Can a Force Act Opposite to Motion? HSC Physics Explained

A crate is sliding down a loading ramp while someone pulls it uphill with a rope. The rope is taut, but the crate is still gaining speed downhill.

Does that mean the rope isn’t pulling hard enough to produce a force? Not at all.

Yes, a force can act in the direction opposite to an object’s motion. In fact, several forces can oppose the motion while the object continues moving forwards or even speeds up. What determines its acceleration is the net force: the combined effect of all the forces acting on it.

Here’s the prediction worth testing: if the person pulls harder uphill, must the crate immediately start moving uphill?

No. A stronger pull might first reduce its downhill acceleration. If the net force eventually points uphill, the crate will slow down while still travelling downhill. It can only reverse direction after its downhill velocity reaches zero, and the forces must still be sufficient to start it moving uphill.

The force direction tells us how velocity is changing, not necessarily which way the object is currently travelling.

01Start by identifying the forces on one object

Before calculating anything, choose the object whose motion you’re investigating.

For the crate, our system is just the crate itself. The rope, Earth, and ramp are outside the system, so their interactions with the crate are external forces.

There are four main forces:

Notice that two forces act uphill, directly opposing the crate’s downhill velocity.

They are real forces even if the crate speeds up downhill.

What about Newton’s third law?

Every interaction produces equal-magnitude, opposite-direction forces on two different objects.

For example, the ramp exerts uphill friction on the sliding crate. The crate exerts an equal downhill friction force on the ramp.

Those forces do not cancel when we analyse the crate because only one acts on the crate. Similarly, the crate’s weight and normal force are not a Newton’s third-law pair: both act on the crate.

This is why choosing your system first matters. You add forces acting on the same object, not forces from different sides of an interaction pair.

Resolve forces along the ramp

Imagine the ramp makes an angle \(\theta\) with the horizontal.

Choose downhill along the ramp as positive. Choose perpendicular away from the ramp as the second positive direction.

Gravity acts vertically, not directly along the ramp. We therefore split the crate’s weight into two components:

• \(mg\sin\theta\), acting downhill along the ramp.
• \(mg\cos\theta\), acting into the ramp.

Here, \(m\) is the crate’s mass in kilograms (kg), and \(g\) is gravitational acceleration, approximately \(9.8\ \mathrm{m\,s^{-2}}\) near Earth’s surface.

For a crate sliding downhill with a rope parallel to the ramp, the net force along the slope is

\[ F_{\parallel}=mg\sin\theta-T-f_k \]

where:

• \(F_{\parallel}\) is the signed net force parallel to the ramp, in newtons (N).
• \(T\) is the magnitude of the uphill rope tension, in newtons.
• \(f_k\) is the magnitude of kinetic friction, in newtons.

The normal force, \(N\), acts perpendicular to the surface. If the crate stays in contact with the ramp and has no acceleration perpendicular to it,

\[ N=mg\cos\theta \]

For the standard constant-coefficient model of sliding friction,

\[ f_k=\mu_k N \]

where \(\mu_k\) is the coefficient of kinetic friction, a dimensionless number describing the two surfaces.

Newton’s second law now gives

\[ F_{\parallel}=ma_{\parallel} \]

where \(a_{\parallel}\) is the signed acceleration along the ramp.

If the crate is moving downhill, a positive acceleration means it is speeding up downhill. A negative acceleration means it is slowing down, at least until its velocity reaches zero.

The sign alone doesn’t tell us the direction of motion. We must consider the velocity as well.

02Testing the idea: why a block can speed up despite friction

A useful experiment starts with a rough block sliding along a level surface.

Give the block a push, then release it. It continues moving briefly before slowing down and stopping.

The explanation is straightforward. On a horizontal surface, weight and normal force balance vertically. With no continuing horizontal push, friction acts opposite the block’s motion and produces an acceleration opposite to its velocity.

That observation supports Newton’s laws, but it doesn’t establish that every force opposing motion must make an object slow down.

The first design has a limitation: there is no continuing forward force competing with friction.

We need to change the experiment.

