1. Physics
  2. Laws of Motion
  3. Friction – Static and Kinetic

Laws of Motion · JEE & NEET Physics

Friction – Static and Kinetic: notes and previous year questions

Static and kinetic friction, the laws of friction, the angle of repose, and friction on slopes, ropes and brakes.

Friction – Static and Kinetic in short

  • Friction opposes relative motion or its tendency; it comes from interlocking surface bumps.
  • Static friction adjusts itself from zero up to μ_s N.
  • Kinetic friction is μ_k N while sliding; usually μ_k < μ_s.
  • Friction is proportional to the normal force and does not depend on the contact area.

1What friction is

Even smooth-looking surfaces are covered in tiny bumps and hollows. Where two surfaces touch, the bumps catch on each other and resist sliding. Friction is this contact force along the surface.

When you walk, your foot tends to slip backward on the ground, so friction on the foot points forward, in the direction you move. Friction opposes relative motion, not "motion" in general.

Two kinds: static friction while nothing slides, and kinetic friction once it slides.

2Static friction

Static friction adjusts itself to stop sliding, up to a limit. Push a 10 kg box (μ_s = 0.4, limit 40 N) with 20 N: friction is 20 N. Push with 35 N: friction is 35 N. Push with 50 N: the limit is beaten and the box slides.

fs≤μsNf_s \le \mu_s N
fmax⁡=μsNf_{\max} = \mu_s NLimiting friction, just before sliding starts. μ_s is the coefficient of static friction.

On a graph of friction against the applied force, friction first rises along f = F, reaches the limit μ_s N, then drops a little to the kinetic value and stays flat.

3Kinetic friction

fk=μkNf_k = \mu_k NAlways this value while sliding, opposite to the sliding velocity; nearly independent of speed.

Usually μk<μs\mu_k < \mu_s: at rest the bumps settle deep into each other; while sliding they have no time to lock in. That is why a heavy box is harder to start than to keep moving. Once sliding, a harder push does not increase the friction; it just gives more acceleration.

Surfacesμ_sμ_k
wood on wood0.4–0.50.2–0.3
steel on steel0.70.6
rubber on concrete1.00.8
ice on ice0.10.03
Teflon on Teflon0.040.04

Example: a 10 kg block sliding with μk=0.2\mu_k = 0.2 feels fk=0.2×100=20f_k = 0.2 \times 100 = 20 N, opposite to its motion.

4Laws of friction

  1. Friction opposes relative motion or its tendency.
  2. Limiting friction is proportional to the normal force: fmax⁡=μsNf_{\max} = \mu_s N.
  3. Kinetic friction is proportional to the normal force: fk=μkNf_k = \mu_k N.
  4. Friction does not depend on the area of contact (same N, same surfaces). A brick flat or on its edge needs the same pull; stack a second brick and the friction doubles.
  5. Kinetic friction hardly depends on speed at ordinary speeds.
μ=fN\mu = \frac{f}{N}A ratio of two forces, so it has no unit. It depends on the materials, usually 0.1 to 1.5.

5Angle of repose

Tilt a plank with a block on it. At first static friction holds the block; at one angle it starts to slide. That angle is the angle of repose.

Just about to slide: mgsin⁡θ=μsNmg\sin\theta = \mu_s N and N=mgcos⁡θN = mg\cos\theta, so μs=tan⁡θ\mu_s = \tan\theta.

tan⁡θR=μs\tan\theta_R = \mu_sThe block stays at rest while tan θ ≤ μ_s. With μ_s = 0.58 it holds at 30° and slides at 31°.

The angle of friction λ\lambda is the angle between the normal force and the total contact force RR (normal plus limiting friction): tan⁡λ=μsN/N=μs\tan\lambda = \mu_s N/N = \mu_s. So the angle of repose equals the angle of friction.

Example: a block starts sliding at 37°, so μs=tan⁡37°=0.75\mu_s = \tan 37° = 0.75.

