How to Calculate a Block’s Stopping Distance on a Rough Incline

A block is projected up a rough ramp at 4.0 m/s. It slows down because gravity pulls it towards the bottom of the ramp, while friction also acts against its upward motion. How far does it travel before stopping?

For a straight incline, with no applied force acting after the block is launched, the stopping distance along the ramp is:

\[
\boxed{s=\frac{u^2}{2g(\sin\theta+\mu_k\cos\theta)}}
\]

Here:

• \(s\) is the distance travelled along the incline, in metres.
• \(u\) is the initial speed up the incline, in \(\mathrm{m\,s^{-1}}\).
• \(g\) is gravitational acceleration, usually \(9.8\ \mathrm{m\,s^{-2}}\).
• \(\theta\) is the angle of the incline to the horizontal.
• \(\mu_k\) is the coefficient of kinetic friction.

The important part is understanding where the two terms in the denominator come from. One describes gravity pulling the block down the slope. The other describes friction resisting its upward motion.

There is also a detail students often miss: the normal reaction is not always \(mg\cos\theta\). That expression only applies under particular conditions. Getting the normal reaction right is essential because it determines the frictional force.

01Resolving the forces along a rough incline

The first decision is choosing your axes.

For an inclined-plane problem, it is usually easiest to choose:

• The \(x\)-axis parallel to the incline, positive uphill.
• The \(y\)-axis perpendicular to the incline, positive away from the surface.

This makes the direction of the block’s motion and the forces much easier to track.

What happens to the block’s weight?

The block’s weight always acts vertically downwards, regardless of the incline’s angle.

Its magnitude is:

\[
W=mg
\]

However, our axes are tilted to match the ramp. We therefore need to resolve the weight into two components.

The two components are:

Parallel to the incline:

\[
W_x=mg\sin\theta
\]

This component acts down the slope.

Perpendicular to the incline:

\[
W_y=mg\cos\theta
\]

This component pushes the block into the surface.

Notice that we use sine for the parallel component and cosine for the perpendicular component. This follows from the geometry of the right-angled triangle formed when the weight is resolved.

What is the normal reaction?

The normal reaction, \(N\), is the contact force exerted by the ramp on the block. It acts perpendicular to the surface.

If the ramp is straight, the block remains in contact with it, and no other forces act perpendicular to the surface, there is no perpendicular acceleration.

Therefore:

\[
\sum F_y=0
\]

\[
N-mg\cos\theta=0
\]

So:

\[
\boxed{N=mg\cos\theta}
\]

This is an important result, but don’t use it automatically in every incline problem. We’ll look at when it changes shortly.

Which direction does friction act?

Imagine sliding a book upwards along a rough desk that has been tilted.

The book moves uphill, but the frictional force acts downhill. Friction opposes the relative motion between the surfaces.

For a block sliding uphill, kinetic friction therefore acts down the incline.

Its magnitude is:

\[
f_k=\mu_kN
\]

Substituting the normal reaction:

\[
\boxed{f_k=\mu_kmg\cos\theta}
\]

We now have two forces acting down the slope:

• The parallel component of weight, \(mg\sin\theta\).
• Kinetic friction, \(\mu_kmg\cos\theta\).

Both oppose the block’s upward motion.

02Finding the acceleration and stopping distance

With positive \(x\) defined as uphill, Newton’s second law gives:

\[
\sum F_x=ma
\]

Both forces point downhill, so:

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

Divide through by the mass:

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

Therefore:

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

The negative sign tells us that the acceleration is downhill, opposite to our positive direction.

We can now find the distance travelled before the block stops.

Use the constant-acceleration equation:

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

At the first stopping point, the final velocity is zero:

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

Substitute the acceleration:

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

Rearranging gives:

\[
\boxed{s=\frac{u^2}{2g(\sin\theta+\mu_k\cos\theta)}}
\]

This is the main equation to remember for a block projected uphill on a straight, rough incline when no additional force acts after launch.

Why does the mass disappear?

You might notice that the block’s mass does not appear in the final equation.

This is not an algebraic accident.

A heavier block experiences a greater downslope gravitational component, but it also experiences greater friction because its normal reaction is larger.

Both forces increase in proportion to mass. When we divide Newton’s second law by mass, the mass cancels.

Consequently, two blocks with the same initial speed, incline angle, and coefficient of kinetic friction have the same stopping distance, even if their masses differ.