Revise the apparatus: tilt the surface

Use the same block and rough surface, but place the surface on an adjustable inclined track.

The high end is on the left, and the low end is on the right. A ruler fixed alongside the track provides a position scale. A camera placed side-on, perpendicular to the plane of motion, records the block’s position at successive times.

Give the block an initial downhill velocity, then release it. The block is no longer being pushed by hand.

It now experiences three external forces: gravity, normal force, and kinetic friction.

The slope changes how much of gravity acts along the direction of motion.

To make a prediction, suppose the block has a mass of \(0.80\ \mathrm{kg}\) and a kinetic friction coefficient of \(0.25\).

How could we estimate that friction coefficient?

One method is to pull the block across the same rough surface while it is horizontal, using a spring force meter. If the block slides at a constant speed, the horizontal pulling force balances kinetic friction.

For example, an illustrative steady-speed force reading of \(1.96\ \mathrm{N}\) gives

\[ \begin{aligned} \mu_k&=\frac{f_k}{N}\\ &=\frac{1.96}{0.80(9.8)}\\ &=0.25 \end{aligned} \]

The calculation assumes the block is sliding steadily, the pulling force is horizontal, and the contact conditions are comparable in the inclined trial.

Now compare two ramps: one tilted at \(10^\circ\), and the other at \(20^\circ\).

Suppose the block starts each timed downhill run at \(0.80\ \mathrm{m\,s^{-1}}\).

The model predicts the following results.

These are calculated, illustrative values, not measurements from a conducted experiment. They assume constant kinetic friction coefficient, negligible air resistance, and the block remaining in sliding contact with the ramp.

The predicted observation is surprising if you think friction must always produce slowing: on the steeper ramp, friction points uphill, yet the block gains speed downhill.

Why does the steeper ramp change the result?

Two effects happen together.

First, the downhill component of gravity becomes larger as the ramp gets steeper.

Second, the normal force gets smaller. Under the constant-coefficient friction model, this also reduces kinetic friction.

The steeper ramp therefore has a stronger downhill driving component and a slightly weaker opposing friction force.

Let’s calculate each case properly.

03When does an opposite force actually slow something down?

The distinction is between an individual force and the net force.

Consider an object moving along a straight line.

The important comparison is between the velocity and the net force, not between the velocity and one conveniently chosen force.

An object can experience a force opposite its velocity without slowing if other forces outweigh that force.

A useful contrasting case: a ball thrown upwards

Imagine throwing a ball straight up.

After it leaves your hand, and ignoring air resistance, the only significant force acting on it is gravity.

Gravity points downwards throughout the flight.

While the ball rises, gravity opposes its upward motion. The ball slows down.

At the highest point, its instantaneous velocity is zero, but gravity still acts downwards.

As the ball falls, the same downward gravitational force acts in the direction of its velocity. The ball speeds up.

The gravitational force hasn’t changed direction. The velocity has.

This is why it is incorrect to identify force direction from the direction an object happens to be travelling.

What if the force is sideways?

In two-dimensional motion, the comparison must be made using the component of the net force along the instantaneous velocity.

A force component opposite the velocity tends to reduce speed. A component in the same direction tends to increase it.

A perpendicular component changes the direction of velocity without changing its magnitude at that instant.

For example, a ball travelling at constant speed around a circular path has an inward net force perpendicular to its velocity. Its direction changes continuously, even though its speed remains constant.

So the more general question is not just “Does the force point left or right?”

It is “Which component of the net force acts along or opposite the current velocity?”

04Questions and solutions

Use \(g=9.8\ \mathrm{m\,s^{-2}}\) for all questions. Unless stated otherwise, ignore air resistance, assume forces are constant within each stated phase, and treat any strings as light.

Question 1

A \(0.60\ \mathrm{kg}\) trolley travels right along a horizontal track at \(2.0\ \mathrm{m\,s^{-1}}\).

A string pulls it left with a force of \(1.8\ \mathrm{N}\), and the track exerts an additional resistive force of \(0.60\ \mathrm{N}\) to the left.

