How to Calculate the Coefficient of Friction from Stopping Distance

Two blocks slide across the same horizontal surface at the same initial speed. One has twice the mass of the other. Which one travels further before stopping?

Under the standard kinetic-friction model, they travel the same distance. The heavier block experiences twice the friction force, but it also has twice the inertia. These effects cancel.

This gives us a useful way to measure friction. If you know an object’s initial speed and the distance it travels before stopping, you can calculate the coefficient of kinetic friction.

For an object sliding on a horizontal surface, with friction as its only horizontal force:

\[ \boxed{\mu_k=\frac{u^2}{2gd}} \]

Where:

• \(\mu_k\) is the coefficient of kinetic friction.
• \(u\) is the initial speed in \(\mathrm{m\,s^{-1}}\).
• \(g\) is gravitational acceleration, approximately \(9.8\,\mathrm{m\,s^{-2}}\).
• \(d\) is the stopping distance in metres.

There is one important catch: you cannot calculate the coefficient from stopping distance alone. You also need the initial speed.

The method also gives you the kinetic coefficient, not the static coefficient. Understanding why requires looking at the forces acting while an object slides.

01Why stopping distance tells us about friction

Imagine pushing a wooden block across a table and letting go. The block continues moving, but gradually slows down.

What causes it to stop?

The friction force between the block and the table acts opposite to its motion. This force produces a negative acceleration, reducing the block’s speed until it reaches zero.

A larger friction force produces a greater deceleration. For the same initial speed, this means a shorter stopping distance.

Deriving the stopping-distance formula

We can derive the formula using Newton’s second law and the constant-acceleration equations.

First, make an important assumption: the block is already sliding, the surface is horizontal, and kinetic friction is the only horizontal force.

For the vertical forces, the block has no vertical acceleration. Therefore:

\[ N=mg \]

Here, \(N\) is the normal force, \(m\) is the block’s mass, and \(g\) is gravitational acceleration.

For this calculation, we use the standard kinetic-friction model:

\[ f_k=\mu_kN \]

Substituting the normal force:

\[ f_k=\mu_kmg \]

Now choose the direction of the block’s initial motion as positive.

Friction acts backwards, so Newton’s second law gives:

\[ \sum F_x=-f_k=ma \]

Substituting:

\[ -\mu_kmg=ma \]

Cancel the mass:

\[ a=-\mu_kg \]

Notice something important: the mass has disappeared. Under this model, the deceleration does not depend on the object’s mass.

Next, use the constant-acceleration equation:

\[ v^2=u^2+2ad \]

The block finishes at rest, so \(v=0\). Substituting our expression for acceleration:

\[ 0=u^2-2\mu_kgd \]

Rearranging gives:

\[ \boxed{\mu_k=\frac{u^2}{2gd}} \]

That’s the formula you need.

What does the formula tell us?

Rearranging the equation also gives:

\[ d=\frac{u^2}{2\mu_kg} \]

This reveals two useful relationships.

• If the initial speed doubles, the stopping distance becomes four times greater, assuming the same friction coefficient.
• If the coefficient of kinetic friction doubles, the stopping distance halves, assuming the same initial speed.

For example, a block travelling at \(6\,\mathrm{m\,s^{-1}}\) does not need twice the stopping distance of a block travelling at \(3\,\mathrm{m\,s^{-1}}\). It needs four times the distance.

This squared relationship is particularly useful when interpreting experimental data.

02How an experiment measures the coefficient

Suppose you want to determine the coefficient of kinetic friction between a wooden block and a particular surface.

You need two measurements:

1. The block’s speed when it begins sliding across the test surface.
2. The distance from that starting position until the block stops.

A suitable experimental arrangement uses a spring launcher, a photogate, and a horizontal test surface.

Following the experiment

Step 1: Launch the block.

The spring launcher gives the block an initial speed. It must release the block before the block reaches the photogate, so that the launcher does not continue applying a force during the measurement.

Step 2: Measure the initial speed.

A photogate records how long a card of known length takes to pass through its beam.

If the interrupt card has length \(0.050\,\mathrm m\) and blocks the beam for \(0.0125\,\mathrm s\), the measured speed is approximately:

\[ u=\frac{\text{card length}}{\text{time}} \]

\[ u=\frac{0.050}{0.0125} =4.0\,\mathrm{m\,s^{-1}} \]

This provides the initial speed at the starting line.