Distance travelled versus vertical height

Be careful about what the question actually asks.

The stopping distance \(s\) is measured along the inclined surface. It is not the vertical height gained by the block.

If you need the vertical rise instead:

\[
h=s\sin\theta
\]

Similarly, the horizontal displacement is:

\[
x=s\cos\theta
\]

These quantities describe different aspects of the same motion. For a refresher on the distinction, see Distance and Displacement in HSC Physics Explained Clearly.

03When does the normal reaction change?

This is where a common shortcut causes trouble.

Students often assume that:

\[
N=mg\cos\theta
\]

must be true whenever a block sits on an incline.

It isn’t.

The normal reaction depends on all the forces acting perpendicular to the surface, as well as any acceleration in that direction.

For a straight incline with no perpendicular acceleration:

\[
N=\text{total force towards the surface balanced by contact}
\]

Let’s compare a few situations.

Here, \(F\) is the applied force and \(\phi\) is its angle relative to the incline.

Case 1: A force parallel to the incline

Suppose someone pulls the block uphill using a rope that runs parallel to the ramp.

The force has no perpendicular component.

Therefore:

\[
N=mg\cos\theta
\]

The normal reaction does not change.

The pulling force affects the acceleration along the incline, but not the normal reaction.

Case 2: A rope pulls the block away from the surface

Now imagine the rope is angled above the incline.

The tension has two components:

• \(F\cos\phi\), acting uphill.
• \(F\sin\phi\), acting away from the surface.

The second component partially lifts the block, reducing the normal reaction.

Resolving perpendicular to the incline:

\[
N+F\sin\phi-mg\cos\theta=0
\]

Therefore:

\[
\boxed{N=mg\cos\theta-F\sin\phi}
\]

Because kinetic friction depends on the normal reaction, the frictional force also decreases.

Case 3: A force pushes the block into the surface

Now reverse the rope’s angle so that the applied force pushes uphill but also into the ramp.

Its perpendicular component increases the contact force.

The normal reaction becomes:

\[
\boxed{N=mg\cos\theta+F\sin\phi}
\]

This increases kinetic friction.

The important distinction is that the normal reaction is determined by the perpendicular force balance, not simply by the block’s weight.

A useful decision rule

Whenever you see an incline problem:

1. Draw the forces acting on the block.
2. Resolve every angled force perpendicular to the incline.
3. Use Newton’s second law in the perpendicular direction to find \(N\).
4. Calculate friction using \(f_k=\mu_kN\).
5. Resolve the forces parallel to the incline to find acceleration.

If the block is being pulled away from the surface so strongly that the calculated normal reaction becomes zero or negative, the block loses contact. The ordinary incline-friction model no longer applies.

Also remember that a changing incline angle changes the perpendicular component of weight. On a curved ramp, perpendicular acceleration can change the normal reaction as well.

04Testing the stopping-distance model with an experiment

There is a useful way to test whether the equation actually describes the motion.

The question is:

If we double the initial speed, does the block travel twice as far, or four times as far?

The equation contains \(u^2\), so it predicts that the stopping distance increases by a factor of four.

We can investigate this with a simple ramp experiment.

Setting up the apparatus

Consider a rough board raised at a fixed angle.

A block is launched from the bottom using a spring launcher. A photogate positioned at the starting line measures the block’s initial speed. A ruler or measuring tape runs along the incline, allowing the stopping distance to be measured from the photogate position.

An angle-measuring instrument records the ramp angle.

The experiment should keep the following constant:

• The same block and contact surface.
• The same incline angle.
• The same ramp surface condition.
• The same coefficient of kinetic friction, as far as the experiment allows.

Only the initial speed should change.

What should the experiment show?

From the stopping-distance equation:

\[
s=\frac{u^2}{2g(\sin\theta+\mu_k\cos\theta)}
\]

If \(g\), \(\theta\), and \(\mu_k\) remain constant, everything in the denominator stays the same.

Therefore:

\[
s\propto u^2
\]

The stopping distance should be proportional to the square of the initial speed.

For example, take an illustrative ramp with:

\[
\theta=30^\circ,\qquad \mu_k=0.20
\]

The predicted stopping distances are:

These are illustrative model predictions, not recorded experimental measurements.

Notice what happens when the initial speed doubles from 1.0 m/s to 2.0 m/s.

