Question 19
A 2 kg box exits a manufacturing process at a velocity of 0.5 m/s and then slides down a ramp onto a horizontal surface as shown.
Determine the distance the box slides along the horizontal surface before coming to a complete stop if the coefficient of kinetic friction between the box and all surfaces is 0.35. Answer to the nearest whole unit.
Only the speed at the bottom of the ramp carries across to the floor
This is the QCAA figure set moving. The ramp, the stopping distance and every force are drawn to scale. Watch the force diagram change when the box reaches the floor. The normal force grows to the full weight and friction is left on its own.
Two stages, two force diagrams. The final velocity on the ramp is the initial velocity on the floor.
Change the ramp and the friction, and watch where the box stops
Set the angle, the ramp length and the coefficient. The ramp and the stopping distance are drawn to scale. Then try the trap: carry the ramp’s $\cos\theta$ onto the floor. The question is 25°, 4 m and 0.35.
This is the question. A net 2.07 N down the ramp takes the box from 0.5 m/s to 2.92 m/s. On the floor friction is 6.86 N, so it stops in 1.24 m, which is 1 m to the nearest whole unit.
This is the chaining trap. On the floor nothing is tilted, so $F_N=mg$. Keeping the ramp’s cosine makes friction too small and the box slides too far. It costs the horizontal friction mark. The guide allows follow-through after that, but the distance you report is wrong.
Now $\tan\theta\le\mu_k$, so friction on the ramp is at least as big as the pull down it. The net force points up the ramp, the 0.5 m/s dies away and the box never reaches the floor.
A steeper or longer ramp sends the box onto the floor faster, and the stopping distance grows with $v^2$. A bigger coefficient works against you twice, on the ramp and again on the floor.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The working below carries the rounded values the marking guide shows, so every line matches the guide.
On the ramp the normal force is the cosine part of the weight. The box is sliding, so this is kinetic friction, and it acts up the ramp against the slide.
$$\begin{aligned}F_f&=\mu F_N=\mu mg\cos\theta\\&=0.35\times2\times9.8\times\cos25^\circ\\&=6.22\text{ N}\end{aligned}$$
Marker: determines the force of friction on the ramp. “Accept force of friction from 6.21–6.22 N inclusive.”
The sine part of the weight pulls the box down the ramp. The guide calls it $F_p$.
$$\begin{aligned}F_\parallel&=mg\sin\theta\\&=2\times9.8\times\sin25^\circ\\&=8.28\text{ N}\end{aligned}$$
Marker: determines parallel force down the ramp. “Accept parallel force from 8.28–8.30 N inclusive.”
The forces along the ramp do not balance. The pull down the ramp wins.
$$\begin{aligned}F_r&=F_\parallel-F_f\\&=8.28-6.22\\&=2.06\text{ N}\end{aligned}$$
Unrounded it is 2.07 N, which is inside the accepted range.
Marker: determines resultant force down the ramp. “Accept resultant force from 2.06–2.09 N inclusive.” FT is allowed for an incorrect friction or parallel force.
Divide the net force by the mass.
$$\begin{aligned}a&=\frac{F}{m}=\frac{2.06}{2}\\&=1.03\text{ m s}^{-2}\end{aligned}$$
Marker: determines ramp acceleration. “Accept ramp acceleration from 1.03–1.05 m/s2 inclusive.” FT is allowed for an incorrect resultant force.
You know $u=0.5$, $a=1.03$ and $s=4$, and you want $v$. There is no time, so use the equation without $t$. The 0.5 m/s from the stem belongs here.
$$\begin{aligned}v^2&=u^2+2as\\&=0.5^2+2\times1.03\times4\\&=8.49\end{aligned}$$
Marker: provides correct formula and substituted values for velocity
Take the square root. This speed is the only thing that crosses to the floor.
$$\begin{aligned}v&=\sqrt{8.49}\\&=2.91\text{ m s}^{-1}\end{aligned}$$
Kept unrounded, $v=2.92$ m/s.
Marker: determines velocity at the bottom of the ramp. “Accept velocity from 2.91–2.94 m/s inclusive.” FT is allowed for an incorrect ramp acceleration.
Start a fresh force diagram. The floor is horizontal, so the normal force is the whole weight and there is no cosine.
$$\begin{aligned}F_f&=\mu F_N=\mu mg\\&=0.35\times2\times9.8\\&=6.86\text{ N}\end{aligned}$$
Marker: determines force of friction on the horizontal surface
Friction is now the only force along the floor, so it sets the deceleration.
$$\begin{aligned}a&=\frac{F}{m}=\frac{6.86}{2}\\&=3.43\text{ m s}^{-2}\end{aligned}$$
Marker: determines deceleration on the horizontal surface. “Allow FT error for incorrect force of friction when determining deceleration.”
Now $u=2.91$, $v=0$ and $a=-3.43$, because the box is slowing down. Rearrange the same equation for $s$.
$$\begin{aligned}s&=\frac{v^2-u^2}{2a}\\&=\frac{0^2-2.91^2}{2\times(-3.43)}\\&=\frac{-8.47}{-6.86}=1.23\text{ m}\end{aligned}$$
Carry full precision through every step and you get $s=1.24$ m. The question asks for the nearest whole unit.
QCAA’s sample response stops at 1.235 m and the guide accepts 1.234 to 1.260 m, even though the stem and the criterion both say the nearest whole unit. Write the decimal line and then the rounded answer, and you are covered either way.
Marker: determines the answer to the nearest whole unit. “Accept answer from 1.234–1.260 m inclusive.” FT is allowed for an incorrect velocity or deceleration.
Each phrase of the question gave you something
The 2023 subject report used this question as a sample of effective practice. Strong responses found the resultant force down the incline from the friction and parallel forces, then the acceleration, the velocity at the bottom of the ramp and the deceleration on the floor, and finished to the nearest whole unit.
QCAA did not publish a band for this question. It reads as complex familiar because each stage is a routine you have practised, and the difficulty is in chaining them through nine marks with nothing but the velocity to link them.
If I had to guess where you might lose marks, it would be the floor. Carry $\cos25^\circ$ across and friction drops to 6.22 N, the deceleration to 3.11 m/s², and the box slides 1.37 m. The other slip is starting the ramp from rest. Leaving out the 0.5 m/s gives 2.87 m/s at the bottom and costs the velocity marks.
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 2023 QCAA Engineering external assessment, © State of Queensland (QCAA) 2023, 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.
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