Input parameters

Block and ramp

Mass

Mass of the block

1.000e-41.000e+6

Plane angle

Inclination from the horizontal (0–90°)

090

Plane length

Available run along the ramp

0.11000

Surfaces and frictionSurface pair (μs, μk)

Static

Resists the start of motion

01.5

Kinetic

Acts during sliding

01.5

Gravity9.807 m/s²

Gravity preset

Gravity

m/s²

Gravitational acceleration of the system

0.011000

Inclined plane & forces

θ = 30.00°g = 9.807 m/s²

Ep 49.03 J

Ec 0 J

Q 0 J

Show forcesInitial position

Kinetic friction5.096 N · uphill

Parallel weight (mg·sin θ)9.807 N

Energy distributionEp 100%Ec 0%Friction Q 0%

Animation duration is adapted for viewing. t is physical time; predictions describe the complete path.

Full-path prediction

EnergySpeedTime

Slides

Acceleration

2.355

m/s²

Final speed

4.853

m/s

Normal force (N): 16.99 N
Critical angle (θc): 26.57 °
Slide time: 2.060 s
Plane height: 2.500 m
Initial potential energy (Ep): 49.03 J
Energy dissipated by friction (Q): 25.48 J
Final kinetic energy (Ec): 23.55 J
Total dissipated energy: 51.96 %

Ideal model: point-like block, dry Coulomb friction with 0 ≤ μk ≤ μs, constant gravity and no air resistance. Nonrelativistic mechanics. Material presets are approximate.

Fundamentals & Explanation

Resolving the weight

On an incline it is convenient to use one axis parallel and one perpendicular to the surface. The weight $mg$ splits into a parallel component, $mg\sin\theta$ , which pulls downhill, and a perpendicular one, $mg\cos\theta$ , which the normal force balances because the block does not leave the plane.

\begin{gathered}N = mg\cos\theta\\ F_{\parallel} = mg\sin\theta\end{gathered}

Friction acts along the contact. While sliding it equals $f_k = \mu_k N$ and opposes the motion. At rest, static friction takes whatever value is needed for balance, bounded by $|f_s| \le \mu_s N$ : it reaches $\mu_s N$ only at the threshold.

Does it slide or stay at rest?

The block stays at rest as long as maximum static friction can balance the parallel component, that is, while $mg\sin\theta \le \mu_s\, mg\cos\theta$ . Dividing by $mg\cos\theta$ , mass and gravity cancel out:

\begin{gathered}\text{slides} \iff \tan\theta > \mu_s\\ \theta_c = \arctan\mu_s\end{gathered}

At exactly the critical angle, static friction reaches its maximum and the block is in limiting equilibrium; any steeper incline makes it slide.

Measuring μs by tilting the surface

The critical-angle criterion suggests a method that needs no force gauge: rest the object on the surface and tilt it slowly until it starts to slide. At that angle, $\mu_s = \tan\theta_c$ , regardless of mass. The tilt must increase slowly and without knocks: a vibration breaks the rest before the threshold and leads to underestimating $\mu_s$ .

Once motion starts, friction becomes kinetic. Since $\mu_k \le \mu_s$ , the block accelerates even if the angle stops increasing.

Downhill and uphill accelerations

\begin{aligned}a_{\downarrow} &= g(\sin\theta - \mu_k\cos\theta)\\ a_{\uparrow} &= -g(\sin\theta + \mu_k\cos\theta)\end{aligned}

Going down, kinetic friction subtracts from the weight component; going up, weight and friction both point downhill and brake together. Mass cancels out of both expressions: with the same coefficients, a 1 kg block and a 100 kg block slide exactly alike. At $\theta = 90°$ the normal force vanishes and free fall, $a = g$ , is recovered.

This creates an asymmetry: if a launched block comes back and friction is nonzero, it takes less time going up than coming down the same stretch; without friction both times are equal. If $\tan\theta \le \mu_s$ , a block that stops at the top stays at rest there.

Energy balance

\begin{aligned}E_p &= mgL\sin\theta\\ Q &= \mu_k N L\\ E_c &= E_p - Q\end{aligned}

With kinetic friction, mechanical energy decreases. $Q \ge 0$ is the energy dissipated in the surfaces, and friction does work $W_f = -Q$ on the block. On the way down, the dissipated fraction is $Q/E_p = \mu_k/\tan\theta$ , independent of mass, length and gravity.

Without friction, $Q = 0$ and mechanical energy is conserved, as in free fall.

Worked example: sliding down with friction

A 2 kg block is released from rest on a 5 m plane with $\theta = 30°$ , $\mu_s = 0.45$ and $\mu_k = 0.30$ . We use $g = 9.80665\ \text{m/s}^2$ , rounded in the steps.

1. Check whether it slides. $\tan 30° = 0.577$ exceeds $\mu_s = 0.45$ , so the block slides. The critical angle is $\theta_c = \arctan 0.45 = 24.23°$ .

2. Normal force and acceleration.

\begin{aligned}N &= 2 \cdot 9.807 \cdot \cos 30°\\ &= 16.99\ \text{N}\\ a &= 9.807\,(0.5 - 0.30\cdot 0.866)\\ &= 2.355\ \text{m/s}^2\end{aligned}

3. Time and speed at the bottom. The motion is uniformly accelerated from rest:

\begin{aligned}t &= \sqrt{2L/a} = \sqrt{2\cdot 5/2.355}\\ &= 2.060\ \text{s}\\ v_f &= \sqrt{2aL} = \sqrt{2\cdot 2.355\cdot 5}\\ &= 4.853\ \text{m/s}\end{aligned}

4. Energy balance.

\begin{aligned}E_p &= 2\cdot 9.807\cdot 5\cdot\sin 30°\\ &= 49.03\ \text{J}\\ Q &= 0.30\cdot 16.99\cdot 5 = 25.48\ \text{J}\\ E_c &= E_p - Q = 23.55\ \text{J}\end{aligned}

Friction dissipates 51.96% of the initial potential energy, which is $\mu_k/\tan 30°$ : more than half ends up as heat in the surfaces. As a check, $\tfrac12 m v_f^2 = \tfrac12\cdot 2\cdot 4.853^2 = 23.55\ \text{J}$ matches $E_c$ .

Scope of the model

Dry Coulomb friction. The coefficients are taken as constant, with $0 \le \mu_k \le \mu_s$ ; in reality they depend on speed, temperature and surface condition.
Point mass. No rolling and no tipping: the block can only slide. Rolling spheres and cylinders require the moment of inertia.
No air resistance. It matters only at high speeds or for very light objects.
Indicative coefficients. The tabulated material pairs are representative; real values vary with humidity, polish and wear.

Sources

OpenStax, University Physics Vol. 1, §6.2 Friction: coefficients in Table 6.1 and the inequality $f_s \le \mu_s N$ .

OpenStax, §5.7 Drawing Free-Body Diagrams: resolving the weight along axes parallel and perpendicular to the plane.

OpenStax, §8.3 Conservation of Energy: mechanical energy and the work of non-conservative forces.

NIST, standard acceleration of gravity: $g_n = 9.80665\ \text{m/s}^2$ .

Inclined Plane with Friction

Input parameters

Inclined plane & forces

Full-path prediction

Fundamentals & Explanation

Resolving the weight

Does it slide or stay at rest?

Measuring μs by tilting the surface

Downhill and uphill accelerations

Energy balance

Worked example: sliding down with friction

Scope of the model

Sources

Resolving the weight

Does it slide or stay at rest?

Measuring μs by tilting the surface

Downhill and uphill accelerations

Energy balance

Worked example: sliding down with friction

Scope of the model

Sources