Calculate the trolley’s acceleration and its velocity after \(0.30\ \mathrm{s}\). Explain why it can still be moving right despite the leftward forces.

Take right as positive.

Solution 1

The trolley accelerates at \(4.0\ \mathrm{m\,s^{-2}}\) to the left, but it is still moving right at \(0.80\ \mathrm{m\,s^{-1}}\) after \(0.30\ \mathrm{s}\).

Both horizontal forces act left, so their signed sum is

\[ \begin{aligned} F_{\mathrm{net}}&=-1.8-0.60\\ &=-2.4\ \mathrm{N}\\ a&=\frac{-2.4}{0.60}\\ &=-4.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

Using \(v=u+at\),

\[ \begin{aligned} v&=2.0+(-4.0)(0.30)\\ &=+0.80\ \mathrm{m\,s^{-1}} \end{aligned} \]

The positive velocity means the trolley is still moving right.

A negative acceleration means its rightward velocity is decreasing. The forces don’t instantaneously reverse its motion.

Question 2

A \(2.0\ \mathrm{kg}\) motorised trolley travels right at \(1.2\ \mathrm{m\,s^{-1}}\) along a level track.

A winch pulls the trolley left with a constant force of \(5.0\ \mathrm{N}\). At the same time, its motor provides a rightward driving force of \(8.0\ \mathrm{N}\).

Whenever the trolley is moving, the track exerts a resistive force of \(1.0\ \mathrm{N}\) opposite its velocity.

After \(0.60\ \mathrm{s}\), the motor switches off, but the winch continues pulling left.

At rest, the track can provide a maximum static resistive force of \(1.5\ \mathrm{N}\).

Determine:

(a) The trolley’s velocity when the motor switches off.

(b) How long it takes to stop after the motor switches off.

(c) Whether it reverses direction after stopping, and its velocity \(0.20\ \mathrm{s}\) after the stopping instant.

Take right as positive. Assume the winch remains taut and maintains its stated force throughout.

Solution 2

The trolley first speeds up to \(1.8\ \mathrm{m\,s^{-1}}\) right, then stops \(0.60\ \mathrm{s}\) after the motor switches off. It subsequently reverses and reaches \(0.40\ \mathrm{m\,s^{-1}}\) left after another \(0.20\ \mathrm{s}\).

The key is that the net force changes when the motor switches off, and resistance changes direction when the trolley reverses.

Part (a): While the motor is operating

The trolley is moving right, so resistance acts left.

\[ \begin{aligned} F_{\mathrm{net}}&=8.0-5.0-1.0\\ &=+2.0\ \mathrm{N}\\ a&=\frac{2.0}{2.0}\\ &=+1.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

Its velocity after \(0.60\ \mathrm{s}\) is

\[ \begin{aligned} v&=1.2+(1.0)(0.60)\\ &=+1.8\ \mathrm{m\,s^{-1}} \end{aligned} \]

The winch has pulled against the trolley’s motion the whole time, but the trolley has sped up because the motor force is larger than the combined opposing forces.

Part (b): After the motor switches off

While the trolley still moves right, the winch and resistance both act left.

\[ \begin{aligned} F_{\mathrm{net}}&=-5.0-1.0\\ &=-6.0\ \mathrm{N}\\ a&=\frac{-6.0}{2.0}\\ &=-3.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

The time to stop from \(1.8\ \mathrm{m\,s^{-1}}\) is

\[ \begin{aligned} t&=\frac{0-1.8}{-3.0}\\ &=0.60\ \mathrm{s} \end{aligned} \]

This is \(1.20\ \mathrm{s}\) after the start of the question.

Part (c): At rest and after reversal

The winch still pulls left with \(5.0\ \mathrm{N}\).

The maximum static resistance is only \(1.5\ \mathrm{N}\), so the track cannot balance the winch’s pull. The trolley therefore starts moving left.

