How to Compare Work Along a Ramp with a Vertical Lift

Lift a 4.0 kg crate through a vertical height of 1.0 m. You can lift it straight up, or pull it up a frictionless ramp that is 4.0 m long. Which method requires more work?

The ramp requires less force, so it might seem like the easier option must require less work. But that’s not quite right. If the crate starts and finishes at the same speed, both methods require exactly 39.2 J of work against gravity.

The ramp trades a smaller force for a longer distance. On a real ramp, friction can increase the work required, even though the normal reaction itself does no work.

That’s the key to comparing work along a ramp with work in a vertical lift:

• Same mass and same vertical height gain mean the same work against gravity.
• A frictionless ramp and a vertical lift require the same applied work if the initial and final speeds are equal.
• A ramp with friction requires additional work to overcome friction.
• The normal reaction affects friction, even though it does no work directly.

01Why does the ramp require the same work?

Let’s begin with the simplest comparison: a frictionless ramp and a vertical lift.

Assume:

• The same object is used in both situations.
• Both methods raise the object through the same vertical height, h.
• The ramp is stationary and frictionless.
• The object moves at constant speed, so its kinetic energy does not change.
• Air resistance is negligible.

Take gravitational field strength as \(g=9.8\ \mathrm{m\,s^{-2}}\).

Lifting vertically

When you lift an object vertically at constant speed, the applied upward force balances its weight.

Weight is:

\[ F_g=mg \]

where:

• \(m\) is the object’s mass in kilograms.
• \(g\) is gravitational field strength.
• \(F_g\) is the magnitude of the object’s weight in newtons.

Work is force multiplied by displacement in the direction of the force.

Therefore, the work done by the lifting force is:

\[ W_{\mathrm{vertical}}=mgh \]

The object gains gravitational potential energy equal to \(mgh\).

Moving along a ramp

Now consider moving the same object up a ramp.

The applied force can be much smaller than the object’s weight because only part of the weight acts down the slope.

Let:

• \(s\) be the distance travelled along the ramp.
• \(h\) be the vertical height gained.
• \(\theta\) be the angle between the ramp and the horizontal.

The ramp forms a right-angled triangle, giving:

\[ h=s\sin\theta \]

For a frictionless ramp, the pulling force required to maintain constant speed is:

\[ F_{\mathrm{pull}}=mg\sin\theta \]

The work done by the pulling force is:

\[ W_{\mathrm{ramp}}=F_{\mathrm{pull}}s \]

Substitute the expression for the pulling force:

\[ W_{\mathrm{ramp}}=mg\sin\theta\times s \]

Since \(s\sin\theta=h\):

\[ \boxed{W_{\mathrm{ramp}}=mgh} \]

This is exactly the same result as the vertical lift.

The ramp reduces the required force, but increases the distance over which that force acts.

The important point is that work depends on both force and displacement, not force alone.

02Resolving weight along the slope

To solve ramp questions properly, you need to resolve the weight into components parallel and perpendicular to the ramp.

This is a standard application of vector resolution. The same underlying skill appears when separating horizontal and vertical motion in projectile motion.

Imagine a block resting on a straight ramp inclined at an angle \(\theta\) to the horizontal.

Choose two axes:

• The parallel axis, pointing up the ramp.
• The perpendicular axis, pointing away from the ramp.

The weight always acts vertically downwards, regardless of the ramp’s orientation.

However, its components change direction relative to our chosen axes.

The two components of weight are:

Parallel component:

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

This component acts down the ramp.

Perpendicular component:

\[ F_{\perp}=mg\cos\theta \]

This component acts into the ramp.

Notice that sine and cosine appear in different components. This is because the angle \(\theta\) is measured from the horizontal, not from the vertical.

What happens to the normal reaction?

The normal reaction, \(N\), is the contact force exerted by the ramp perpendicular to its surface.

If the object is on a straight, stationary ramp, the pulling force is parallel to the ramp, and there is no acceleration perpendicular to the surface, then:

\[ N=mg\cos\theta \]

This is an important result, but it is not universally true.

