Now you put the inclined-plane toolkit to work. Whether a block holds steady at constant velocity (friction exactly balancing gravity) or speeds up as it slides, the same components and formulas crack the problem. You will also design a real experiment to test how the acceleration changes with the slope angle, and see your derived formula come to life in measured data.
A crate slides down a ramp at a perfectly constant velocity, neither speeding up nor slowing down.
Before reading: what must be true about the friction force compared with the down-slope component of weight? Write your prediction and one sentence of reasoning.
When friction exactly cancels the down-slope pull
An object can slide down a ramp at a steady speed if friction up the slope perfectly matches the weight component pulling it down, the two cancel, so there is no acceleration.
Constant velocity means zero acceleration, so by Newton's first law the net force along the slope is zero. The down-slope weight component $mg\sin\theta$ is exactly balanced by the friction force up the slope:
$$mg\sin\theta = f \quad (\text{constant velocity}) \qquad \Rightarrow \qquad \mu = \tan\theta$$
If the friction is kinetic, $f = \mu mg\cos\theta$, so setting it equal to $mg\sin\theta$ gives $\mu = \tan\theta$, a neat way to find the coefficient of friction by finding the angle at which an object slides at constant speed.
For constant velocity down a slope the net force is zero: friction up the slope equals $mg\sin\theta$ down the slope. With kinetic friction $\mu mg\cos\theta = mg\sin\theta$, which simplifies to $\mu = \tan\theta$.
Pause, copy the highlighted condition into your book before moving on.
Did you get this? True or false: if a block slides down a ramp at constant velocity, the friction force equals the down-slope component of its weight.
Using the master formula and the equations of motion together
We just saw the constant-velocity case. That raises a question: when the down-slope pull beats friction, how fast does the object end up moving? This card answers it → find the acceleration with $a = g(\sin\theta - \mu\cos\theta)$, then chain it into the equations of motion.
When gravity wins, you find the acceleration from the incline formula and then use the motion equations to get speed or distance, two familiar tools working in sequence.
For an accelerating slide, use $a = g(\sin\theta - \mu\cos\theta)$, then feed that constant acceleration into the equations of uniformly accelerated motion (such as $v^2 = u^2 + 2as$) to find a final speed or distance.
A block is released from rest at the top of a $4.0\ \text{m}$ ramp inclined at $30°$, with coefficient of kinetic friction $\mu = 0.15$. Using $g = 9.8\ \text{m s}^{-2}$, find the acceleration and the block's speed at the bottom.
For an accelerating slide: find $a = g(\sin\theta - \mu\cos\theta)$, then use the equations of motion (e.g. $v^2 = u^2 + 2as$) for speed or distance. Here $a = 9.8(\sin 30° - 0.15\cos 30°) = 3.6\ \text{m s}^{-2}$ and $v = \sqrt{2 \times 3.6 \times 4.0} = 5.4\ \text{m s}^{-1}$.
Add the highlighted two-step method to your notes before the check below.
Quick check: A block slides from rest down a frictionless slope of $20°$. Using $g = 9.8\ \text{m s}^{-2}$, its acceleration is closest to…
Pattern C, Design, measure, analyse
Design a laboratory experiment to investigate how the acceleration of a trolley (or cart) down a ramp depends on the ramp angle. In your book: (1) state the independent variable (angle $\theta$), the dependent variable (acceleration $a$), and two variables you would control (e.g. same trolley/mass, same surface); (2) describe how you would measure the acceleration, for example, with a motion sensor or by timing the trolley over a measured distance and using $s = \tfrac{1}{2}at^2$; (3) explain how repeating each angle three times improves reliability; (4) sketch the predicted graph of acceleration against $\sin\theta$ for negligible friction, then use $a=g(\sin\theta-\mu\cos\theta)$ to explain how a nonzero constant $\mu$ changes that prediction. State the friction assumption whenever you interpret a straight-line graph.
Odd one out: three of these describe an object sliding down a slope at CONSTANT velocity. Which one does not?
Dot point: Solve problems involving objects with constant velocity and objects accelerating on inclined planes
Compare the frictionless linear prediction with the friction model
We just designed the experiment. That raises a question: what graph follows from the acceleration formula? This card answers it → negligible friction gives a straight line against $\sin\theta$; constant nonzero friction adds a changing cosine term.
Plotting acceleration against $\sin\theta$ gives a straight line only for the frictionless model (or when friction is negligible over the measured range).
For a frictionless ramp, $a = g\sin\theta$, so acceleration ($y$-axis) against $\sin\theta$ ($x$-axis) is a straight line through the origin with gradient $g \approx 9.8\ \text{m s}^{-2}$. For a block sliding down with constant kinetic-friction coefficient $\mu>0$, $a=g\sin\theta-\mu g\cos\theta$. Since $\cos\theta=\sqrt{1-\sin^2\theta}$, the second term changes with the horizontal variable: the predicted $a$-versus-$\sin\theta$ graph is curved, not a straight line with a reduced gradient. Its zero-acceleration angle satisfies $\tan\theta=\mu$, while its graph x-coordinate is $\sin\theta=\mu/\sqrt{1+\mu^2}$. A straight-line fit with gradient $g$ therefore tests the negligible-friction model; do not interpret a curved frictional result as that same line.
Plot acceleration against $\sin\theta$: theory predicts $a = g\sin\theta$ (frictionless), so the graph should be a straight line through the origin with gradient $g \approx 9.8\ \text{m s}^{-2}$. Kinetic friction lowers acceleration relative to the frictionless prediction. Only a negligible-friction straight-line trend tests the $a = g\sin\theta$ model; constant nonzero kinetic friction predicts a curved graph.
