Question 9
Two vertical forces act on a skydiver with a mass of 85 kg, as shown.
When the magnitude of the air resistance is 62 N, the magnitude of the acceleration of the skydiver is
Not quite. $0.73$ is $\frac{62}{85}$, the air resistance on its own divided by the mass. The weight acts on the skydiver too, so the net force is $833-62=771\text{ N}$. In 2021, 20.8% of Queensland students chose (A).
Not quite. $2.65$ is $\frac{(85-62)\times9.8}{85}$, which treats the $62$ as a mass in kilograms. It is a force in newtons, so take it off the weight: $85\times9.8-62=771\text{ N}$. In 2021, 5.1% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2021, 68.8% of Queensland students chose (C).
Not quite. $12.44$ is $\frac{771}{62}$. The net force is right, but it was divided by the air resistance. Newton’s second law divides by the mass: $\frac{771}{85}\approx9.07\text{ m s}^{-2}$. In 2021, 4.7% of Queensland students chose (D).
The arrows are drawn to scale. Blue is the air resistance, black is the weight and orange is the net force.
This is the question. The air resistance takes only a little off the weight, so the acceleration is a bit less than $g$: about $9.07\text{ m s}^{-2}$.
With no air resistance the only force is the weight, so the acceleration is $g=9.8\text{ m s}^{-2}$.
The two forces balance, so there is no acceleration. The skydiver falls at a constant speed, the terminal velocity.
The air resistance is now bigger than the weight, so the net force points up. The skydiver is still falling but slowing down, as happens when a parachute opens.
The weight is bigger than the air resistance, so the net force points down and the skydiver speeds up. Set the air resistance to $62\text{ N}$ for the question.
Find the weight first. It acts straight down. Use $g=9.8\text{ m s}^{-2}$.
$$W=mg=85\times9.8=833\text{ N}$$
The air resistance acts up and the weight acts down, so they work against each other. Take down as positive.
$$F_{\text{net}}=833-62=771\text{ N}$$
Newton’s second law, $F=ma$, turns the net force into an acceleration.
$$a=\frac{771}{85}\approx9.07\text{ m s}^{-2}$$
It is a little less than $g$, which makes sense with a small upward force.
Take down as positive. The net force is the weight minus the air resistance.
$$\begin{aligned}F_{\text{net}}&=85\times9.8-62=771\text{ N}\\a&=\frac{F_{\text{net}}}{m}=\frac{771}{85}\approx9.07\text{ m s}^{-2}\end{aligned}$$
With $g=9.81$ you get $9.08\text{ m s}^{-2}$, which still points to the same option.
How Queensland students answered in 2021
From the QCAA Specialist Mathematics subject report for 2021. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it