Why a More Powerful Motor Does Not Necessarily Use Less Energy

An electric motor lifts a 20 kg crate through a height of 2.0 m. Motor A takes 4.0 seconds. Motor B takes just 2.0 seconds.

Both motors have an efficiency of 80%.

Which motor uses more electrical energy?

You might predict that Motor B uses twice as much because it is twice as powerful. But that prediction is wrong. Both motors use exactly the same amount of electrical energy: 490 J.

A more powerful motor does not necessarily use less energy, or more energy. Power measures how quickly energy is transferred. The total energy consumed depends on the work being done, the motor’s efficiency, and how long it operates.

The trick is to stop treating power and energy as interchangeable quantities.

01Power is a rate, not an amount of energy

Imagine driving a car to a destination 60 km away.

If you travel at 60 km/h, the journey takes one hour. If you travel at 120 km/h, it takes half an hour.

The faster car travels at twice the speed, but it doesn’t cover twice the distance. Both journeys cover 60 km.

Motor power works in the same way. A motor with greater power can transfer energy more quickly, but that doesn’t automatically mean it transfers more energy overall.

Power is defined as the rate of energy transfer:

\[ P=\frac{E}{t} \]

Where:

• \(P\) is power, measured in watts (W).
• \(E\) is energy transferred, measured in joules (J).
• \(t\) is time, measured in seconds (s).

Rearranging:

\[ E=Pt \]

This equation is particularly important for HSC Physics.

One watt is equivalent to one joule of energy transferred per second:

\[ 1\ \mathrm{W}=1\ \mathrm{J\,s^{-1}} \]

So, a motor operating at 200 W transfers energy at a rate of 200 J every second.

If it operates at constant power for 10 seconds, it transfers 2,000 J.

Notice that the power stays at 200 W. The total energy increases as the operating time increases.

Returning to the two motors

Let’s calculate the energy required to lift our original crate.

The useful mechanical work done against gravity is equal to the increase in gravitational potential energy:

\[ E_{\mathrm{useful}}=mgh \]

Where:

• \(m=20\ \mathrm{kg}\) is the crate’s mass.
• \(g=9.8\ \mathrm{m\,s^{-2}}\) is gravitational acceleration.
• \(h=2.0\ \mathrm{m}\) is the vertical height gained.

Therefore:

\[ E_{\mathrm{useful}}=20\times9.8\times2.0 \]

\[ E_{\mathrm{useful}}=392\ \mathrm{J} \]

Both motors lift the same crate through the same height. They must therefore deliver the same 392 J of useful mechanical energy.

However, they deliver that energy at different rates.

Motor B delivers useful mechanical energy twice as quickly.

Now consider their electrical energy consumption.

Since both motors are 80% efficient, they each require 490 J of electrical energy to deliver 392 J of useful mechanical energy.

Their electrical input powers are different:

Motor B draws twice the electrical power, but operates for half the time.

The total electrical energy consumed is identical.

This is the central idea:

A higher power rating tells you how quickly a motor can transfer energy. It does not tell you how much energy it will consume during an entire task.

02How motor efficiency changes energy consumption

So far, we have assumed that both motors have identical efficiency.

Real motors can have different efficiencies. This changes the comparison.

An electric motor converts electrical energy into useful mechanical energy, usually delivered through a rotating shaft.

However, not all the electrical energy becomes useful mechanical output.

Some energy is transferred to the surroundings through:

• Heating of the motor windings and other components.
• Friction between moving parts.
• Sound and vibration.
• Other non-useful energy transfers.

Energy is conserved. The motor does not destroy energy. It transfers some of its input energy into forms that are not useful for its intended task.

Understanding the efficiency equation

Efficiency describes the proportion of input energy converted into useful output energy.

It is calculated using:

\[ \eta=\frac{E_{\mathrm{useful}}}{E_{\mathrm{input}}} \]

Where:

• \(\eta\) is efficiency, expressed as a decimal.
• \(E_{\mathrm{useful}}\) is useful energy transferred by the motor.
• \(E_{\mathrm{input}}\) is total electrical energy supplied to the motor.

