Friction and Forces
A block slides on a surface with adjustable static and kinetic friction coefficients — explore how friction affects motion and stopping distance.
Objective
Confirm that a block sliding on a horizontal surface decelerates uniformly under kinetic friction and verify the closed-form stopping-distance relation d = v₀² / (2·μk·g), where v₀ is the initial speed, μk is the coefficient of kinetic friction, and g = 9.81 m/s². Read the friction force directly as f_k = μk·m·g, and observe how halving or doubling μk transforms both the stopping time and the stopping distance. The surface is uniform and the block is rigid — friction is the only horizontal force at play.
Setup
- Press Reset to return the block to its starting position at the left of the track. The Time, Speed, Friction, and Distance readouts all return to 0.00, and the trail clears.
- Set the Static Friction (μs) slider to its default value of 0.50. The block in this experiment is launched directly at 8 m/s, so μs is recorded for reference rather than acting as a release threshold.
- Set the Kinetic Friction (μk) slider to its default value of 0.30. This is the coefficient that will decelerate the block; it sets both the friction force and the rate of slowing.
- Press Start. The block accelerates instantly to its launch speed of 8 m/s and slides rightward; a red arrow shows the friction force pointing backward, opposite to motion.
- Wait until the Speed readout reaches 0.00 m/s and the simulation halts. Record the final values shown for Time, Friction, and Distance — these are the numbers compared against the prediction.
Analytical Prediction
On a horizontal surface with no applied force after launch, kinetic friction is the only horizontal force on the block. Newton's second law gives a constant deceleration a = −μk·g. The block stops after t_stop = v₀ / (μk·g) and travels d = v₀² / (2·μk·g). The kinetic friction force itself is f_k = μk·m·g and stays constant while the block moves. With v₀ = 8 m/s, μk = 0.30, m = 5 kg, g = 9.81 m/s²:
These four values — deceleration, stopping time, stopping distance, and friction force — are the targets to verify against the readouts.
Results Analysis
When the run halts, the readouts settle at fixed values that can be compared one by one to the prediction. With the default μk = 0.30 and v₀ = 8 m/s, the Time readout should land at ≈ 2.72 s, Distance at ≈ 10.87 m, and Friction at ≈ 14.72 N (= 0.30 · 5 · 9.81); Speed reads 0.00 m/s at termination. Agreement within roughly 0.5% confirms that the simulated motion obeys a = −μk·g. A sharper test: press Reset and double μk to 0.60 while leaving everything else fixed. The prediction d = 64 / (2 · 0.60 · 9.81) ≈ 5.44 m is exactly half the default. The Friction readout doubles to ≈ 29.43 N, and the stopping time halves to ≈ 1.36 s. Halving μk to 0.15 should instead double the distance to ≈ 21.75 m. These scaling checks demonstrate the inverse-linear dependence of d on μk that the formula encodes.
Source of Error
What this sim does NOT model: air drag, surface temperature variations, pressure-dependent friction, deformation of the block or the surface, or any rolling component (the block slides as a point mass). Static and kinetic friction coefficients are treated as constants independent of speed. The closed forms f_s ≤ μs·N and f_k = μk·m·g assume the same idealizations, so they cancel rather than contributing to the residual stopping distance or speed. The remaining gap between prediction and readouts is therefore purely numerical, not physical, for this sim.
Further Exploration
- Run the experiment at μk = 0.10, 0.20, 0.30, 0.50, 0.80, and 1.00, recording the Distance readout at each. Plot d versus 1/μk on paper. Does the data fall on a straight line through the origin, as d = v₀² / (2·μk·g) predicts?
- The mass slider is fixed at 5 kg in this simulation. Using the formula, explain why doubling the mass would leave the stopping distance unchanged even though the friction force itself would double. Which two terms cancel?
- Set μs = 1.50 and μk = 0.10 — the largest gap between static and kinetic friction the sliders allow. The block still launches at 8 m/s and slides freely. Why does μs not affect the run, and under what physical scenario would it dominate instead?
- Predict the stopping time analytically for μk = 0.25, then run the sim and compare with the Time readout. The formula t_stop = v₀ / (μk·g) gives ≈ 3.26 s. How close does the readout come, and which of the error sources in the previous section likely accounts for the gap?
- If the block were sliding on ice with μk ≈ 0.03, predict the stopping distance with v₀ = 8 m/s. The slider minimum is 0.10, so this value lies outside the experimental range — but the formula still gives an answer. What does d ≈ 108.7 m tell you about why winter braking distances grow so dramatically?