The stopping distance increases from approximately 0.076 m to 0.303 m. That’s four times the original distance.

A graph of stopping distance against initial speed squared should therefore be approximately a straight line through the origin.

What could go wrong with the original design?

Suppose you simply push the block by hand and measure how far it travels.

A shorter stopping distance might mean the surface has more friction. But it might also mean you gave the block a smaller initial speed.

You cannot distinguish these explanations just by measuring the stopping distance.

A photogate fixes this problem by measuring the actual initial speed on every trial. Repeated trials, consistent launching, and careful distance measurements improve the reliability of the comparison.

The experiment tests the constant-friction model. Small differences between predicted and measured distances may result from variations in the surface, the launch, or the measurement process.

05Worked examples

06Questions and solutions

The following questions move from a straightforward stopping-distance calculation to situations where you need to combine force resolution, friction, and motion through different stages.

Question 1

A \(1.0\ \mathrm{kg}\) block is projected up a rough incline at \(3.0\ \mathrm{m/s}\). The incline makes an angle of \(30^\circ\) with the horizontal, and the coefficient of kinetic friction is \(0.20\).

Calculate the distance travelled before the block first stops.

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

Solution 1

The block travels approximately \(0.682\ \mathrm{m}\) up the incline before stopping.

Because no additional force acts on the block, we can use the standard stopping-distance equation.

\[
s=\frac{u^2}{2g(\sin\theta+\mu_k\cos\theta)}
\]

Substitute the given values:

\[
s=\frac{3.0^2}
{2(9.8)(\sin30^\circ+0.20\cos30^\circ)}
\]

Using:

\[
\sin30^\circ=0.5
\]

\[
\cos30^\circ=0.866
\]

We obtain:

\[
s=\frac{9.0}
{19.6(0.5+0.20(0.866))}
\]

\[
s=\frac{9.0}{13.195}
\]

\[
\boxed{s=0.682\ \mathrm{m}}
\]

The block travels \(0.682\ \mathrm{m}\) along the ramp.

The key is recognising that both gravity and friction act downhill while the block moves uphill.

Question 2

A \(2.0\ \mathrm{kg}\) block is projected up a rough incline at \(2.5\ \mathrm{m/s}\). The incline is at \(25^\circ\) to the horizontal, and the coefficient of kinetic friction is \(0.20\).

A rope pulls the block uphill with a constant force of \(5.0\ \mathrm{N}\). The rope is parallel to the incline.

Calculate:

(a) The normal reaction.

(b) The acceleration of the block.

(c) The distance travelled before the block stops.

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

Solution 2

The normal reaction is approximately \(17.76\ \mathrm{N}\), and the block travels approximately \(0.914\ \mathrm{m}\) before stopping.

(a) Normal reaction

The rope is parallel to the incline, so it has no perpendicular component.

Therefore:

\[
N=mg\cos25^\circ
\]

\[
N=2.0(9.8)\cos25^\circ
\]

\[
\boxed{N=17.76\ \mathrm{N}}
\]

(b) Acceleration

First calculate friction:

\[
f_k=\mu_kN
\]

\[
f_k=0.20(17.76)
\]

\[
f_k=3.55\ \mathrm{N}
\]

The gravitational component down the slope is:

\[
mg\sin25^\circ
=2.0(9.8)\sin25^\circ
\]

\[
=8.28\ \mathrm{N}
\]

Taking uphill as positive:

\[
\sum F_x=F-mg\sin25^\circ-f_k
\]

\[
=5.0-8.28-3.55
\]

\[
=-6.83\ \mathrm{N}
\]

Using Newton’s second law:

\[
-6.83=2.0a
\]

\[
\boxed{a=-3.42\ \mathrm{m\,s^{-2}}}
\]

The negative acceleration means the block is slowing down.

(c) Stopping distance

Use:

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

At the stopping point:

\[
0=2.5^2+2(-3.42)s
\]

\[
0=6.25-6.84s
\]

Therefore:

\[
\boxed{s=0.914\ \mathrm{m}}
\]

The block travels approximately \(0.914\ \mathrm{m}\) uphill before stopping.

The important detail is that the rope changes the net force parallel to the incline, but it does not change the normal reaction because it acts parallel to the surface.