To express efficiency as a percentage:

\[ \eta=\frac{E_{\mathrm{useful}}} {E_{\mathrm{input}}}\times100\% \]

For example, an efficiency of 80% means that 80% of the input energy becomes useful output energy.

The remaining 20% is transferred through non-useful pathways.

You can also calculate efficiency using power:

\[ \eta=\frac{P_{\mathrm{useful}}} {P_{\mathrm{input}}} \]

This power relationship applies when comparing the input and output powers over the same operating interval, such as during steady operation.

The percentage convention is important:

Always convert a percentage into a decimal before substituting it into the efficiency equation.

The equation that connects energy and efficiency

Rearranging the energy efficiency equation gives:

\[ E_{\mathrm{input}} =\frac{E_{\mathrm{useful}}}{\eta} \]

This is particularly useful when comparing motors performing the same task.

If two motors lift the same load through the same height, they deliver the same useful mechanical energy.

The motor with the higher efficiency requires less electrical input energy.

For example, suppose two motors each need to deliver 1,000 J of useful mechanical energy.

Motor B uses less electrical energy because a greater proportion of its input becomes useful mechanical output.

Notice that we haven’t needed to know how powerful either motor is.

For the same useful task, efficiency determines the total electrical energy required, provided the efficiencies remain constant throughout the comparison.

What about power losses?

For a motor operating steadily, with no net change in its stored energy, the input power is divided between useful output and non-useful transfers:

\[ P_{\mathrm{input}} =P_{\mathrm{useful}}+P_{\mathrm{non-useful}} \]

For example, if a motor receives 500 W of electrical power and delivers 400 W of useful mechanical power:

\[ P_{\mathrm{non-useful}}=500-400 \]

\[ P_{\mathrm{non-useful}}=100\ \mathrm{W} \]

The motor transfers 100 J of energy per second through non-useful pathways.

This does not mean the energy has disappeared. It has been transferred into forms such as thermal energy and sound.

For a motor that is accelerating or changing its operating state, some energy may also be temporarily stored as kinetic or other forms of energy. The simple power balance above assumes no net change in stored energy.

03When does a more powerful motor use more or less energy?

The answer depends on what you’re comparing.

There are three particularly important situations.

Situation A: Both motors perform the same task with the same efficiency

Suppose two motors lift an identical load through the same height.

Motor A takes 10 seconds. Motor B takes 5 seconds.

Both have an efficiency of 80%.

They require the same electrical energy because:

• The useful mechanical work is identical.
• Their efficiencies are identical.
• The faster motor simply completes the task in less time.

Motor B has twice the useful output power, and twice the electrical input power, but it consumes the same total energy.

Situation B: Both motors perform the same task, but have different efficiencies

Now suppose Motor A has an efficiency of 60%, while Motor B has an efficiency of 90%.

Both must deliver 900 J of useful mechanical energy.

Motor A requires:

\[ E_{\mathrm{input,A}}=\frac{900}{0.60} =1500\ \mathrm{J} \]

Motor B requires:

\[ E_{\mathrm{input,B}}=\frac{900}{0.90} =1000\ \mathrm{J} \]

Motor B consumes less electrical energy.

It might also be more powerful, but its greater efficiency is what reduces the total energy needed for the task.

Situation C: Both motors operate for the same amount of time

This comparison produces a different result.

Suppose two motors actually draw the following electrical input powers:

• Motor A: 500 W.
• Motor B: 900 W.

Both operate for 60 seconds.

Their electrical energy consumption is:

For Motor A:

\[ E_A=500\times60=30\,000\ \mathrm{J} \]

For Motor B:

\[ E_B=900\times60=54\,000\ \mathrm{J} \]

Motor B consumes more energy.

Why? Because it draws more electrical power for exactly the same duration.

However, this comparison does not establish which motor is more efficient. They might be performing completely different tasks.

The important distinction is that the quoted powers here are actual electrical input powers, not simply maximum mechanical output ratings.

Comparing the three situations

There is another detail worth remembering.

A motor’s rated power is not necessarily the power it draws at every moment.

Its actual electrical input depends on its operating conditions, including the mechanical load and how it is being controlled.