Once it moves left, the kinetic resistance points right.

\[ \begin{aligned} F_{\mathrm{net}}&=-5.0+1.0\\ &=-4.0\ \mathrm{N}\\ a&=\frac{-4.0}{2.0}\\ &=-2.0\ \mathrm{m\,s^{-2}} \end{aligned} \]

Starting from zero velocity,

\[ \begin{aligned} v&=0+(-2.0)(0.20)\\ &=-0.40\ \mathrm{m\,s^{-1}} \end{aligned} \]

The trolley is moving left at \(0.40\ \mathrm{m\,s^{-1}}\).

The tempting mistake is to use the same \(-3.0\ \mathrm{m\,s^{-2}}\) acceleration after reversal. That would leave the resistive force pointing left even though the trolley is now moving left. Sliding resistance must reverse with the sliding direction.

Question 3

An investigator is testing two explanations for the force opposing a block as it slides downhill on an inclined track.

The block has a mass of \(1.00\ \mathrm{kg}\).

Two models are proposed:

Model A: The opposing force has a constant magnitude of \(2.41\ \mathrm{N}\), independent of the ramp angle.

Model B: The opposing force is kinetic friction with \(\mu_k=0.25\), so its magnitude depends on the normal force.

The investigator records the block’s acceleration while it is sliding downhill on two different slopes.

The following measurements are illustrative problem data.

Downhill is positive for both trials.

Determine which model is consistent with both measurements. Calculate the predicted acceleration for each model at each angle, and explain what this comparison does and does not establish about the opposing force.

Solution 3

Model B is consistent with both measurements. Model A fits the shallow-ramp result but predicts too little acceleration on the steeper ramp.

The important difference is that the normal force changes as the track becomes steeper.

Model A: constant opposing force

With downhill positive,

\[ a=g\sin\theta-\frac{2.41}{m} \]

At \(10^\circ\),

\[ \begin{aligned} a&=9.8\sin10^\circ-2.41\\ &=-0.71\ \mathrm{m\,s^{-2}} \end{aligned} \]

At \(50^\circ\),

\[ \begin{aligned} a&=9.8\sin50^\circ-2.41\\ &=+5.10\ \mathrm{m\,s^{-2}} \end{aligned} \]

The \(50^\circ\) prediction differs noticeably from the recorded \(5.93\ \mathrm{m\,s^{-2}}\).

Model B: normal-force-dependent friction

Here,

\[ f_k=\mu_k mg\cos\theta \]

So,

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

At \(10^\circ\),

\[ \begin{aligned} a&=9.8(\sin10^\circ\\ &\quad-0.25\cos10^\circ)\\ &=-0.71\ \mathrm{m\,s^{-2}} \end{aligned} \]

At \(50^\circ\),

\[ \begin{aligned} a&=9.8(\sin50^\circ\\ &\quad-0.25\cos50^\circ)\\ &=+5.93\ \mathrm{m\,s^{-2}} \end{aligned} \]

These match the illustrative measurements to the stated precision.

Why doesn’t the constant-force model work?

It assumes the same uphill resistance at every angle. But when the slope increases from \(10^\circ\) to \(50^\circ\), the block presses less strongly against the ramp.

Under Model B, that decreases the friction magnitude while the downhill gravitational component increases.

The experiment therefore favours the normal-force-dependent model.

However, two trials do not prove that the kinetic friction law is exact in every situation. Real friction can depend on contact conditions, speed, surface contamination, and other effects. A stronger investigation would collect repeated measurements over several angles, estimate uncertainties, and check how closely acceleration follows the model.

The main experimental lesson is that one successful prediction is not always enough to distinguish two plausible explanations.

Question 4

Two blocks, A and B, slide downhill on the same \(20^\circ\) ramp.

Block A is higher up the ramp, behind block B. They each have mass \(1.00\ \mathrm{kg}\).

They are connected by a light, flexible, inextensible cord of length \(1.00\ \mathrm{m}\), initially straight and just taut along the ramp.

The coefficients of kinetic friction are:

• Block A: \(\mu_{k,A}=0.10\)
• Block B: \(\mu_{k,B}=0.40\)

At the instant of release, both blocks are moving downhill at \(1.00\ \mathrm{m\,s^{-1}}\). No further applied forces act on them.

