Question 4
A box slides down a slope as shown. If the coefficient of friction between the box and the slope is 0.2, what is the acceleration of the box?
Not quite. One way to reach 8.0 m/s² is $(mg-F_f)\div m=(98-17.76)\div10$, which takes friction away from the whole weight. Only the component $mg\sin25^\circ$ pulls the box down the slope. In 2021, 9.2% of Queensland students chose (A).
Not quite. 4.1 m/s² is $g\sin25^\circ$, the acceleration with no friction at all. The coefficient is 0.2, so friction takes some of that pull away. In 2021, 25.4% of Queensland students chose (B).
That is it. Step through the hints below to see why. In 2021, 45.4% of Queensland students chose (C).
Not quite. 1.8 m/s² is $\mu g\cos25^\circ$, the friction force divided by the mass. That is how much friction slows the box, and it still has to come off the pull down the slope. In 2021, 19.1% of Queensland students chose (D).
The box is already sliding, so the 0.2 is a coefficient of kinetic friction. Friction opposes the sliding, so it points up the slope, against the pull of the weight.
The forces along the slope do not balance. Whatever is left over accelerates the box down the slope.
Resolve the weight, $mg=98\text{ N}$, then find the friction from the normal force.
$$\begin{aligned}F_N&=10\times9.8\times\cos25^\circ=88.82\text{ N}\\F_f&=\mu_k F_N=0.2\times88.82=17.76\text{ N}\\F_\parallel&=10\times9.8\times\sin25^\circ=41.42\text{ N}\end{aligned}$$
Dividing the friction alone by the mass gives $1.78\text{ m s}^{-2}$, which is (D).
Take friction off the pull down the slope, then use $F=ma$.
$$\begin{aligned}F_{net}&=F_\parallel-F_f\\&=41.42-17.76\\&=23.66\text{ N}\\a&=\frac{F_{net}}{m}\\&=\frac{23.66}{10}\\&=2.37\text{ m s}^{-2}\end{aligned}$$
Leaving friction out gives $g\sin25^\circ=4.14\text{ m s}^{-2}$, which is (B).
$$\begin{aligned}a&=\frac{mg\sin25^\circ-\mu_k mg\cos25^\circ}{m}\\&=\frac{41.42-17.76}{10}\\&=2.37\text{ m s}^{-2}\approx2.4\text{ m s}^{-2}\end{aligned}$$
Friction points up the slope because the box slides down. The net force is the weight component minus friction, and dividing by the mass gives the acceleration.
The most common wrong answer was (B), from leaving friction out. The stem gives you a coefficient of friction, so friction has to appear in your net force.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and any figures are from the 2021 QCAA Engineering external assessment, © State of Queensland (QCAA) 2021, 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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