The normal reaction equals \(mg\cos\theta\) only under those conditions.

For example, consider what happens as the ramp becomes steeper.

As the ramp gets steeper, the perpendicular component of weight decreases. Consequently, the normal reaction decreases, provided no other perpendicular forces act.

When does the normal reaction change?

There are three particularly important situations.

1. The ramp angle changes.

For a parallel pulling force and no perpendicular acceleration:

\[ N=mg\cos\theta \]

Changing \(\theta\) changes the normal reaction.

2. The applied force is angled relative to the ramp.

Suppose a rope pulls the object upwards at an angle \(\alpha\) above the ramp.

The rope now has a component perpendicular to the surface:

\[ T_{\perp}=T\sin\alpha \]

This component pulls the object away from the ramp, reducing the normal reaction.

Using perpendicular force balance:

\[ N+T\sin\alpha-mg\cos\theta=0 \]

Therefore:

\[ \boxed{N=mg\cos\theta-T\sin\alpha} \]

Here, \(T\) is the rope tension.

If you instead push the object into the ramp at an angle, the normal reaction increases.

3. The object accelerates perpendicular to the surface or loses contact.

If the object follows a curved track, its perpendicular acceleration can affect the normal reaction.

If the object loses contact with the ramp, the normal reaction becomes zero. A surface can push an object, but it cannot pull it back towards itself.

Does the normal reaction do work?

This is a common source of confusion.

For an object moving along a fixed ramp, its displacement is parallel to the surface. The normal reaction is perpendicular to that displacement.

Work is:

\[ W=Fd\cos\phi \]

where \(\phi\) is the angle between force and displacement.

For the normal reaction, \(\phi=90^\circ\).

Therefore:

\[ W_N=Ns\cos90^\circ \]

\[ \boxed{W_N=0} \]

The normal reaction does no work.

However, this does not mean it is irrelevant to the work calculation.

The normal reaction determines the magnitude of friction:

\[ f_k=\mu_kN \]

where:

• \(f_k\) is the kinetic friction force.
• \(\mu_k\) is the coefficient of kinetic friction.
• \(N\) is the normal reaction.

So the normal reaction can change the amount of work required indirectly by changing the friction force.

That’s an important distinction: a force can do zero work itself while still affecting other forces that do work.

03How friction changes the comparison

So far, we’ve assumed a frictionless ramp.

Real surfaces have friction, which opposes the object’s movement.

When an object moves up a ramp, kinetic friction acts down the ramp.

For a block pulled parallel to the surface at constant speed:

\[ F_{\mathrm{pull}}=mg\sin\theta+f_k \]

If the ramp is straight and the pulling force is parallel to it:

\[ N=mg\cos\theta \]

Therefore:

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

The pulling force becomes:

\[ F_{\mathrm{pull}} =mg\sin\theta+\mu_kmg\cos\theta \]

The applied work is:

\[ W_{\mathrm{pull}}=F_{\mathrm{pull}}s \]

Substituting:

\[ W_{\mathrm{pull}} =mgs\sin\theta+\mu_kmgs\cos\theta \]

Since \(s\sin\theta=h\):

\[ \boxed{ W_{\mathrm{pull}} =mgh+\mu_kmg\cos\theta\,s } \]

This equation is useful when comparing a real ramp with a vertical lift.

The first term, \(mgh\), is the work required to increase gravitational potential energy.

The second term is the additional work done against friction.

For a vertical lift without friction, the work remains \(mgh\), provided the initial and final speeds are equal.

How could we test this experimentally?

Imagine setting up an investigation using the same block in two arrangements.

First, place the block on a ramp raised above a bench. Attach a force sensor to the block using a string parallel to the ramp. Mark the distance travelled along the surface.

Next, use the same block and force sensor to lift it vertically through the same height.

In both trials, move the block at constant speed. Record the applied force and the displacement in the direction of movement.

What should the experiment show?

In the ideal frictionless model, the ramp requires a smaller force but a longer distance. The calculated work should match the vertical lift.