Determine whether the cord remains taut or becomes slack immediately after release.

Calculate the blocks’ accelerations and their separation \(0.40\ \mathrm{s}\) later.

Explain why simply assuming both blocks have the same acceleration could produce an impossible result.

Ignore the cord’s mass, collisions, and air resistance. Assume both blocks remain sliding downhill during the stated interval.

Solution 4

The cord becomes slack immediately after release. Block A accelerates downhill at \(2.43\ \mathrm{m\,s^{-2}}\), while block B has an uphill acceleration of \(0.33\ \mathrm{m\,s^{-2}}\) despite still moving downhill. Their separation decreases by \(0.221\ \mathrm{m}\) over the first \(0.40\ \mathrm{s}\).

The central constraint is that a flexible cord can pull but cannot push.

First, consider each block without tension.

With downhill positive,

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

For block A,

\[ \begin{aligned} a_A&=9.8(\sin20^\circ\\ &\quad-0.10\cos20^\circ)\\ &=+2.43\ \mathrm{m\,s^{-2}} \end{aligned} \]

For block B,

\[ \begin{aligned} a_B&=9.8(\sin20^\circ\\ &\quad-0.40\cos20^\circ)\\ &=-0.332\ \mathrm{m\,s^{-2}} \end{aligned} \]

Block A tends to accelerate downhill towards B. Block B is slowing because its friction exceeds the downhill gravitational component.

Therefore, the distance between the blocks naturally decreases.

The cord cannot prevent them from moving closer together: doing so would require it to push them apart.

Test the assumption that the cord remains taut.

If it remained taut, both blocks would need the same acceleration.

Adding the forces on both blocks and dividing by their combined mass gives

\[ \begin{aligned} a_{\mathrm{taut}} &=9.8\sin20^\circ\\ &\quad-0.25(9.8\cos20^\circ)\\ &=1.05\ \mathrm{m\,s^{-2}} \end{aligned} \]

This uses the average friction coefficient because the masses are equal.

For the upper block A, a taut cord would pull downhill towards B. Let \(T\) be that downhill tension.

\[ \begin{aligned} ma_{\mathrm{taut}}&=ma_A+T\\ T&=m(a_{\mathrm{taut}}-a_A)\\ &=1.00(1.05-2.43)\\ &=-1.38\ \mathrm{N} \end{aligned} \]

A negative result means that maintaining the assumed common acceleration would require the cord to push A uphill instead of pulling it downhill.

A flexible cord cannot supply that force.

So the taut-cord assumption is physically impossible, and tension becomes zero.

Calculate the change in separation.

The blocks start with equal velocities, so their relative velocity is initially zero.

Their relative acceleration is

\[ \begin{aligned} a_B-a_A&=-0.332-2.43\\ &=-2.76\ \mathrm{m\,s^{-2}} \end{aligned} \]

After \(0.40\ \mathrm{s}\), their separation is

\[ \begin{aligned} d&=1.00+\frac12(-2.76)(0.40)^2\\ &=0.779\ \mathrm{m} \end{aligned} \]

The separation decreases by approximately \(0.221\ \mathrm{m}\), leaving the cord slack.

Both blocks are still moving downhill at this time. Block B’s speed has decreased to approximately \(0.867\ \mathrm{m\,s^{-1}}\), while A’s has increased to approximately \(1.97\ \mathrm{m\,s^{-1}}\).

The tempting method was to treat the two blocks as one constrained system immediately.

That method only works if the assumed constraint can actually be maintained. A string can transmit tension but not compression, so a physically impossible negative tension tells us to revise the model.

Question 5

A \(1.00\ \mathrm{kg}\) block rests on a rough ramp inclined at \(20^\circ\).

A winch connected uphill pulls the block with a constant force of \(5.00\ \mathrm{N}\), parallel to the ramp. The force remains the same whether the block moves or stays at rest.

The friction coefficients are

\[ \mu_k=0.10,\qquad \mu_s=0.40 \]

The winch remains active throughout every trial.

