5.2Displacement and Force in Two Dimensions
Friction — American System Physics Revision Notes & Topical Questions
10 min read 7 parts Lesson 4 of 5 1 quiz
Describe friction as a contact force opposing relative motion and explain its effects.Kinetic frictionFk = μkFN (N).Static friction0 ≤ Fs ≤ μsFN (N).On a level surface with only vertical forces weight and normal: FN = mg, so Fk = μkmg.Coefficients have no unitμ = Ff/FN (N/N).Push…
- Pushed at constant velocity
- Static friction is a 'responsive' force
- Friction is essential
- Friction is also a nuisance
- A 10 kg box is pulled horizontally by 60 N. μk = 0.40. Find its acceleration
- A car moving at 20 m/s skids to a stop. μk = 0.70. Find the stopping distance
Part 1 of 7Key ideas
Part 2 of 7Learning objectives
Part 3 of 7Essential knowledge
Part 4 of 7Process & sequence
Pushing a heavy boxwrite these steps in this order; each step is a marking point.
Part 5 of 7Worked examples
- FN = mg = 25 × 9.8 = 245 N.
Fk = μkFN = 0.30 × 245 = 73.5 N.
Constant velocity ⇒ push = friction = 74 N (2 s.f.).
- Fk = 0.40 × 10 × 9.8 = 39.2 N.
Fnet = 60 − 39.2 = 20.8 N.
a = 20.8/10 = 2.1 m/s2.
- a = −μkg = −0.70 × 9.8 = −6.86 m/s2.
v2 = u2 + 2as ⇒ 0 = 400 + 2(−6.86)s.
s = 400/13.72 = 29 m.What the symbols mean: v = final velocity · u = initial velocity (m s−1) · a = acceleration (m s−2) · s = displacement (m)
Part 6 of 7Summary
Part 7 of 7Compare: Compare
| Feature | Static friction | Kinetic friction |
|---|---|---|
| Surfaces | Not sliding | Sliding |
| Size | Variable, 0 up to μsFN | Fixed: μkFN |
| Equation | Fs ≤ μsFN | Fk = μkFN |
| Typical coefficient | Larger | Smaller |
| Example | Box that won't budge; tyres gripping road | Box sliding across floor; skidding car |
| Surfaces | μs | μk |
|---|---|---|
| Rubber on dry concrete | 0.80 | 0.65 |
| Wood on wood | 0.50 | 0.20 |
| Steel on steel (dry) | 0.74 | 0.57 |
| Steel on ice | 0.10 | 0.06 |
Values vary with surface condition; always use the data given in a question.
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Q1. A box on a level floor has FN = 200 N and μk = 0.25. The kinetic friction is
- A. 800 N
- B. 50 N
- C. 200 N
- D. 0.25 N
Show answer
B. Fk = 0.25 × 200 = 50 N.
Q2. A 30 N push does not move a box (μsFN = 80 N). The static friction is
- A. 80 N
- B. 0 N
- C. 30 N
- D. 110 N
Show answer
C. Static friction matches the applied force until its maximum.
Q3. The unit of the coefficient of friction is
- A. N
- B. N/kg
- C. kg
- D. none
Show answer
D. It is a ratio of two forces.
Q4. Why is it harder to start a box sliding than to keep it sliding?
- A. μs > μk
- B. The normal force decreases once moving
- C. Kinetic friction increases with speed
- D. Inertia disappears once moving
Show answer
A. Maximum static friction is larger than kinetic friction.
Q5. Which change reduces friction on a sliding block?
Show answer
B. A lubricant separates the surfaces.
Q6. A puck slides on ice with μk = 0.050. Its deceleration is about
- A. 0.49 m/s2
- B. 0.05 m/s2
- C. 4.9 m/s2
- D. 9.8 m/s2
Show answer
A. a = μkg = 0.050 × 9.8 = 0.49 m/s2.
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