A first attempt that compares only the force sensor readings would be misleading. The ramp reading is lower, but that doesn’t establish that less work has been done.

The revised investigation must measure both force and displacement.

It must also keep the force sensor aligned with the direction of movement. If the string pulls at an angle, it changes the normal reaction and therefore the friction force.

Illustrative experimental data

Consider a 3.0 kg block raised through 0.80 m on a ramp inclined at \(30^\circ\).

The ramp distance is:

\[ s=\frac{h}{\sin30^\circ} \]

\[ s=\frac{0.80}{0.5}=1.60\ \mathrm{m} \]

The weight is:

\[ mg=3.0\times9.8=29.4\ \mathrm{N} \]

For a frictionless ramp, the pulling force is:

\[ F_{\mathrm{pull}}=29.4\sin30^\circ \]

\[ F_{\mathrm{pull}}=14.7\ \mathrm{N} \]

The vertical lifting force is 29.4 N.

The following values are illustrative calculations, not measurements from an actual experiment.

For the frictional ramp, the normal reaction is:

\[ N=29.4\cos30^\circ=25.5\ \mathrm{N} \]

The friction force is:

\[ f_k=0.15\times25.5=3.82\ \mathrm{N} \]

This increases the required pulling force to approximately 18.5 N.

The extra work is:

\[ W_{\mathrm{friction}}=f_ks \]

\[ W_{\mathrm{friction}}=3.82\times1.60 \]

\[ W_{\mathrm{friction}}=6.11\ \mathrm{J} \]

The ramp now requires approximately 29.6 J rather than 23.5 J.

The result illustrates the distinction:

• The gravitational potential energy gained is unchanged.
• The additional energy is transferred to the surroundings through friction.

If the pulling force varies during the movement, you cannot simply multiply one instantaneous force reading by the total distance. Instead, determine the work from the area under the force-displacement graph, or add the work over smaller sections.

04Worked examples

05Choosing the right equation

Before starting a ramp question, identify the conditions. The following table helps you select the appropriate calculation.

For an angled rope, remember that you must calculate the normal reaction using all perpendicular forces before calculating friction.

For an accelerating object, also remember that the applied work does not necessarily equal the increase in gravitational potential energy. Some of the work changes kinetic energy.

06Questions and solutions

Question 1

A 2.5 kg block is pulled up a frictionless ramp inclined at \(30^\circ\) to the horizontal. It rises through a vertical height of 0.60 m at constant speed.

Calculate:

a) The distance travelled along the ramp.

b) The pulling force required parallel to the ramp.

c) The work done by the pulling force.

d) The work required to lift the same block vertically through 0.60 m.

Solution 1

The ramp requires 14.7 J of work, exactly the same as the vertical lift. Its pulling force is smaller because the block travels a greater distance.

a) Distance along the ramp

Use:

\[ h=s\sin\theta \]

Rearrange:

\[ s=\frac{h}{\sin\theta} \]

\[ s=\frac{0.60}{\sin30^\circ} \]

\[ \boxed{s=1.20\ \mathrm{m}} \]

b) Pulling force

Resolve the weight parallel to the ramp:

\[ F_{\mathrm{pull}}=mg\sin\theta \]

\[ F_{\mathrm{pull}} =2.5\times9.8\times\sin30^\circ \]

\[ \boxed{F_{\mathrm{pull}}=12.25\ \mathrm{N}} \]

The pulling force balances the component of weight down the ramp.

c) Work along the ramp

\[ W=Fs \]

\[ W=12.25\times1.20 \]

\[ \boxed{W=14.7\ \mathrm{J}} \]

d) Vertical lift

\[ W=mgh \]

\[ W=2.5\times9.8\times0.60 \]

\[ \boxed{W=14.7\ \mathrm{J}} \]

The two methods require equal work because the block gains the same gravitational potential energy, with no friction or change in kinetic energy.

Question 2

A 5.0 kg crate is pulled 2.0 m up a ramp inclined at \(35^\circ\) to the horizontal.

The coefficient of kinetic friction between the crate and ramp is 0.18. The pulling force acts parallel to the ramp, and the crate moves at constant speed.