Consider three separate trials, each beginning at the same location.

Trial A: The block is initially sliding downhill at \(0.90\ \mathrm{m\,s^{-1}}\).

Trial B: The block starts from rest.

Trial C: The block is initially sliding uphill at \(0.20\ \mathrm{m\,s^{-1}}\).

Determine:

(a) For Trial A, the acceleration, time to stop, and downhill stopping distance. Does the block subsequently move uphill?

(b) For Trial B, whether the block begins moving and which direction static friction acts.

(c) For Trial C, the acceleration and velocity after \(1.00\ \mathrm{s}\).

(d) The complete range of constant uphill pulling forces for which a block starting from rest could remain stationary, yet the same block already sliding uphill would accelerate uphill.

Explain why the three trials can produce different outcomes even though the mass, ramp, friction coefficients, and pulling force are identical.

Assume ideal static and kinetic friction models, and no air resistance. Take uphill as positive throughout.

Solution 5

The downhill-moving block stops and stays at rest. The block initially at rest also remains stationary. Yet the block already moving uphill accelerates uphill.

This is possible because kinetic friction changes direction with sliding, while static friction adjusts its magnitude to maintain equilibrium, up to a maximum.

The same applied force does not produce the same net force in all three trials.

First, establish the common forces.

The downhill component of weight is

\[ \begin{aligned} mg\sin20^\circ &=1.00(9.8)\sin20^\circ\\ &=3.352\ \mathrm{N} \end{aligned} \]

The normal force is

\[ \begin{aligned} N&=1.00(9.8)\cos20^\circ\\ &=9.209\ \mathrm{N} \end{aligned} \]

Therefore, the magnitude of kinetic friction is

\[ \begin{aligned} f_k&=0.10(9.209)\\ &=0.921\ \mathrm{N} \end{aligned} \]

The maximum available static friction is

\[ \begin{aligned} f_{s,\max}&=0.40(9.209)\\ &=3.684\ \mathrm{N} \end{aligned} \]

These values apply across the trials, but the actual friction force and its direction depend on the motion state.

Part (a): Trial A – initially moving downhill

The block is sliding downhill, so kinetic friction acts uphill.

The winch also pulls uphill.

Taking uphill as positive,

\[ \begin{aligned} F_{\mathrm{net}} &=5.00+0.921-3.352\\ &=+2.569\ \mathrm{N} \end{aligned} \]

Hence,

\[ a=+2.569\ \mathrm{m\,s^{-2}} \]

The block is initially moving downhill, but its acceleration points uphill.

Its initial signed velocity is \(-0.90\ \mathrm{m\,s^{-1}}\).

The time to stop is

\[ \begin{aligned} t&=\frac{0-(-0.90)}{2.569}\\ &=0.350\ \mathrm{s} \end{aligned} \]

To calculate the signed displacement, use

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

where \(s\) is the displacement along the chosen axis.

\[ \begin{aligned} s&=\frac{0^2-(-0.90)^2}{2(2.569)}\\ &=-0.158\ \mathrm{m} \end{aligned} \]

So the block travels \(0.158\ \mathrm{m}\) downhill before stopping.

But does it then reverse?

We cannot answer that by extending the kinetic-friction calculation past zero velocity.

At rest, the winch pulls uphill with \(5.00\ \mathrm{N}\), while gravity pulls downhill with \(3.352\ \mathrm{N}\).

Without friction, the remaining force would be

\[ 5.00-3.352=1.648\ \mathrm{N} \]

uphill.

Static friction can oppose that tendency by acting downhill with \(1.648\ \mathrm{N}\).

Since

\[ 1.648\ \mathrm{N}<3.684\ \mathrm{N} \]

the available static friction is sufficient.

The block therefore stays at rest instead of reversing uphill.

Part (b): Trial B – initially at rest

The block remains stationary.

The winch creates an uphill tendency of \(1.648\ \mathrm{N}\), but static friction acts downhill with exactly \(1.648\ \mathrm{N}\).

The net force is zero.

