Learn7357Paper 2 · Pure and MechanicsForces and Newton's lawsMotion on inclined planes

Forces and Newton's laws · Subtopic 8 of 8

Motion on Inclined Planes

Gravity pulls along the slope with $mg\sin\alpha$ ; friction, if any, opposes the motion with $\mu mg\cos\alpha$ . Put them into $F=ma$ along the slope, then let suvat finish the job.

Cambridge 9709 Paper 4Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Write $F=ma$ along an inclined plane, with $R$ from resolving perpendicular to it.
Find accelerations up and down smooth and rough slopes.
Find $\mu$ from a measured acceleration.
Combine slope dynamics with suvat, including a particle projected up a slope.

Pause and think

A block slides down a rough slope. Then it is pushed up the same slope with the same speed and allowed to slow down. Is the size of its acceleration the same both ways?

Section 2 of 14

Core idea

Start hereSledging

A sledge slides down a snowy hill. Steeper hills give bigger accelerations; rougher snow slows it. Combine "weight along the slope" with friction and $F=ma$ to predict exactly how fast.

The big idea

Along the slope: (forces down) − (forces up) $=ma$ . Perpendicular: no acceleration, so forces balance, giving $R$ . On a smooth slope $a=g\sin\alpha$ .

1. Sliding down

Forces on a block on a 30° slope (sliding down)

Split the weight into two parts:

mg\sin\alpha

pulling down the slope and

mg\cos\alpha

pushing into it. Friction acts up the slope, opposing the motion (or the tendency to move).

Worked example2 kg on $30^\circ$ , $\mu=0.25$

Find the acceleration down the slope.

Goal: $a$ .
Strategy: $R$ then friction then $F=ma$ .

$R=20\cos30^\circ=17.32$
Perpendicular.
$F=0.25\times17.32=4.33$
Sliding.
$20\sin30^\circ-4.33=2a\Rightarrow a=2.83$
Along.

Answer

2.83

m s

^{-2}

✓ CheckLess than the smooth-slope value

5

✓.

Acceleration down a smooth slope as the angle grows

Flat (

0^\circ

): no acceleration. Vertical (

90^\circ

): free fall at

10

m/s². Friction reduces these values.

2. Projected up

Worked exampleProjected up at 8 m/s

Same slope. How far up, and does it slide back?

Goal: Distance; yes/no.
Strategy: Friction now acts down the slope too.

$10+4.33=2a\Rightarrow a=7.17\text{ (deceleration)}$
Both forces down the slope.
$0=64-2(7.17)s\Rightarrow s=4.47$
$v^2=u^2+2as$ .
$10>4.33$
Down-slope pull beats max friction: slides back.

Answer

4.47

m; yes.

✓ CheckUp deceleration (

7.17

) > down acceleration (

2.83

): friction always opposes motion ✓.

Common misconceptionFriction direction Friction flips when the motion reverses.

Pause and think

Smooth $20^\circ$ slope acceleration?

Summary

Perpendicular for $R$ ; along for $ma$ .
Friction opposes current motion.

Resolve: Find a component.

HistoryDiluting gravity. Galileo's ramps worked because a smooth slope "dilutes" gravity: a ball rolling down accelerates at

g\sin\alpha

instead of

g

. With a gentle slope he could slow free fall down enough to time it with water dripping from a tank, and so discovered the laws of falling bodies.

Fun factSkis are almost frictionless. Waxed skis on snow have

\mu

as low as about

0.05

, because a thin film of melted water forms under them. On a

20^\circ

slope that gives an acceleration of

10(\sin20^\circ-0.05\cos20^\circ)\approx2.9\ \text{m s}^{-2}

, almost as much as a smooth slope. Air resistance is what really limits a skier's speed.

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