Question 3

A \(1.5\ \mathrm{kg}\) block is projected up a rough incline at \(3.0\ \mathrm{m/s}\). The incline is at \(30^\circ\) to the horizontal, and the coefficient of kinetic friction is \(0.25\).

A \(6.0\ \mathrm{N}\) force acts uphill at an angle of \(30^\circ\) to the incline.

Consider two separate trials:

• Trial A: The force pulls the block at \(30^\circ\) above the incline.
• Trial B: The force pushes the block at \(30^\circ\) below the incline, towards the surface.

For each trial, calculate the distance travelled before the block stops.

Assume the applied force remains constant, the block stays in contact with the ramp, and \(g=9.8\ \mathrm{m\,s^{-2}}\).

Solution 3

The stopping distance is approximately \(1.47\ \mathrm{m}\) in Trial A and \(1.11\ \mathrm{m}\) in Trial B.

The difference comes from the normal reaction, which changes the amount of friction.

First, calculate the components of the block’s weight:

\[
mg=1.5(9.8)=14.7\ \mathrm{N}
\]

Parallel to the incline:

\[
mg\sin30^\circ=7.35\ \mathrm{N}
\]

Perpendicular to the incline:

\[
mg\cos30^\circ=12.73\ \mathrm{N}
\]

The applied force has the same parallel component in both trials:

\[
F\cos30^\circ=6.0(0.866)
\]

\[
=5.20\ \mathrm{N}
\]

Trial A: Force pulling away from the surface

The perpendicular component of the applied force reduces the normal reaction.

\[
N=mg\cos30^\circ-F\sin30^\circ
\]

\[
N=12.73-6.0(0.5)
\]

\[
N=9.73\ \mathrm{N}
\]

Calculate kinetic friction:

\[
f_k=0.25(9.73)
\]

\[
f_k=2.43\ \mathrm{N}
\]

The net force uphill is:

\[
\sum F_x=5.20-7.35-2.43
\]

\[
=-4.58\ \mathrm{N}
\]

Therefore:

\[
a=\frac{-4.58}{1.5}
\]

\[
a=-3.06\ \mathrm{m\,s^{-2}}
\]

Use the kinematic equation:

\[
0=3.0^2+2(-3.06)s
\]

\[
s=\frac{9.0}{6.12}
\]

\[
\boxed{s_A=1.47\ \mathrm{m}}
\]

Trial B: Force pushing into the surface

The perpendicular component now increases the normal reaction.

\[
N=mg\cos30^\circ+F\sin30^\circ
\]

\[
N=12.73+3.00
\]

\[
N=15.73\ \mathrm{N}
\]

Kinetic friction becomes:

\[
f_k=0.25(15.73)
\]

\[
f_k=3.93\ \mathrm{N}
\]

The net force uphill is:

\[
\sum F_x=5.20-7.35-3.93
\]

\[
=-6.08\ \mathrm{N}
\]

Therefore:

\[
a=\frac{-6.08}{1.5}
\]

\[
a=-4.06\ \mathrm{m\,s^{-2}}
\]

Using the kinematic equation:

\[
0=3.0^2+2(-4.06)s
\]

\[
s=\frac{9.0}{8.12}
\]

\[
\boxed{s_B=1.11\ \mathrm{m}}
\]

Comparison

Both applied forces have the same uphill component.

However, they produce different normal reactions. The pulling force reduces friction, while the pushing force increases it.

This is why you cannot calculate friction correctly without first resolving all perpendicular forces.

Question 4

A \(1.0\ \mathrm{kg}\) block is projected up a rough incline at \(3.6\ \mathrm{m/s}\). The incline makes an angle of \(25^\circ\) with the horizontal.

The block stops after travelling \(0.95\ \mathrm{m}\) along the incline.

Assume no additional force acts on the block and the coefficient of kinetic friction remains constant.

(a) Calculate the coefficient of kinetic friction.

(b) A second block has twice the mass of the first block but the same initial momentum. It is projected up the same incline, with the same surface conditions. Calculate its stopping distance.

Solution 4

The coefficient of kinetic friction is approximately \(0.302\), and the second block travels \(0.238\ \mathrm{m}\) before stopping.