Calculate:

a) The normal reaction.

b) The friction force.

c) The required pulling force.

d) The work done by the pulling force.

e) The increase in gravitational potential energy.

Explain why the applied work is greater than the gravitational potential energy gained.

Solution 2

The pulling force is approximately 35.3 N, and the work done is 70.7 J. Only 56.2 J becomes gravitational potential energy; the remaining 14.4 J is transferred through friction.

a) Normal reaction

Because the pulling force is parallel to the ramp, it has no perpendicular component.

Therefore:

\[ N=mg\cos\theta \]

\[ N=5.0\times9.8\times\cos35^\circ \]

\[ \boxed{N=40.1\ \mathrm{N}} \]

b) Friction force

Use:

\[ f_k=\mu_kN \]

\[ f_k=0.18\times40.1 \]

\[ \boxed{f_k=7.22\ \mathrm{N}} \]

Friction acts down the ramp because the crate moves up it.

c) Pulling force

The crate moves at constant speed, so the net force parallel to the ramp is zero.

Therefore:

\[ F_{\mathrm{pull}}=mg\sin\theta+f_k \]

\[ F_{\mathrm{pull}} =5.0\times9.8\times\sin35^\circ+7.22 \]

\[ F_{\mathrm{pull}}=28.1+7.22 \]

\[ \boxed{F_{\mathrm{pull}}=35.3\ \mathrm{N}} \]

d) Work done by the pulling force

The pulling force is parallel to the displacement.

\[ W=Fs \]

\[ W=35.3\times2.0 \]

\[ \boxed{W=70.7\ \mathrm{J}} \]

e) Increase in gravitational potential energy

First calculate the vertical height gained:

\[ h=s\sin35^\circ \]

\[ h=2.0\times0.5736 \]

\[ h=1.147\ \mathrm{m} \]

Now calculate the increase in gravitational potential energy:

\[ \Delta E_p=mgh \]

\[ \Delta E_p=5.0\times9.8\times1.147 \]

\[ \boxed{\Delta E_p=56.2\ \mathrm{J}} \]

The difference is:

\[ W_{\mathrm{friction}}=70.7-56.2 \]

\[ \boxed{W_{\mathrm{friction}}=14.5\ \mathrm{J}} \]

This agrees with the friction calculation:

\[ W_{\mathrm{friction}}=7.22\times2.0 =14.4\ \mathrm{J} \]

The small difference is due to rounding.

The tempting mistake is to assume that all applied work becomes gravitational potential energy. Friction transfers some of the energy to the surroundings instead.

Question 3

A 3.0 kg object is moved from rest to a final speed of 2.0 m/s while being raised through a vertical height of 1.20 m.

Compare two frictionless arrangements:

• A vertical lift.
• A ramp inclined at an angle to the horizontal.

In both cases, the object begins at rest and finishes at 2.0 m/s.

Calculate:

a) The increase in gravitational potential energy.

b) The increase in kinetic energy.

c) The total work done by the applied force in each arrangement.

d) The work done by gravity.

A student claims that the applied work must be exactly \(mgh\) because both methods raise the object through the same height. Explain the error.

Solution 3

Both arrangements require 41.3 J of applied work, not 35.3 J. The additional 6.0 J increases the object’s kinetic energy.

a) Gravitational potential energy

\[ \Delta E_p=mgh \]

\[ \Delta E_p=3.0\times9.8\times1.20 \]

\[ \boxed{\Delta E_p=35.28\ \mathrm{J}} \]

b) Kinetic energy

The object starts from rest, so its initial kinetic energy is zero.

The final kinetic energy is:

\[ E_k=\frac12mv^2 \]

\[ E_k=\frac12\times3.0\times2.0^2 \]

\[ E_k=6.0\ \mathrm{J} \]

Therefore:

\[ \boxed{\Delta E_k=6.0\ \mathrm{J}} \]

c) Applied work

Use the work-energy theorem:

\[ W_{\mathrm{net}}=\Delta E_k \]

For either arrangement, the normal reaction does no work, and friction is absent.