6Sliding on slopes

Friction always opposes the motion along the slope:

CaseFrictionEquation
At restopposite to the way it would slidef≤μsmgcos⁡θf \le \mu_s mg\cos\theta
Sliding downup the slopea=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta)
Moving up (pushed with F)down the slopeF−mgsin⁡θ−μkmgcos⁡θ=maF - mg\sin\theta - \mu_k mg\cos\theta = ma
Moving down while pushed up with Fup the slopemgsin⁡θ−F−μkmgcos⁡θ=mamg\sin\theta - F - \mu_k mg\cos\theta = ma
a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta)

Example: 30° with μk=0.2\mu_k = 0.2: a=10(0.5−0.173)=3.27a = 10(0.5 - 0.173) = 3.27 m/s². At 37° with μk=0.5\mu_k = 0.5: a=10(0.6−0.4)=2a = 10(0.6 - 0.4) = 2 m/s². A block sent up a 30° slope slows at g(sin⁡θ+μkcos⁡θ)=6.73g(\sin\theta + \mu_k\cos\theta) = 6.73 m/s².

7Pulling and braking

Pull a crate with a rope at angle θ\theta above the horizontal. The vertical part Fsin⁡θF\sin\theta lifts it a little, so N=mg−Fsin⁡θN = mg - F\sin\theta. At the limit Fcos⁡θ=μs(mg−Fsin⁡θ)F\cos\theta = \mu_s(mg - F\sin\theta):

F=μsmgcos⁡θ+μssin⁡θF = \frac{\mu_s mg}{\cos\theta + \mu_s\sin\theta}20 kg, μ_s = 0.5, θ = 30°: F = 100/1.116 ≈ 89.6 N.

The pull is smallest when tan⁡θ=μ\tan\theta = \mu, that is, at the angle of friction λ\lambda above the horizontal:

Fmin⁡=μmg1+μ2=mgsin⁡λF_{\min} = \frac{\mu mg}{\sqrt{1 + \mu^2}} = mg\sin\lambdaWith μ = 0.75 and 10 kg: θ ≈ 37°, F_min = 75/1.25 = 60 N.

A box on a truck accelerating at aa stays put only if friction can give it mama: μmg≥ma\mu mg \ge ma, so μ≥a/g\mu \ge a/g. At 3 m/s², μ≥0.3\mu \ge 0.3. The same limit means a car cannot accelerate faster than μsg\mu_s g without wheelspin.

Braking: friction is the only horizontal force, so the car decelerates at μkg\mu_k g. From v2=2asv^2 = 2as:

s=v22μkgs = \frac{v^2}{2\mu_k g}72 km/h = 20 m/s with μ_k = 0.6: s = 400/12 ≈ 33.3 m. 10 m/s with μ_k = 0.5: 10 m. To stop from 20 m/s in 25 m needs μ_k = 0.8.

Summary

Key ideas

  • Friction opposes relative motion or its tendency; it comes from interlocking surface bumps.
  • Static friction adjusts itself from zero up to μ_s N.
  • Kinetic friction is μ_k N while sliding; usually μ_k < μ_s.
  • Friction is proportional to the normal force and does not depend on the contact area.
  • μ is a ratio of forces and has no unit.
  • The angle of repose satisfies tan θ_R = μ_s and equals the angle of friction.
  • On a slope friction opposes the motion: up the slope when sliding down, down the slope when moving up.
  • Sliding down a rough slope: a = g(sin θ − μ_k cos θ).
  • Pulling at the angle of friction needs the least force.
  • Braking distance is v²/(2μ_k g); a box on a truck needs μ ≥ a/g.

Every equation

Static friction
fs≤μsNf_s \le \mu_s N
Limiting friction
fmax⁡=μsNf_{\max} = \mu_s N
Kinetic friction
fk=μkNf_k = \mu_k N
Coefficient
μ=f/N\mu = f/N
Least horizontal push
F=μsmgF = \mu_s mg
Angle of repose
tan⁡θR=μs\tan\theta_R = \mu_s
Angle of friction
tan⁡λ=μs\tan\lambda = \mu_s
Rest on a slope
tan⁡θ≤μs\tan\theta \le \mu_s
Sliding down
a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta)
Pull at an angle
F=μsmgcos⁡θ+μssin⁡θF = \frac{\mu_s mg}{\cos\theta + \mu_s\sin\theta}
Least pull
Fmin⁡=μmg1+μ2F_{\min} = \frac{\mu mg}{\sqrt{1 + \mu^2}}
Box on a truck
μ≥a/g\mu \ge a/g
Largest acceleration
amax⁡=μsga_{\max} = \mu_s g
Stopping distance
s=v22μkgs = \frac{v^2}{2\mu_k g}
Wedge held by friction
a=gsin⁡θ−μcos⁡θcos⁡θ+μsin⁡θa = g\frac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta}
Belt on a pulley
T1/T2=eμθT_1/T_2 = e^{\mu\theta}

Previous year questions with solutions

Real JEE and NEET questions on friction – static and kinetic. Try each one before you open the solution.

