $h = 4\ \mathrm{m}$ per $l = 100\ \mathrm{m}$. The coefficient of friction (rolling resistance) is $\mu = 0.08$. Find the work and power of the engine if the car covers $s = 3\ \mathrm{km}$ in $t = 4\ \mathrm{min}$ at constant speed.

Explanation of Method

Engine work and power on a hill.

Step 1: Forces

\[ \tan\theta = 0.04, \ \theta \approx 2.29^\circ, \]

\[ mg\sin\theta \approx 784\ \mathrm{N}, \]

\[ \mu mg\cos\theta \approx 1568\ \mathrm{N}. \]

\[ F = mg\,(\sin\theta + \mu\cos\theta) \approx 2353\ \mathrm{N}. \]

Step 2: Work and power

\[ W = Fs = 2353 \cdot 3000, \]

\[ P = \frac{W}{t} = \frac{W}{240\ \mathrm{s}}. \]

\[W \approx 7.06 \cdot 10^6\ \mathrm{J}, \qquad P \approx 29.4\ \mathrm{kW}\]

$F = 2353\ \mathrm{N}$ $\mu mg\cos\theta = 1568$ $mg$ $\text{slope } 4 : 100 \text{ (angle exaggerated)}$ $mg\sin\theta = 784\ \mathrm{N}$

At constant speed the engine's force balances the slope component and friction. The slope is drawn four times steeper than it really is.

Result

\[\boxed{\,W \approx 7.06 \cdot 10^6\ \mathrm{J}, \qquad P \approx 29.4\ \mathrm{kW}\,}\]

Remark

About the original solution. The method is right, but $mg\sin\theta$ was evaluated as 684.9 N instead of $19\,620 \cdot 0.04 \approx 784\ \mathrm{N}$ (and the sum 684.9 + 1567.9 was also written as 2252.8). The correct results are $W \approx 7.06\ \mathrm{MJ}$ and $P \approx 29.4\ \mathrm{kW}$, not 6.76 MJ and 28.17 kW.