Pause, copy the highlighted analysis into your book before the check below.
Last check: Decide if the statement is true or false.
For a frictionless ramp, a graph of acceleration against $\sin\theta$ is a straight line through the origin with gradient $g$.
In the angle-versus-acceleration practical, match each choice with its purpose:
Dot point: Conduct a laboratory experiment to analyse the relationship between the angle of an inclined plane and the acceleration of an object
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 6(5 marks) 1. A block is released from rest and slides $3.0\ \text{m}$ down a ramp inclined at $25°$ with coefficient of kinetic friction $\mu = 0.10$. Using $g = 9.8\ \text{m s}^{-2}$, calculate (a) the acceleration and (b) the speed at the bottom of the ramp.
1 mark: uses $a = g(\sin\theta - \mu\cos\theta)$ · 1 mark: $a = 9.8(\sin 25° - 0.10\cos 25°) \approx 3.3\ \text{m s}^{-2}$ · 1 mark: selects $v^2 = u^2 + 2as$ with $u = 0$ · 1 mark: $v^2 = 2 \times 3.3 \times 3.0$ · 1 mark: $v \approx 4.4\ \text{m s}^{-1}$
ApplyBand 6(4 marks) 2. A $12\ \text{kg}$ box slides down a $25^\circ$ incline with a coefficient of kinetic friction of $0.20$. Taking $g = 9.8\ \text{m s}^{-2}$, calculate its acceleration, then determine the angle at which it would instead slide at constant velocity.
1 mark: uses $a = g(\sin\theta - \mu\cos\theta)$ · 1 mark: $a = 2.4\ \text{m s}^{-2}$ · 1 mark: constant velocity requires $\tan\theta = \mu$ · 1 mark: $\theta = 11^\circ$
Dot point: Solve problems involving objects with constant velocity and objects accelerating on inclined planes
EvaluateBand 6(3 marks) 3. A student investigates how the acceleration of a block down a ramp depends on the ramp angle. Identify the independent, dependent and one controlled variable, and state what plotting $a$ against $\sin\theta$ would show if friction were negligible.
1 mark: independent is the ramp angle, dependent is the acceleration · 1 mark: a valid controlled variable, such as the same block and surface so $\mu$ is fixed, or the same measured distance · 1 mark: a straight line through the origin of gradient $g$
Dot point: Conduct a laboratory experiment to analyse the relationship between the angle of an inclined plane and the acceleration of an object
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (5 marks): (a) Using the incline formula, $a = g(\sin\theta - \mu\cos\theta) = 9.8(\sin 25° - 0.10\cos 25°) = 9.8(0.423 - 0.091) = 9.8 \times 0.332 \approx 3.3\ \text{m s}^{-2}$. (b) The block starts from rest ($u = 0$) and accelerates uniformly, so use $v^2 = u^2 + 2as = 0 + 2 \times 3.3 \times 3.0 = 19.8$, giving $v = \sqrt{19.8} \approx 4.4\ \text{m s}^{-1}$. The block reaches about $4.4\ \text{m s}^{-1}$ at the bottom.
Q2 (4 marks): Using the incline result, $$a = g(\sin\theta - \mu\cos\theta) = 9.8(\sin 25^\circ - 0.20\cos 25^\circ)$$ $$= 9.8(0.4226 - 0.20 \times 0.9063) = 9.8(0.4226 - 0.1813) = 9.8 \times 0.2413 = 2.4\ \text{m s}^{-2}$$ down the slope. The mass is not needed, because it cancels from the derivation. Here $\mu=0.20$ is the kinetic-friction coefficient, and the block is already sliding. For constant velocity its acceleration must be zero, which requires $\sin\theta = \mu_k\cos\theta$, that is $$\tan\theta = \mu_k = 0.20 ;\Rightarrow; \theta = \tan^{-1}(0.20) = 11^\circ$$ At this angle a sliding block continues at constant speed under the stated kinetic-friction model. At a lower angle it decelerates if it is already sliding; whether a stationary block starts moving depends on the separate coefficient of static friction $\mu_s$.
Q3 (3 marks): The independent variable is the angle of the ramp, which the student sets. The dependent variable is the acceleration of the block, which is measured at each angle. Controlled variables include using the same block on the same ramp surface throughout, so that the coefficient of friction stays constant, and measuring over the same marked distance each time. If friction is negligible then $a = g\sin\theta$, so plotting acceleration against $\sin\theta$ gives the predicted straight line through the origin with gradient $g$. If a nonzero constant kinetic-friction coefficient is included, $a=g\sin\theta-\mu g\cos\theta$ predicts a curved graph against $\sin\theta$. The zero-acceleration angle still satisfies $\tan\theta=\mu$, but this is a threshold method rather than a reduced-gradient straight-line fit.
You can now solve constant-velocity and accelerating incline problems, and design a valid experiment relating angle to acceleration. Next lesson looks at how Aboriginal and Torres Strait Islander Peoples used inclined planes to move heavy objects.
↩ Back to focus-area hubAt the start you predicted the friction needed for constant velocity. The answer: friction must exactly equal the down-slope weight component $mg\sin\theta$, so the net force is zero and the speed stays constant, which also gives the neat result $\mu = \tan\theta$. If you thought friction had to exceed the weight component, revisit Card 1.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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