2023 A/L Physics Past Paper · Question 45 (English Medium)

Answer: (2) $34\,\text{kW}$

Given information

What we know:

The car has a mass $m = 1200\,\text{kg}$ and a constant speed $v = 20\,\text{m s}^{-1}$.

On the level road the engine power is $P_1 = 22\,\text{kW}$.

The second road is identical but inclined at $3^{\circ}$. We take $\pi = 3$, $\sin\theta = \theta$ and $g = 10\,\text{m s}^{-2}$.

What the question asks

We want: the engine power $P_2$ on the slope, at the same speed.

What each letter means:

$F_r$ ($\text{N}$): the resistance to the motion on the level road.

$\theta$: the angle of the slope, in radians.

$F$ ($\text{N}$): the driving force of the engine on the slope.

$P_2$ ($\text{W}$): the engine power on the slope.

Relevant theory

power = force × velocity

Newton's first law states that a resultant force can change the state of motion of an object. Newton's second law states about this resultant force.

The first quote gives the power of a force that moves a body at a velocity: $$power = force \times velocity$$ $P = Fv$. The second quote gives Newton's second law, $F = ma$. At a constant speed, $a = 0$, so the forces along the road balance: the driving force of the engine equals the resistance, plus the weight part along a slope.

For example, a cyclist who rides at the same speed up a hill has to push the pedals harder than on a flat road.

Required equations

$P = Fv$

$F_r = \dfrac{P_1}{v}$

$F = F_r + mg\sin\theta$

$\sin\theta = \theta = \dfrac{3\pi}{180}$

$P_2 = Fv$

Here $\text{N} \times \text{m s}^{-1}$ gives $\text{W}$, so a force in newtons and a speed in $\text{m s}^{-1}$ give a power in watts.

Applying it to this question

$1100$

$F_r = \dfrac{P_1}{v} = \dfrac{22000}{20}$

$0.05$

$\theta = \dfrac{3 \times 3}{180}$

$600$

$mg\sin\theta = 1200 \times 10 \times 0.05$

$1700$

$F = F_r + mg\sin\theta = 1100 + 600$

$34000$

$P_2 = Fv = 1700 \times 20$

$34$

$P_2 = \dfrac{34000}{1000}$

Reasoning and calculation

$(22000 / v + m * g * 3 * 3 / 180) * v / 1000$

Now match it with the five options.

Option (2) is $34\,\text{kW}$. It matches.

Option (3), $35\,\text{kW}$, is what $\pi = 3.14$ would give. The other options are $25$, $42$ and $47\,\text{kW}$, which need an extra force of $150$, $1000$ and $1250\,\text{N}$ and not $600\,\text{N}$.

Checks

Check the units. $F_r = \dfrac{P_1}{v}$ is $\text{W} \div \text{m s}^{-1} = \text{N}$. $mg\sin\theta$ is $\text{kg} \times \text{m s}^{-2} = \text{N}$.

Then $Fv$ is $\text{N} \times \text{m s}^{-1} = \text{W}$, a power, as it must be.

Does it make sense?

Climbing needs more power than the level road, and $34\,\text{kW}$ is more than $22\,\text{kW}$.

The slope is gentle, with $\sin\theta = 0.05$, which is a rise of $5\,\text{m}$ in $100\,\text{m}$. The extra $12\,\text{kW}$ is the power to lift the car: $mg \times 20 \times 0.05 = 12000\,\text{W}$.

Conclusion

$P_2 = 34\,\text{kW}$

The engine power on the slope is $P_2 = 34\,\text{kW}$.

The answer is (2).

The quotes give power as force times velocity and Newton's second law. That the resistance is the same on the identical road, and that we take $g = 10\,\text{m s}^{-2}$, which the question does not state, are our reasoning and our choice.

Teacher's note

At a constant speed the driving force balances the resistance and the weight part along the slope. Find the force first, then use $P = Fv$.

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