Step 3: Measure the stopping distance.

Measure the distance from the photogate’s starting line to the block’s final resting position.

The reference point on the block must be measured consistently. Do not measure from the photogate in one trial and from the launcher in another.

Step 4: Calculate the coefficient.

Use:

\[ \mu_k=\frac{u^2}{2gd} \]

Repeat the experiment several times and calculate the mean coefficient.

Why the original measurement method needs improving

Imagine an experiment where students simply compress a spring by the same amount each time, launch a block, and measure its stopping distance.

It seems reasonable to assume that identical spring compression produces identical initial speeds.

However, the actual speed can vary because of differences in release, alignment, and energy transferred during the launch.

There is also a more fundamental problem: if you only measure stopping distance, you cannot separate the effect of initial speed from the effect of friction.

The revision is simple. Add a photogate to measure the actual initial speed in every trial.

Now each stopping distance can be matched to its corresponding initial speed.

Interpreting experimental results

Consider the following illustrative data. These are example values, not results from an actual experiment.

The calculated coefficients are very similar.

Their mean is approximately:

\[ \overline{\mu_k}=0.324 \]

This supports the assumption that the kinetic-friction coefficient remains approximately constant across these trials.

You can also investigate the relationship between initial speed and stopping distance by plotting \(d\) against \(u^2\).

From the equation:

\[ d=\frac{1}{2\mu_kg}u^2 \]

The gradient of this graph is:

\[ \text{gradient}=\frac{1}{2\mu_kg} \]

Therefore:

\[ \mu_k=\frac{1}{2g(\text{gradient})} \]

This provides another way to determine the coefficient.

It also gives you a useful experimental check: if the relationship is approximately linear, the constant-friction model is reasonably consistent with the data.

03Static friction versus kinetic friction

Here’s a distinction that can cost marks in a Physics exam.

The coefficient calculated from stopping distance is the coefficient of kinetic friction. It is not the coefficient of static friction.

The two describe different situations.

Static friction: before sliding begins

Imagine a block resting on a horizontal table. You gradually increase a horizontal pushing force.

Initially, the block doesn’t move. Static friction adjusts to oppose your push.

If you push with \(2\,\mathrm N\), static friction can provide \(2\,\mathrm N\) in the opposite direction. If you increase your push to \(5\,\mathrm N\), static friction can increase to \(5\,\mathrm N\).

However, static friction has a maximum value:

\[ \boxed{f_s\leq\mu_sN} \]

Here:

• \(f_s\) is the actual static-friction force.
• \(\mu_s\) is the coefficient of static friction.
• \(N\) is the normal force.

The maximum, or limiting, static friction is:

\[ f_{s,\max}=\mu_sN \]

At the instant the block is about to slip, static friction reaches this maximum.

The important point is that static friction is an inequality. It does not automatically equal \(\mu_sN\).

Kinetic friction: after sliding begins

Once the block starts sliding, we switch to the kinetic-friction model:

\[ \boxed{f_k=\mu_kN} \]

For the standard HSC model, we treat the coefficient as constant during the motion.

This is the friction relevant to our stopping-distance calculation.

Can stopping distance give you the coefficient of static friction?

No.

A stopping-distance experiment measures the friction acting while the block is sliding.

To measure static friction, you need an experiment that identifies the point at which sliding begins.

One possible method is to gradually tilt a surface until the block just starts slipping.

At the critical angle \(\theta_c\):

\[ mg\sin\theta_c=\mu_smg\cos\theta_c \]

Rearranging:

\[ \boxed{\mu_s=\tan\theta_c} \]

This measures the limiting static-friction coefficient under the stated assumptions.

Keep the two methods separate:

One final limitation: the standard kinetic-friction equation is a model. Real surfaces can behave differently as their condition, temperature, or sliding speed changes. A coefficient calculated from stopping distance should therefore be understood as an estimate for the particular contact conditions.

04Worked examples

05Questions and solutions

The following questions build from a standard stopping-distance calculation to more involved situations involving changing mass, experimental measurements, inclined planes, and the transition between kinetic and static friction.