Therefore:

\[ W_{\mathrm{applied}}+W_g=\Delta E_k \]

Gravity acts opposite to the upward displacement, so:

\[ W_g=-mgh \]

Rearranging:

\[ W_{\mathrm{applied}}=mgh+\Delta E_k \]

Substitute:

\[ W_{\mathrm{applied}}=35.28+6.0 \]

\[ \boxed{W_{\mathrm{applied}}=41.28\ \mathrm{J}} \]

This applies to both the vertical lift and the ramp.

d) Work done by gravity

\[ W_g=-mgh \]

\[ \boxed{W_g=-35.28\ \mathrm{J}} \]

The student’s mistake is confusing work done against gravity with total applied work.

The quantity \(mgh\) accounts for the increase in gravitational potential energy. It does not account for the additional kinetic energy gained.

The two routes still require equal work because their initial and final kinetic energies are identical, and their vertical height gains are identical.

Question 4

A 3.0 kg block is pulled at constant speed along a ramp made of two straight sections.

The first section is 1.0 m long and inclined at \(30^\circ\). The second section is 1.0 m long and inclined at \(60^\circ\).

The coefficient of kinetic friction is 0.20 throughout the ramp. The pulling force acts parallel to each section.

Assume the junction is smoothly connected over a negligible distance.

Calculate:

a) The normal reaction on each section.

b) The friction force on each section.

c) The total vertical height gained.

d) The total work done by the pulling force.

e) The work required for a frictionless vertical lift through the same height.

Explain why the normal reaction changes even though the mass stays constant.

Solution 4

The total work done by the pulling force is approximately 48.2 J. A frictionless vertical lift through the same height would require only 40.2 J.

The difference is the work done against friction.

a) Normal reaction on each section

For a parallel pulling force:

\[ N=mg\cos\theta \]

For the first section:

\[ N_1=3.0\times9.8\times\cos30^\circ \]

\[ \boxed{N_1=25.5\ \mathrm{N}} \]

For the second section:

\[ N_2=3.0\times9.8\times\cos60^\circ \]

\[ \boxed{N_2=14.7\ \mathrm{N}} \]

The normal reaction decreases because the second section is steeper.

b) Friction force on each section

Use:

\[ f_k=\mu_kN \]

First section:

\[ f_1=0.20\times25.5 \]

\[ \boxed{f_1=5.09\ \mathrm{N}} \]

Second section:

\[ f_2=0.20\times14.7 \]

\[ \boxed{f_2=2.94\ \mathrm{N}} \]

The steeper section has a smaller normal reaction, so it also has less friction.

c) Total vertical height gained

Calculate the height gained on each section separately.

First section:

\[ h_1=1.0\sin30^\circ \]

\[ h_1=0.50\ \mathrm{m} \]

Second section:

\[ h_2=1.0\sin60^\circ \]

\[ h_2=0.866\ \mathrm{m} \]

Total height:

\[ h=h_1+h_2 \]

\[ \boxed{h=1.366\ \mathrm{m}} \]

d) Total work done by the pulling force

The pulling force must overcome both the downslope component of weight and friction on each section.

For the first section:

\[ F_1=mg\sin30^\circ+f_1 \]

\[ F_1=14.7+5.09 \]

\[ F_1=19.79\ \mathrm{N} \]

Work on the first section:

\[ W_1=F_1s_1 \]

\[ W_1=19.79\times1.0 \]

\[ W_1=19.79\ \mathrm{J} \]

For the second section:

\[ F_2=mg\sin60^\circ+f_2 \]

\[ F_2=25.46+2.94 \]

\[ F_2=28.40\ \mathrm{N} \]

Work on the second section:

\[ W_2=F_2s_2 \]

\[ W_2=28.40\times1.0 \]

\[ W_2=28.40\ \mathrm{J} \]