It would be incorrect to automatically use the maximum static friction of \(3.684\ \mathrm{N}\). That is the greatest force static friction can provide, not the force it must always provide.

At rest, static friction takes the value needed to prevent sliding, provided the required magnitude does not exceed its maximum.

Part (c): Trial C – initially moving uphill

Now the block is sliding uphill.

Kinetic friction reverses direction and acts downhill.

Gravity also has a downhill component.

The net force is

\[ \begin{aligned} F_{\mathrm{net}} &=5.00-0.921-3.352\\ &=+0.727\ \mathrm{N} \end{aligned} \]

Therefore,

\[ a=+0.727\ \mathrm{m\,s^{-2}} \]

The block is moving uphill and accelerating uphill, so it speeds up.

After \(1.00\ \mathrm{s}\),

\[ \begin{aligned} v&=0.20+(0.727)(1.00)\\ &=+0.927\ \mathrm{m\,s^{-1}} \end{aligned} \]

The block’s uphill speed increases to \(0.927\ \mathrm{m\,s^{-1}}\).

This is the surprising contrast: the same \(5.00\ \mathrm{N}\) winch cannot start the stationary block moving uphill, yet it can accelerate a block that is already sliding uphill.

Part (d): Find the pulling-force range

Let \(P\) be the magnitude of the constant uphill pulling force in newtons.

For a block already sliding uphill to accelerate uphill, the pulling force must exceed gravity’s downhill component plus kinetic friction.

Therefore,

\[ P>mg\sin\theta+\mu_kmg\cos\theta \]

Substituting the known values,

\[ \begin{aligned} P&>3.352+0.921\\ P&>4.273\ \mathrm{N} \end{aligned} \]

Now consider the block at rest.

For static friction to prevent the winch from pulling the block uphill, the winch force cannot exceed the downhill component of gravity plus the maximum available downhill static friction.

Therefore,

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

Substituting,

\[ \begin{aligned} P&\leq3.352+3.684\\ P&\leq7.036\ \mathrm{N} \end{aligned} \]

Combining the constraints,

\[ 4.27\ \mathrm{N}<P\leq7.04\ \mathrm{N} \]

Within this range, a stationary block can remain stationary, while a block already sliding uphill accelerates uphill.

The lower boundary is strict because a block sliding uphill would have zero acceleration if the pull exactly balanced gravity and kinetic friction. The upper boundary includes the ideal static-friction limit; real systems near that limit can be sensitive to small disturbances.

The given pull of \(5.00\ \mathrm{N}\) lies inside the range.

This result is more interesting than a simple disagreement about signs.

The system has different possible behaviour under the same external pull because its friction law depends on whether it is stationary, sliding uphill, or sliding downhill.

In Trial A, the winch and kinetic friction initially oppose the motion, causing slowing. At rest, the friction model changes and prevents reversal.

In Trial B, static friction prevents movement from beginning.

In Trial C, the winch is strong enough to overcome gravity and kinetic friction, so upward motion becomes faster.

The subtle mistake would be treating friction as a single fixed force that always points in the same direction, regardless of the block’s motion.

05What to decide before solving the next force question

When an HSC Physics question gives you an object moving in one direction with a force pointing the other way, don’t immediately label it as decelerating.

First, identify the system and the external interactions. Decide which forces act on that object, including friction, tension, gravity, and contact forces. Choose a positive direction, resolve the forces into components, and calculate the net force.

Only then compare the acceleration with the object’s current velocity.

If the net force is opposite the velocity, the object initially slows. If they point in the same direction, it speeds up. If it reaches zero velocity, check whether the forces or constraints change before predicting what happens next.

For a wider explanation of how these decisions appear in displacement-time, velocity-time, and acceleration-time graphs, see Force, Motion and Graph Representations for HSC Physics.

The same distinction becomes useful in electric fields. A charged particle can experience an electric force opposite its velocity, with its acceleration determined by the net force. To work out the direction of that electric interaction, see Coulomb’s Law for HSC Physics: Force and Direction.

The principle carries across both situations: a force tells you how motion changes. It does not, by itself, tell you which way something is already moving.