Question 10
A 40 kg object is pushed 3 m up a 25º incline at a uniform velocity.
If the frictional force opposing motion is 163 N and the weight force component acting down the incline is 165.66 N, the work done is
Not quite. $163\times3=489$ only pushes against friction. The weight component also pulls the object down the incline, so the push has to beat both. In 2023, 9.9% of Queensland students chose (A).
Not quite. $165.66\times3=497$ only pushes against the weight component. It is the potential energy the object gains, but friction takes another 489 J on the way up. In 2023, 20.4% of Queensland students chose (B).
That is it. Step through the hints below to see why. In 2023, 60.0% of Queensland students chose (C).
Not quite. $1066$ is the normal force, $mg\cos25^\circ=355.27$ N, times 3 m. The normal force is at right angles to the motion, so it does no work at all. In 2023, 8.4% of Queensland students chose (D).
Uniform velocity means the forces along the incline balance. The push up the incline equals everything acting down it, which is friction and the weight component.
The stem hands you both forces, so you do not need $\sin25^\circ$ at all. QCAA printed 165.66 N. It is 165.67 N to two decimal places, and either value gives 986 J.
The push acts up the incline and the object moves 3 m up the incline, so the whole 3 m counts.
The 2019 formula book that 2023 students had printed this as $W=Fd$. The 2026 book writes $W=Fs$.
Each distractor is one force times 3 m. Split the work by where it goes and two of them appear.
489 J becomes heat at the surface and 497 J becomes potential energy. You need both. The normal force, behind (D), is at right angles to the motion and does no work.
At uniform velocity the push balances friction and the weight component together. Multiply that push by the 3 m it moves.
The most common wrong answer was (B), from pushing against the weight component and forgetting friction. That is the work done against gravity alone, so it is the right answer only for a frictionless incline.
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 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.
Spot a mistake on this page? Report it