Question 1

A \(0.80\,\mathrm{kg}\) block slides across a level horizontal surface at an initial speed of \(3.6\,\mathrm{m\,s^{-1}}\). It comes to rest after travelling \(1.5\,\mathrm m\).

Assuming kinetic friction is the only horizontal force, calculate the coefficient of kinetic friction.

Use \(g=9.8\,\mathrm{m\,s^{-2}}\).

Solution 1

The coefficient of kinetic friction is approximately \(0.44\).

Step 1: Identify the known values.

\[ u=3.6\,\mathrm{m\,s^{-1}} \]

\[ v=0,\qquad d=1.5\,\mathrm m \]

Step 2: Apply the stopping-distance formula.

\[ \mu_k=\frac{u^2}{2gd} \]

Substitute:

\[ \mu_k= \frac{(3.6)^2}{2(9.8)(1.5)} \]

\[ \mu_k=\frac{12.96}{29.4} \]

\[ \boxed{\mu_k=0.44} \]

The block experiences a coefficient of kinetic friction of approximately \(0.44\).

The mass is irrelevant to this calculation under the stated model because it cancels when Newton’s second law is applied.

Question 2

A student investigates how mass and initial speed affect stopping distance on the same horizontal surface.

The same block is used in both trials, with additional mass added for the second trial.

The following results are recorded:

Assume kinetic friction is the only horizontal force.

(a) Calculate the coefficient of kinetic friction for each trial.

(b) Explain whether the results support the assumption of a constant coefficient of kinetic friction.

(c) Explain why increasing the mass by 50% does not change the stopping distance predicted for a given initial speed.

Solution 2

Both trials give a coefficient of approximately \(0.40\). The results support the constant-coefficient model, and the increased mass does not change the predicted stopping distance at a fixed initial speed.

(a) Calculate the coefficient for each trial.

For Trial A:

\[ \mu_k=\frac{u^2}{2gd} \]

\[ \mu_k= \frac{(2.5)^2}{2(9.8)(0.80)} \]

\[ \mu_k=\frac{6.25}{15.68} \]

\[ \mu_k=0.3986 \]

Therefore:

\[ \mu_{k,A}\approx0.399 \]

For Trial B:

\[ \mu_k= \frac{(5.0)^2}{2(9.8)(3.20)} \]

\[ \mu_k=\frac{25}{62.72} \]

\[ \mu_k=0.3986 \]

Therefore:

\[ \mu_{k,B}\approx0.399 \]

(b) Assess the results.

Both trials produce approximately the same coefficient.

This supports the assumption that kinetic friction can be modelled using a constant coefficient for these conditions.

It also demonstrates the squared relationship between speed and stopping distance.

The initial speed doubles from \(2.5\) to \(5.0\,\mathrm{m\,s^{-1}}\), while the stopping distance increases from \(0.80\) to \(3.20\,\mathrm m\).

The stopping distance becomes four times greater.

(c) Explain the effect of mass.

The kinetic-friction model gives:

\[ f_k=\mu_kN \]

On a horizontal surface:

\[ f_k=\mu_kmg \]

Newton’s second law gives:

\[ a=-\frac{f_k}{m} \]

Substituting:

\[ a=-\frac{\mu_kmg}{m} \]

\[ a=-\mu_kg \]

The mass cancels.

Therefore, increasing mass increases the friction force, but it also increases the object’s inertia by the same proportion.

The resulting deceleration remains unchanged, so the predicted stopping distance at a given initial speed is also unchanged.

Question 3

A student uses two photogates to investigate the motion of a block sliding across a level horizontal surface.

Photogate A records an initial speed of \(4.00\,\mathrm{m\,s^{-1}}\).

Photogate B is positioned \(0.50\,\mathrm m\) further along the track and records a speed of \(3.58\,\mathrm{m\,s^{-1}}\).

The block then travels another \(2.00\,\mathrm m\) before stopping.

Assume kinetic friction is the only horizontal force and remains constant throughout the motion.

(a) Calculate the coefficient of kinetic friction between A and B.

(b) Calculate the coefficient between B and the stopping position.

(c) Determine whether the results are consistent with a constant coefficient, allowing for rounding.

