Car on a hill: power, force and acceleration

In the last lesson you saw that an engine working at power $P$ while moving at speed $v$ pushes with a driving force

$D = \frac{P}{v}$

On a slope, three forces act along the road:

1. The driving force $D = \dfrac{P}{v}$, forwards.
2. The resistance $R$, always against the direction of motion.
3. Part of the weight, $mg\sin\alpha$, always pointing down the slope.

Why $mg\sin\alpha$? Look at the diagram below. Split the weight $mg$ into a part along the slope and a part pressing into it. The angle between the weight (straight down) and the line at right angles to the slope is $\alpha$, so the part along the slope is the side opposite that angle: $mg\sin\alpha$. On a flat road $\alpha = 0$ and $\sin 0 = 0$, so this force disappears, as it should.

(The other part of the weight, $mg\cos\alpha$, presses into the road and is balanced by the road's push. It doesn't affect motion along the slope.)

Newton's second law says resultant force $=$ mass $\times$ acceleration. Pick the direction of motion as positive, add up the forces along the slope with signs, and set the total equal to $ma$.

Going up the hill: the driving force is forwards (up), while the resistance and the weight component both point down the slope:

$\frac{P}{v} - mg\sin\alpha - R = ma$

Going down the hill: now the weight component points the same way as the motion, so it helps. Only the resistance opposes:

$\frac{P}{v} + mg\sin\alpha - R = ma$

If the engine is off (the car coasts), $P = 0$ and the first term disappears.

Constant speed or top speed means $a = 0$, so the forces balance: put $0$ on the right-hand side.