Add the two contributions:

\[ W_{\mathrm{total}}=W_1+W_2 \]

\[ W_{\mathrm{total}}=19.79+28.40 \]

\[ \boxed{W_{\mathrm{total}}=48.2\ \mathrm{J}} \]

e) Compare with a vertical lift

A frictionless vertical lift through the same height requires:

\[ W_{\mathrm{vertical}}=mgh \]

\[ W_{\mathrm{vertical}} =3.0\times9.8\times1.366 \]

\[ \boxed{W_{\mathrm{vertical}}=40.2\ \mathrm{J}} \]

The extra work is approximately:

\[ W_{\mathrm{friction}}=48.2-40.2 \]

\[ \boxed{W_{\mathrm{friction}}=8.0\ \mathrm{J}} \]

The important reasoning is that the normal reaction is not constant along the path.

The mass and gravitational field strength stay the same, but the ramp angle changes. This changes the perpendicular component of weight, the normal reaction, and therefore the friction force.

You cannot use one normal reaction for both sections.

Question 5

A 5.0 kg block is pulled up a straight ramp inclined at \(30^\circ\) to the horizontal.

The ramp is 2.0 m long, the coefficient of kinetic friction is 0.20, and the block moves at constant speed.

A rope pulls the block at an angle \(\alpha\) above the ramp surface.

Compare two situations:

• Situation A: The rope is parallel to the ramp, so \(\alpha=0^\circ\).
• Situation B: The rope is angled \(30^\circ\) above the ramp.

Calculate the tension, normal reaction, and work done by the rope in both situations.

Then consider increasing \(\alpha\).

a) Determine the limiting rope angle at which the normal reaction becomes zero.

b) Calculate the work done by the rope at this limiting angle.

c) Explain why increasing the rope angle cannot reduce the required work below \(mgh\) while the block remains in contact with the ramp and moves at constant speed.

Assume the rope is light, the ramp is fixed, and the block remains in contact with the ramp up to the limiting condition.

Solution 5

Increasing the rope angle from \(0^\circ\) to \(30^\circ\) reduces the work from approximately 66.0 J to 59.1 J.

However, the rope cannot be angled beyond \(60^\circ\) while maintaining contact with the ramp. At this limiting angle, the normal reaction is zero, friction vanishes, and the work is exactly 49.0 J, equal to \(mgh\).

This question brings together force resolution, friction, work, and the physical limits of contact.

Situation A: Rope parallel to the ramp

When \(\alpha=0^\circ\), the normal reaction is:

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

\[ N=5.0\times9.8\times0.866 \]

\[ \boxed{N=42.4\ \mathrm{N}} \]

The friction force is:

\[ f_k=0.20\times42.4 \]

\[ f_k=8.49\ \mathrm{N} \]

The pulling force must overcome both friction and the component of weight down the ramp:

\[ T=mg\sin30^\circ+f_k \]

\[ T=24.5+8.49 \]

\[ \boxed{T=33.0\ \mathrm{N}} \]

Since the rope is parallel to the displacement:

\[ W_A=Ts \]

\[ W_A=33.0\times2.0 \]

\[ \boxed{W_A=66.0\ \mathrm{J}} \]

Situation B: Rope angled \(30^\circ\) above the ramp

Now the rope has a component perpendicular to the ramp.

The normal reaction becomes:

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

The friction force is:

\[ f_k=\mu_kN \]

Along the ramp, the block moves at constant speed:

\[ T\cos30^\circ=mg\sin30^\circ+f_k \]

Substitute the friction expression:

\[ T\cos30^\circ =mg\sin30^\circ +0.20(mg\cos30^\circ-T\sin30^\circ) \]

Collect the tension terms:

\[ T(\cos30^\circ+0.20\sin30^\circ) \]

\[ =mg(\sin30^\circ+0.20\cos30^\circ) \]

Substitute:

\[ T= \frac{49(0.5+0.20\times0.866)} {0.866+0.20\times0.5} \]

\[ \boxed{T=34.1\ \mathrm{N}} \]

Notice something interesting: the tension is actually greater than in Situation A.

But does that mean the work is greater?

No. The rope is no longer parallel to the displacement.