(d) A student calculates the coefficient using \(4.00\,\mathrm{m\,s^{-1}}\) as the initial speed and \(2.00\,\mathrm m\) as the stopping distance. Explain the error.

Use \(g=9.8\,\mathrm{m\,s^{-2}}\).

Solution 3

The coefficient is approximately \(0.326\). The two intervals give consistent results within rounding, but the student’s calculation in part (d) incorrectly combines measurements from different starting positions.

(a) Calculate the coefficient between A and B.

First, use the constant-acceleration equation:

\[ v^2=u^2+2a\Delta x \]

Rearranging:

\[ a=\frac{v^2-u^2}{2\Delta x} \]

Substitute the measurements:

\[ a= \frac{(3.58)^2-(4.00)^2} {2(0.50)} \]

\[ a=\frac{12.8164-16.00}{1.00} \]

\[ a=-3.1836\,\mathrm{m\,s^{-2}} \]

On a horizontal surface, with friction as the only horizontal force:

\[ a=-\mu_kg \]

Therefore:

\[ \mu_k=\frac{|a|}{g} \]

\[ \mu_k=\frac{3.1836}{9.8} \]

\[ \boxed{\mu_{k,AB}=0.325} \]

(b) Calculate the coefficient between B and the stopping position.

At B, the block’s speed is \(3.58\,\mathrm{m\,s^{-1}}\).

It stops after another \(2.00\,\mathrm m\).

Apply:

\[ \mu_k=\frac{u^2}{2gd} \]

\[ \mu_k= \frac{(3.58)^2}{2(9.8)(2.00)} \]

\[ \mu_k=\frac{12.8164}{39.2} \]

\[ \boxed{\mu_{k,B}=0.327} \]

(c) Assess consistency.

The two estimates are:

\[ \mu_{k,AB}=0.325 \]

\[ \mu_{k,B}=0.327 \]

Their small difference is consistent with rounding in the recorded speed.

A reasonable overall estimate is:

\[ \boxed{\mu_k\approx0.326} \]

This supports the assumption of approximately constant kinetic friction throughout the track.

(d) Explain the student’s error.

The student’s calculation combines the speed at A with the distance measured from B.

These measurements do not describe the same interval.

If the student wants to use the speed at A, the corresponding stopping distance must include the distance from A to B as well:

\[ d_{\text{total}}=0.50+2.00 \]

\[ d_{\text{total}}=2.50\,\mathrm m \]

The correct calculation from A is:

\[ \mu_k= \frac{(4.00)^2}{2(9.8)(2.50)} \]

\[ \boxed{\mu_k=0.327} \]

The key experimental rule is simple: the initial speed and stopping distance must refer to the same starting position.

Question 4

A \(2.0\,\mathrm{kg}\) block slides up a straight incline at an initial speed of \(5.0\,\mathrm{m\,s^{-1}}\).

The incline makes an angle of \(25^\circ\) with the horizontal. The block comes to rest after travelling \(2.0\,\mathrm m\) along the incline.

Determine the coefficient of kinetic friction between the block and the incline.

Assume kinetic friction is constant, no other forces act along the incline, and \(g=9.8\,\mathrm{m\,s^{-2}}\).

Solution 4

The coefficient of kinetic friction is approximately \(0.24\).

The important difference from the earlier questions is that gravity now contributes to the block’s deceleration.

Step 1: Resolve the forces along the incline.

Choose uphill as the positive direction.

The component of gravitational force parallel to the incline is:

\[ F_{g,\parallel}=mg\sin\theta \]

The normal force is:

\[ N=mg\cos\theta \]

Therefore, kinetic friction is:

\[ f_k=\mu_kmg\cos\theta \]

Both gravity’s parallel component and friction act downhill, opposite to the block’s motion.

Step 2: Apply Newton’s second law along the incline.

\[ \sum F_{\parallel}=ma \]

\[ -mg\sin\theta-\mu_kmg\cos\theta=ma \]

Divide by the mass:

\[ a=-g(\sin\theta+\mu_k\cos\theta) \]

Step 3: Use the stopping-distance equation.

