Question 22
A child on a sled slides down a snow-covered hill with a 15° incline at an initial velocity of 4 m/s. The combined mass of the sled and the child is 40 kg. The coefficient of kinetic friction between the sled and the snow is 0.35.
Friction up the hill is bigger than the pull down it, so the sled slows to a stop
The forces are drawn to scale. Compare the two forces along the slope, then watch the slide in real time. The orange marks are equal steps in time, and they bunch up as the sled slows.
The net force points up the hill while the sled moves down it, so the sled loses speed at 0.78 m/s every second. From 4 m/s that takes 5.15 s and 10.3 m.
Change the friction and the slope and see when the sled stops
Set the coefficient of kinetic friction and the angle of the hill, and choose the method for the last two marks. The diagram is drawn to scale and the bar shows how far the sled goes. The forces are carried at full precision and shown to two decimal places, and each distance line uses the rounded value it shows, as the guide’s working does. The question is 0.35 and 15°.
This is the question. Since $\tan15^\circ=0.27$ is less than 0.35, friction (132.53 N) beats the pull down the hill (101.46 N) and the sled slows. With $a$ rounded to $-0.78$ the motion equation gives 10.26 m, and work and energy gives 10.30 m. Both are the guide’s 10.3 m, and the gap between them is the cost of rounding early.
Friction is still bigger than the pull down the slope, so the net force points up the hill and the sled stops. Less friction or a steeper hill shrinks that gap, and a small net force means a long slide. Both methods give the same distance, apart from rounding.
Now $\mu_k$ is no bigger than $\tan\theta$, so the pull down the slope is at least as big as friction. The net force is zero or points down the hill, and the sled never stops. At exactly $\mu_k=\tan\theta$ it slides at constant velocity, the balanced case from 2024 Question 20.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The guide accepts two methods. Steps 1 to 6 are the same for both, and steps 7 and 8 show them side by side.
Draw the force diagram first. Three forces act on the sled: the weight $F_g$ straight down, the normal force $F_N$ perpendicular to the hill and kinetic friction $F_f$ up the hill, against the sliding. There is no push. Arrowheads and labels on all three.
Marker: includes an appropriate free-body diagram correctly showing all forces. “Accept free-body diagram without the slope”, so three labelled vectors from one point are enough. A missing force is not.
Say why the normal force is what it is before you calculate it. The sled does not move into or away from the hill, so the forces perpendicular to the hill balance:
$$F_N=F_\perp=mg\cos\theta$$
This statement is a mark on its own, separate from the number in the next step.
Marker: correctly identifies that normal force is equal to the force perpendicular to the hill
Use the combined mass of the sled and the child.
$$\begin{aligned}F_N&=40\times9.8\times\cos15^\circ\\&=378.64\text{ N}\end{aligned}$$
Marker: correctly calculates the normal force
The sled is sliding, so use kinetic friction. It acts up the hill.
$$\begin{aligned}F_f&=\mu_kF_N\\&=0.35\times378.6429\ldots\\&=132.53\text{ N}\end{aligned}$$
Keep the unrounded normal force in your calculator here. The guide’s range starts at 132.525 N, and $0.35\times378.64$ from the rounded value is 132.524 N, which rounds to 132.52 N and falls just outside it. The guide’s own working writes $0.35\times378.64=132.53$, so it carried the unrounded value without saying so.
Marker: calculates the frictional force. “Allow FT marks for errors in prior working. Allow implied mark for calculating normal force within the calculation for frictional force. $F_f=\mu_kmg\cos\theta$. Accept range of 132.525–133 N inclusive.”
The component of the weight along the hill pulls the sled down it.
$$\begin{aligned}F_\parallel&=mg\sin\theta\\&=40\times9.8\times\sin15^\circ\\&=101.46\text{ N}\end{aligned}$$
Marker: calculates the parallel force
Take down the hill as positive, the direction the sled moves. Friction points the other way, so it goes in with a minus sign.
$$\begin{aligned}F_{net}&=F_\parallel-F_f\\&=101.46-132.53\\&=-31.07\text{ N}\end{aligned}$$
The negative sign says the net force points up the hill, against the motion. The guide writes this line as $F_{net}=F_f-F_\parallel$ and then substitutes $-132.53+101.46$, which is $F_\parallel-F_f$. Its number is right and its label has the signs reversed.
Marker: calculates the net force. “Allow implied mark for calculating parallel force within the calculation for net force. $F_{net}=F_f-mg\sin\theta$”
Newton’s second law gives the acceleration. Keep the sign.
$$\begin{aligned}a&=\frac{F_{net}}{m}=\frac{-31.07}{40}\\&=-0.78\text{ m s}^{-2}\end{aligned}$$
The calculator shows $-0.7767$, which rounds to $-0.78$, not $-0.77$.
Marker: calculates the acceleration. “Accept positive number if response states deceleration.”
Name the principle. The work done by the net force equals the change in kinetic energy, from $W=Fs$ and $KE=\tfrac{1}{2}mv^2$ in the formula book.
$$F_{net}\,d=\tfrac{1}{2}mv^2-\tfrac{1}{2}mu^2$$
Marker: identifies that work is equal to the change in kinetic energy
You know $u=4$, $v=0$ and $a$, and you want $s$. There is no $t$, so use $v^2=u^2+2as$.
$$\begin{aligned}s&=\frac{v^2-u^2}{2a}\\&=\frac{0^2-4^2}{2(-0.78)}\\&=10.3\text{ m}\end{aligned}$$
Marker: calculates the distance travelled downhill. “Accept distance as negative number.”
The sled ends at rest, so the change in kinetic energy is negative.
$$\begin{aligned}-31.07\,d&=0-\tfrac{1}{2}(40)(4^2)\\-31.07\,d&=-320\\d&=10.3\text{ m}\end{aligned}$$
The guide drops the minus on the right, writes $-31.07\,d=320$ and gets $-10.3$ m, then accepts the positive value if you say the sled went downhill. The signs above are the consistent version.
Marker: calculates the distance travelled downhill. “Accept distance as positive number if response states the distance is travelled downhill.”
The guide prints 10.3 m and gives no range for this line. Its general rule says the nearest whole unit, which would be 10 m, while the 2025 subject report says not to round to a whole unit unless told to. Write 10.3 m, the value the guide shows.
Each phrase of the question gave you something
The 2025 subject report asks for force vectors that are clearly labelled, with arrows, in free-body diagrams, and for rounding left until the end of a multi-step calculation. Its rounding example, 0.776 rounds to 0.78 and not 0.77, matches this question’s acceleration.
QCAA did not publish a band for this question. It reads as complex familiar because it chains forces on an incline into a motion or energy calculation over eight marks, with every value supplied and the same shape as 2024 Question 22, mirrored. The twist is the sign. The net force points against the motion, so the acceleration is negative and the sled stops.
If I had to guess where you might lose marks, it would be the signs. Point friction down the hill and the sled speeds up forever, and a lost minus sign gives a negative distance. Then there is rounding. Rounding $F_N$ before you find friction puts you at 132.52 N, just under the range, and $-0.7767$ must become $-0.78$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording, any figures and marking-guide steps are from the 2025 QCAA Engineering external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted, except where a third-party source is named. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it