Q1JEE Main 2025One correct option

A cubic block of mass mm is sliding down on an inclined plane at 60∘{60}^{\circ } with an acceleration of g2\frac{g}{2}, the value of coefficient of kinetic friction is

  1. A32\frac{\sqrt{3}}{2}
  2. B23\frac{\sqrt{2}}{3}
  3. C1−321-\frac{\sqrt{3}}{2}
  4. D3−1\sqrt{3}-1
Show answer and solution

Answer: Option D

Same equation, steeper slope: gsin⁡60∘−μgcos⁡60∘=g2g\sin 60^{\circ} - \mu g\cos 60^{\circ} = \dfrac{g}{2}. Divide by gg and it reads 32−μ2=12\dfrac{\sqrt{3}}{2} - \dfrac{\mu}{2} = \dfrac{1}{2}, so μ=3−1≈0.73\mu = \sqrt{3} - 1 \approx 0.73. Watch what the steep angle did to the cosine: at 60∘60^{\circ} the block presses into the slope with only half its weight, so friction has very little to work with, and μ\mu has to be large to slow the block as much as it does. Repose check: sliding needs μ<tan⁡60∘=1.73\mu < \tan 60^{\circ} = 1.73, and 0.730.73 clears it comfortably.

Q2NEET 2024One correct option

A box of mass 5kg5 \mathrm{kg} is pulled by a cord, up along a frictionless plane inclined at 30∘{30}^{\circ } with the horizontal. The tension in the cord is 30N30 N. The acceleration of the box is (Take g=10ms−2g=10 ms^{-2})

  1. A2ms−22 ms^{-2}
  2. BZero
  3. C0.1ms−20.1 ms^{-2}
  4. D1ms−21 ms^{-2}
Show answer and solution

Answer: Option D

Tilt the axes and only one direction has anything in it. Up the slope the cord pulls with T=30 NT = 30\ \mathrm{N}; down the slope the weight pulls back with mgsin⁡30∘=5×10×12=25 Nmg\sin 30^{\circ} = 5 \times 10 \times \tfrac{1}{2} = 25\ \mathrm{N}; the plane is frictionless, so nothing else lies along the slope at all. T−mgsin⁡θ=maT - mg\sin\theta = ma gives 30−25=5a30 - 25 = 5a, so a=1 m/s2a = 1\ \mathrm{m/s^{2}} up the slope. The normal reaction never entered the sum, because it is perpendicular to the only axis the motion is on — which is the entire reason for tilting the axes. Had the tension been exactly 25 N25\ \mathrm{N} the box would have gone up at a steady speed, and below that it would have slid back.

Q3JEE Advanced 2011Numerical answer

A block is moving on an inclined plane making an angle 45∘{45}^{\circ } with the horizontal and the coefficient of friction is μ\mu. The force required to just push it up the inclined plane is 3 times the force required to just

prevent it from sliding down. If we define N = 10 μ\mu, then N is

Show answer and solution

Answer: 5

The same two expressions at 45∘45^{\circ}: F1=mg2(1+μ)F_{1} = \dfrac{mg}{\sqrt{2}}(1+\mu) to push it up, F2=mg2(1−μ)F_{2} = \dfrac{mg}{\sqrt{2}}(1-\mu) to just hold it. F1=3F2F_{1} = 3F_{2} gives 1+μ=3−3μ1 + \mu = 3 - 3\mu, so 4μ=24\mu = 2 and μ=0.5\mu = 0.5, making N=10μ=5N = 10\mu = 5. Only the multiplier changed from the doubling version, and there is a general result hiding in it: at 45∘45^{\circ} the ratio is 1+μ1−μ\dfrac{1+\mu}{1-\mu}, so a multiplier of kk always gives μ=k−1k+1\mu = \dfrac{k-1}{k+1}. Here k=3k = 3 and μ=12\mu = \dfrac{1}{2}; with k=2k = 2 it would have been 13\dfrac{1}{3}.