The block stops, so \(v=0\).

\[ v^2=u^2+2as \]

Substituting the acceleration:

\[ 0=u^2-2gs(\sin\theta+\mu_k\cos\theta) \]

Rearranging:

\[ \mu_k= \frac{\frac{u^2}{2gs}-\sin\theta} {\cos\theta} \]

Step 4: Substitute the values.

\[ \mu_k= \frac{ \frac{(5.0)^2}{2(9.8)(2.0)} -\sin25^\circ } {\cos25^\circ} \]

Evaluate the terms:

\[ \frac{25}{39.2}=0.6378 \]

\[ \sin25^\circ=0.4226 \]

\[ \cos25^\circ=0.9063 \]

Therefore:

\[ \mu_k= \frac{0.6378-0.4226}{0.9063} \]

\[ \boxed{\mu_k\approx0.24} \]

The coefficient of kinetic friction is approximately \(0.24\).

Why can’t we use the horizontal formula?

If you used:

\[ \mu_k=\frac{u^2}{2gd} \]

you would obtain approximately \(0.64\).

That result is incorrect because it assumes friction is responsible for all the deceleration.

On the incline, gravity also pulls the block downhill. Some of the block’s kinetic energy is converted into gravitational potential energy, so not all the energy lost is due to friction.

Question 5

This final problem combines stopping distance, energy transfer, kinetic friction, and limiting static friction.

A \(2.0\,\mathrm{kg}\) block is launched at \(6.0\,\mathrm{m\,s^{-1}}\) along a horizontal rough surface, section A.

Section A is \(3.0\,\mathrm m\) long and has a coefficient of kinetic friction of \(0.21\).

The block then enters section B, a rough incline at \(30^\circ\) to the horizontal. Assume its speed is unchanged as it passes from A to B.

The block slides \(1.5\,\mathrm m\) up section B before momentarily coming to rest.

The coefficient of static friction on section B is \(0.55\).

Assume the kinetic-friction coefficient on B is constant, there are no other forces, and \(g=9.8\,\mathrm{m\,s^{-2}}\).

(a) Calculate the block’s speed as it enters section B.

(b) Determine the coefficient of kinetic friction on section B.

(c) Does the block remain at rest when it reaches its stopping position, or does it slide back down the incline? Justify your answer using limiting static friction.

(d) If the block starts sliding back down, calculate its initial acceleration.

Solution 5

The coefficient of kinetic friction on section B is approximately \(0.35\). The block does not remain at rest: its static friction is insufficient to hold it on the incline, so it slides back down with an initial acceleration of approximately \(1.9\,\mathrm{m\,s^{-2}}\).

This is the important distinction: having a static-friction coefficient greater than the kinetic-friction coefficient does not automatically mean an object can remain stationary on an incline.

(a) Calculate the speed at the entrance to section B.

We cannot immediately apply the stopping-distance formula to section B because we don’t yet know the block’s speed when it reaches the incline.

We first need to account for the energy lost to friction along section A.

Using the constant-acceleration equation on A:

\[ v_A^2=u_A^2+2a_AL_A \]

The acceleration on the horizontal surface is:

\[ a_A=-\mu_{kA}g \]

Therefore:

\[ v_A^2=u_A^2-2\mu_{kA}gL_A \]

Substitute:

\[ v_A^2= (6.0)^2-2(0.21)(9.8)(3.0) \]

\[ v_A^2=36-12.348 \]

\[ v_A^2=23.652 \]

Taking the square root:

\[ v_A=\sqrt{23.652} \]

\[ \boxed{v_A=4.86\,\mathrm{m\,s^{-1}}} \]

The block enters section B at approximately \(4.86\,\mathrm{m\,s^{-1}}\).

(b) Calculate the coefficient of kinetic friction on section B.

Now consider the block travelling uphill along the incline.

Choose uphill as the positive direction.

The forces opposing its motion along the incline are:

• The component of gravity parallel to the incline, \(mg\sin30^\circ\).
• Kinetic friction, \(\mu_{kB}mg\cos30^\circ\).

The normal force is:

\[ N=mg\cos30^\circ \]

Newton’s second law gives:

\[ ma=-mg\sin30^\circ -\mu_{kB}mg\cos30^\circ \]

Divide by mass:

\[ a=-g(\sin30^\circ+\mu_{kB}\cos30^\circ) \]

Apply the constant-acceleration equation over the \(1.5\,\mathrm m\) uphill distance.