(a) Finding the coefficient of kinetic friction

Start with the stopping-distance equation:

\[
s=\frac{u^2}{2g(\sin\theta+\mu_k\cos\theta)}
\]

Rearrange to make \(\mu_k\) the subject:

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

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

Therefore:

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

Substitute the values:

\[
\mu_k=
\frac{\frac{3.6^2}{2(9.8)(0.95)}-\sin25^\circ}
{\cos25^\circ}
\]

\[
\mu_k=
\frac{0.696-0.4226}{0.9063}
\]

\[
\boxed{\mu_k=0.302}
\]

The coefficient of kinetic friction is approximately \(0.302\). It is dimensionless.

(b) Finding the second block’s stopping distance

This part requires a little more thought.

The second block has twice the mass but the same initial momentum.

Momentum is:

\[
p=mv
\]

Let the first block have mass \(m\) and speed \(3.6\ \mathrm{m/s}\).

The second block has mass \(2m\).

Because their momenta are equal:

\[
m(3.6)=2mv_2
\]

Cancel \(m\):

\[
3.6=2v_2
\]

\[
\boxed{v_2=1.8\ \mathrm{m/s}}
\]

Now consider the stopping-distance equation.

The two blocks have the same incline angle and coefficient of kinetic friction. Their stopping distances therefore depend on the square of their initial speeds.

\[
\frac{s_2}{s_1}=\frac{v_2^2}{v_1^2}
\]

Substitute:

\[
\frac{s_2}{0.95}=\frac{1.8^2}{3.6^2}
\]

\[
\frac{s_2}{0.95}=\frac{1}{4}
\]

\[
s_2=0.95(0.25)
\]

\[
\boxed{s_2=0.238\ \mathrm{m}}
\]

The second block travels approximately \(0.238\ \mathrm{m}\).

The tempting assumption is that doubling the mass should double the stopping distance or leave it unchanged. Neither is correct here.

The mass cancels from the acceleration equation, but equal momentum does not mean equal initial speed. The second block starts at half the speed, so it travels one-quarter of the distance.

Question 5

A \(2.0\ \mathrm{kg}\) block is projected up a rough incline at \(4.0\ \mathrm{m/s}\). The incline is at \(25^\circ\) to the horizontal, and the coefficient of kinetic friction is \(0.20\).

After the block has travelled \(0.50\ \mathrm{m}\), a winch begins pulling it uphill with a constant force of \(8.0\ \mathrm{N}\). The rope is directed at \(30^\circ\) above the incline.

The rope continues pulling until the block stops. At that point, it remains attached and continues exerting the same force.

The coefficient of static friction between the block and the incline is \(0.25\).

Assume the tension increases over a negligible distance without producing an impulse, the block remains in contact with the incline, and \(g=9.8\ \mathrm{m\,s^{-2}}\).

Calculate:

(a) The block’s speed when the winch begins pulling.

(b) The additional distance travelled after the winch starts pulling.

(c) The total uphill distance travelled before the block first stops.

(d) Whether the block remains at rest after stopping or slides back down the incline.

Solution 5

The block travels approximately \(2.95\ \mathrm{m}\) uphill before first stopping. It then remains at rest because static friction is sufficient to prevent it sliding back down.

This question requires us to treat the motion in two stages. The forces change when the winch starts pulling, so the acceleration changes as well.

(a) Speed when the winch begins pulling

Before the winch starts, there is no additional applied force.

Therefore:

\[
N_1=mg\cos25^\circ
\]

\[
N_1=2.0(9.8)\cos25^\circ
\]

\[
N_1=17.76\ \mathrm{N}
\]

The kinetic friction is:

\[
f_{k1}=0.20(17.76)
\]

\[
f_{k1}=3.55\ \mathrm{N}
\]

The downslope gravitational component is:

\[
mg\sin25^\circ
=2.0(9.8)\sin25^\circ
\]

\[
=8.28\ \mathrm{N}
\]

The net force is:

\[
\sum F_x=-8.28-3.55
\]

\[
=-11.83\ \mathrm{N}
\]

Using Newton’s second law:

\[
-11.83=2.0a_1
\]

\[
a_1=-5.92\ \mathrm{m\,s^{-2}}
\]

The block travels \(0.50\ \mathrm{m}\) before the winch starts.

Use:

\[
v_1^2=u^2+2a_1s_1
\]

\[
v_1^2=4.0^2+2(-5.92)(0.50)
\]

\[
v_1^2=16-5.92
\]

\[
v_1^2=10.08
\]

Taking the positive square root:

\[
\boxed{v_1=3.18\ \mathrm{m/s}}
\]

The block is still moving uphill when the winch begins pulling.