First calculate the normal reaction:

\[ N=49\cos30^\circ-34.1\sin30^\circ \]

\[ \boxed{N=25.4\ \mathrm{N}} \]

The friction force is:

\[ f_k=0.20\times25.4 \]

\[ f_k=5.07\ \mathrm{N} \]

The normal reaction has decreased, so friction has decreased too.

Now calculate the work done by the rope:

\[ W_B=Ts\cos30^\circ \]

\[ W_B=34.1\times2.0\times0.866 \]

\[ \boxed{W_B=59.1\ \mathrm{J}} \]

Even though the tension is larger, its component along the direction of movement is smaller.

The reduction in friction also contributes to the lower work requirement.

This is why you must consider both the magnitude and direction of an applied force when calculating work.

a) Find the limiting rope angle

As the rope angle increases, the rope pulls the block further away from the ramp.

Eventually, the normal reaction reaches zero.

At this point:

\[ N=mg\cos\theta-T\sin\alpha=0 \]

Therefore:

\[ T\sin\alpha=mg\cos\theta \]

At the same time, friction is zero because \(N=0\).

The parallel force balance becomes:

\[ T\cos\alpha=mg\sin\theta \]

Divide the parallel equation by the perpendicular equation:

\[ \frac{T\cos\alpha}{T\sin\alpha} = \frac{mg\sin\theta}{mg\cos\theta} \]

Simplifying:

\[ \cot\alpha=\tan\theta \]

Therefore:

\[ \boxed{\alpha=90^\circ-\theta} \]

For a ramp angle of \(30^\circ\):

\[ \alpha=90^\circ-30^\circ \]

\[ \boxed{\alpha=60^\circ} \]

This is the limiting angle.

At angles greater than \(60^\circ\), the equations would require a negative normal reaction to maintain contact. That is physically impossible because the ramp cannot pull the block towards itself.

The block would lose contact.

b) Work at the limiting angle

At \(\alpha=60^\circ\), the rope is vertical.

The normal reaction is zero, so there is no friction.

The parallel force balance gives:

\[ T\cos60^\circ=mg\sin30^\circ \]

\[ T\times0.5=49\times0.5 \]

\[ \boxed{T=49\ \mathrm{N}} \]

The work done by the rope is:

\[ W=Ts\cos\alpha \]

\[ W=49\times2.0\times\cos60^\circ \]

\[ \boxed{W=49.0\ \mathrm{J}} \]

c) Why can’t the work fall below \(mgh\)?

The vertical height gained is:

\[ h=s\sin30^\circ \]

\[ h=2.0\times0.5=1.0\ \mathrm{m} \]

The gravitational potential energy gained is:

\[ \Delta E_p=mgh \]

\[ \Delta E_p=5.0\times9.8\times1.0 \]

\[ \boxed{\Delta E_p=49.0\ \mathrm{J}} \]

At the limiting angle, the rope does exactly 49.0 J of work.

There is no friction, and the block moves at constant speed, so there is no change in kinetic energy.

The applied work must therefore equal the increase in gravitational potential energy.

The tempting prediction is that increasing the rope angle will keep reducing the normal reaction, eventually allowing the work to fall below 49.0 J.

But that prediction ignores the contact condition.

Once the normal reaction reaches zero, the block is at the point of losing contact with the ramp. The ramp cannot provide a negative normal reaction to keep it attached.

You cannot keep reducing friction while assuming the block remains constrained to the ramp.

The energy requirement has not disappeared. The limiting case simply removes the frictional contribution, leaving the minimum work required to gain gravitational potential energy.

07Where to go next

The most useful next step is to connect these calculations to the work-energy theorem.

When an object moves along a ramp, you can account for the applied work by separating it into three contributions:

• The increase in gravitational potential energy.
• The increase or decrease in kinetic energy.
• Energy transferred through friction.

For a ramp with friction, the normal reaction determines the friction force. For a frictionless ramp, the path changes the force and distance but not the work required against gravity.

Once you can identify those contributions separately, you can approach much more complicated HSC Physics energy questions without confusing force, work, and energy transfer.