The final speed is zero:

\[ 0=v_A^2+2as \]

Rearranging for the coefficient:

\[ \mu_{kB}= \frac{ \frac{v_A^2}{2gs}-\sin30^\circ } {\cos30^\circ} \]

Substitute:

\[ \mu_{kB}= \frac{ \frac{23.652}{2(9.8)(1.5)}-0.500 } {0.866} \]

\[ \mu_{kB}= \frac{0.8045-0.500}{0.866} \]

\[ \boxed{\mu_{kB}=0.352} \]

The coefficient of kinetic friction on section B is approximately \(0.35\).

(c) Determine whether the block remains at rest.

This is where we must switch from kinetic friction to static friction.

The block has stopped moving uphill, but that does not guarantee it will stay there.

Gravity still has a component pulling it downhill.

For the block to remain stationary, static friction must be able to balance this component.

The required static-friction force is:

\[ f_s=mg\sin30^\circ \]

Substitute:

\[ f_s=2.0(9.8)(0.5) \]

\[ f_s=9.8\,\mathrm N \]

Now calculate the maximum available static friction.

\[ f_{s,\max}=\mu_sN \]

Since:

\[ N=mg\cos30^\circ \]

We obtain:

\[ f_{s,\max} =0.55(2.0)(9.8)\cos30^\circ \]

\[ f_{s,\max}=9.34\,\mathrm N \]

Compare the two values:

\[ f_s=9.8\,\mathrm N \]

\[ f_{s,\max}=9.34\,\mathrm N \]

The required static-friction force is greater than the maximum available force.

Therefore, static friction cannot hold the block in place.

\[ \boxed{\text{The block slides back down the incline.}} \]

You can also reach this conclusion using the limiting-static-friction condition.

For an object to remain at rest on an incline:

\[ mg\sin\theta\leq\mu_smg\cos\theta \]

Cancel the mass and rearrange:

\[ \tan\theta\leq\mu_s \]

In this problem:

\[ \tan30^\circ=0.577 \]

But:

\[ \mu_s=0.55 \]

Since:

\[ 0.577>0.55 \]

The available static friction is insufficient.

Notice the subtlety: \(\mu_s=0.55\) is greater than \(\mu_k=0.352\), but the block still slides. The relationship between the two coefficients does not tell you whether static friction is strong enough to balance gravity on a particular incline.

(d) Calculate the initial acceleration as the block slides down.

Once the block begins sliding, kinetic friction replaces static friction.

The direction of motion is now downhill, so kinetic friction acts uphill.

Gravity’s component parallel to the incline acts downhill.

Choose downhill as the positive direction.

Newton’s second law gives:

\[ ma=mg\sin30^\circ-f_k \]

Using:

\[ f_k=\mu_{kB}mg\cos30^\circ \]

We obtain:

\[ ma=mg\sin30^\circ -\mu_{kB}mg\cos30^\circ \]

Cancel the mass:

\[ a=g(\sin30^\circ-\mu_{kB}\cos30^\circ) \]

Substitute:

\[ a=9.8(0.500-0.352\times0.866) \]

\[ a=9.8(0.1953) \]

\[ \boxed{a\approx1.9\,\mathrm{m\,s^{-2}}} \]

The block initially accelerates downhill at approximately \(1.9\,\mathrm{m\,s^{-2}}\).

This completes the problem: kinetic friction brings the block to rest on its upward journey, but static friction cannot hold it there, so it begins sliding back down.

The main decision is knowing which friction model applies at each stage.

• While an object is sliding, use \(f_k=\mu_kN\) to model kinetic friction.
• When an object is stationary, use \(f_s\leq\mu_sN\) to check whether static friction can hold it in place.
• When the object starts moving again, return to the kinetic-friction model.

For stopping-distance questions, always check that your initial speed and measured distance refer to the same starting position, and that friction really is the only horizontal force before substituting into the formula.

If you need to revise the underlying distinction between distance and displacement, see Distance and Displacement in HSC Physics Explained Clearly. For motion involving changes in direction across a plane, see Distance and Displacement on a Plane: HSC Physics Guide.