Practice questions, easy to hard

Three questions from the friction – static and kinetic practice ladder: one easy, one medium, one hard.

Q4Numerical answer

Now let it move. On a smooth slope the only force along the slope is mgsin⁡θmg\sin\theta — the normal reaction is perpendicular and contributes nothing there — so ma=mgsin⁡θma = mg\sin\theta and a=gsin⁡θa = g\sin\theta. The mass has gone from the answer entirely: a marble and a cannonball released together on the same smooth slope stay level all the way down.

A smooth slope is inclined at 30∘30^{\circ}. What is the acceleration of a block released on it, in m/s2\mathrm{m/s^{2}}? (Take g=10 m/s2g = 10\ \mathrm{m/s^{2}})

Show answer and solution

Answer: 5 m/s²

a=gsin⁡30∘=10×12=5 m/s2a = g\sin 30^{\circ} = 10 \times \tfrac{1}{2} = 5\ \mathrm{m/s^{2}}, down the slope. Check the two ends again: at θ=0\theta = 0 it is zero, and at θ=90∘\theta = 90^{\circ} it is gg, plain free fall. A smooth slope is a way of diluting gravity by a factor of sin⁡θ\sin\theta, which is exactly what Galileo used one for — a ball creeping down at a fifth of gg can be timed by hand, and a falling one cannot.

Q5One correct option

Project a body up a rough slope and it decelerates at g(sin⁡θ+μcos⁡θ)g(\sin\theta + \mu\cos\theta) until it stops. Whether it comes back is a separate question: it slides down again only if gravity can beat static friction, that is, only if tan⁡θ>μ\tan\theta > \mu. If it does, it returns at g(sin⁡θ−μcos⁡θ)g(\sin\theta - \mu\cos\theta), the smaller of the two, over the same distance.

For a fixed distance LL with one end of the journey at rest, L=12at2L = \tfrac{1}{2}at^{2} gives t∝1at \propto \dfrac{1}{\sqrt{a}}. What is tuptdown\dfrac{t_{\mathrm{up}}}{t_{\mathrm{down}}}?

  1. Asin⁡θ−μcos⁡θsin⁡θ+μcos⁡θ\dfrac{\sin\theta - \mu\cos\theta}{\sin\theta + \mu\cos\theta}
  2. Bsin⁡θ+μcos⁡θsin⁡θ−μcos⁡θ\dfrac{\sin\theta + \mu\cos\theta}{\sin\theta - \mu\cos\theta}
  3. Csin⁡θ+μcos⁡θsin⁡θ−μcos⁡θ\sqrt{\dfrac{\sin\theta + \mu\cos\theta}{\sin\theta - \mu\cos\theta}}
  4. Dsin⁡θ−μcos⁡θsin⁡θ+μcos⁡θ\sqrt{\dfrac{\sin\theta - \mu\cos\theta}{\sin\theta + \mu\cos\theta}}
Show answer and solution

Answer: Option D

Decelerating from uu to rest over LL takes exactly as long as accelerating from rest over LL at the same rate, so both legs are t=2Lat = \sqrt{\dfrac{2L}{a}} and tuptdown=adownaup\dfrac{t_{\mathrm{up}}}{t_{\mathrm{down}}} = \sqrt{\dfrac{a_{\mathrm{down}}}{a_{\mathrm{up}}}} — the accelerations invert, because a gentler one means a longer time. Since aupa_{\mathrm{up}} is the larger, tup<tdownt_{\mathrm{up}} < t_{\mathrm{down}}: going up is always the quicker leg. The form to carry is the squared one, tup2tdown2=sin⁡θ−μcos⁡θsin⁡θ+μcos⁡θ\dfrac{t_{\mathrm{up}}^{2}}{t_{\mathrm{down}}^{2}} = \dfrac{\sin\theta - \mu\cos\theta}{\sin\theta + \mu\cos\theta}, which turns every question about a time of ascent against a time of descent into one line of algebra.