(b) Additional distance after the winch starts pulling

Now the forces perpendicular to the incline change.

The rope pulls away from the surface, reducing the normal reaction.

\[
N_2=mg\cos25^\circ-F\sin30^\circ
\]

\[
N_2=17.76-8.0(0.5)
\]

\[
\boxed{N_2=13.76\ \mathrm{N}}
\]

The new kinetic friction is:

\[
f_{k2}=\mu_kN_2
\]

\[
f_{k2}=0.20(13.76)
\]

\[
\boxed{f_{k2}=2.75\ \mathrm{N}}
\]

Notice that friction has decreased from \(3.55\ \mathrm{N}\) to \(2.75\ \mathrm{N}\).

The rope also provides an uphill component:

\[
F_x=8.0\cos30^\circ
\]

\[
F_x=6.93\ \mathrm{N}
\]

The downslope gravitational component remains:

\[
mg\sin25^\circ=8.28\ \mathrm{N}
\]

Now calculate the net force:

\[
\sum F_x=6.93-8.28-2.75
\]

\[
=-4.10\ \mathrm{N}
\]

Therefore:

\[
-4.10=2.0a_2
\]

\[
\boxed{a_2=-2.05\ \mathrm{m\,s^{-2}}}
\]

The block is still slowing down, but its deceleration is smaller than before.

Use the speed at the beginning of this stage:

\[
v_1^2=10.08\ \mathrm{m^2\,s^{-2}}
\]

At the first stopping point, \(v=0\):

\[
0=v_1^2+2a_2s_2
\]

\[
0=10.08+2(-2.05)s_2
\]

\[
4.10s_2=10.08
\]

\[
\boxed{s_2=2.45\ \mathrm{m}}
\]

The block travels an additional \(2.45\ \mathrm{m}\) after the winch starts pulling.

(c) Total uphill distance

The total distance is the sum of the two stages:

\[
s_{\mathrm{total}}=s_1+s_2
\]

\[
s_{\mathrm{total}}=0.50+2.45
\]

\[
\boxed{s_{\mathrm{total}}=2.95\ \mathrm{m}}
\]

The block travels approximately \(2.95\ \mathrm{m}\) uphill before first stopping.

The important point is that we cannot use one acceleration for the entire journey. The winch changes both the parallel force balance and the normal reaction, which changes friction.

(d) Does the block slide back down?

This is the conceptual part of the question.

A common mistake is to assume that kinetic friction continues acting downhill after the block stops.

That is incorrect.

Once the block reaches zero velocity, it is no longer sliding. We must consider static friction.

The rope continues pulling uphill, while gravity has a larger component down the incline.

The net force along the incline, excluding friction, is:

\[
F_{\mathrm{net,without\ friction}}
=F\cos30^\circ-mg\sin25^\circ
\]

\[
=6.93-8.28
\]

\[
=-1.35\ \mathrm{N}
\]

There is therefore a tendency for the block to slide downhill.

To remain stationary, static friction must act uphill with a magnitude of:

\[
f_s=1.35\ \mathrm{N}
\]

But can the surface provide enough static friction?

The maximum possible static friction is:

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

Using the normal reaction after the rope starts pulling:

\[
f_{s,\max}=0.25(13.76)
\]

\[
f_{s,\max}=3.44\ \mathrm{N}
\]

Compare the two values:

\[
1.35\ \mathrm{N}<3.44\ \mathrm{N}
\]

The required static friction is smaller than the maximum available static friction.

Therefore, the block remains at rest.

\[
\boxed{\text{The block stops and stays stationary.}}
\]

Notice that static friction acts uphill here, even though kinetic friction acted downhill while the block was travelling uphill.

Static friction adjusts to oppose the tendency of motion. It does not automatically equal its maximum value.

This distinction is important whenever a question asks what happens after a moving block reaches its stopping point.

The key decision in these problems is to separate the motion from the force balance. First find the normal reaction, then calculate friction, then determine acceleration, and finally use the appropriate kinematic equation to find distance.

Once the block stops, switch your thinking from kinetic friction and motion to static friction and equilibrium. If you need to determine whether a stationary block can be held on an incline, the next useful problem is How to Calculate the Force Needed to Hold a Block on a Slope.