A motor rated for a high output power might operate at a much lower power when driving a light load.

Also, check what a quoted power rating represents. Some specifications state electrical input power, while others state mechanical shaft output power. These are not interchangeable.

04A reliable method for HSC motor questions

When a question compares the power or energy consumption of two motors, work through these decisions in order.

Step 1: Identify what the power represents.

Is it electrical input power or useful mechanical output power?

Don’t substitute one for the other.

Step 2: Identify the useful mechanical task.

For a vertical lift at constant speed, calculate the useful work using:

\[ E_{\mathrm{useful}}=mgh \]

For another mechanical task, use the appropriate work equation.

Step 3: Account for efficiency.

If efficiency is supplied, calculate the electrical input energy:

\[ E_{\mathrm{input}} =\frac{E_{\mathrm{useful}}}{\eta} \]

Step 4: Calculate power if required.

Once you know the energy and operating time:

\[ P=\frac{E}{t} \]

Make sure you are calculating input power or output power, as requested.

Step 5: Check whether the power changes during operation.

If the input power varies, don’t multiply one instantaneous or maximum power value by the entire operating time.

Calculate the energy for each interval and add the results.

This last step becomes particularly important in more challenging HSC questions.

For a more detailed treatment of the mechanical power required to raise a load, see How to Calculate Lifting Power at Constant Speed.

If you need to rearrange the efficiency relationship to calculate useful output power, see How to Find Useful Output Power from Efficiency and Input Power.

05Worked examples

06Questions and solutions

The following questions progress from standard HSC calculations to more demanding problems involving changing operating conditions.

Question 1

An electric motor lifts a 12 kg crate vertically through a height of 2.5 m in 5.0 seconds at constant speed.

The motor has an efficiency of 80%.

Calculate:

a) The useful mechanical energy transferred to the crate.

b) The useful output power.

c) The electrical energy consumed.

d) The electrical input power.

e) The energy transferred through non-useful pathways.

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

Solution 1

The motor consumes 367.5 J of electrical energy, delivers 294 J of useful mechanical energy, and has an electrical input power of 73.5 W.

Part a: Useful mechanical energy

The crate gains gravitational potential energy:

\[ E_{\mathrm{useful}}=mgh \]

\[ =12\times9.8\times2.5 \]

\[ \boxed{E_{\mathrm{useful}}=294\ \mathrm{J}} \]

Part b: Useful output power

\[ P_{\mathrm{useful}}=\frac{E_{\mathrm{useful}}}{t} \]

\[ =\frac{294}{5.0} \]

\[ \boxed{P_{\mathrm{useful}}=58.8\ \mathrm{W}} \]

Part c: Electrical input energy

Convert the efficiency into decimal form:

\[ \eta=0.80 \]

Rearrange the efficiency equation:

\[ E_{\mathrm{input}} =\frac{E_{\mathrm{useful}}}{\eta} \]

\[ =\frac{294}{0.80} \]

\[ \boxed{E_{\mathrm{input}}=367.5\ \mathrm{J}} \]

Part d: Electrical input power

\[ P_{\mathrm{input}}=\frac{E_{\mathrm{input}}}{t} \]

\[ =\frac{367.5}{5.0} \]

\[ \boxed{P_{\mathrm{input}}=73.5\ \mathrm{W}} \]

Part e: Non-useful energy transfers

Subtract the useful energy from the total input energy:

\[ E_{\mathrm{non-useful}} =E_{\mathrm{input}}-E_{\mathrm{useful}} \]

\[ =367.5-294 \]

\[ \boxed{E_{\mathrm{non-useful}}=73.5\ \mathrm{J}} \]

The motor transfers 73.5 J through non-useful pathways.

Notice that the useful output power and electrical input power are different, just as the useful mechanical energy and electrical input energy are different.

Question 2

Two electric pump motors operate in separate water-transfer tests.

Their measured operating conditions are shown below.

Calculate:

a) The electrical energy consumed by each pump.

b) The useful mechanical energy delivered by each pump.

c) Which pump delivers more useful mechanical energy.

d) Explain why Pump B’s higher electrical input power does not mean it consumes more electrical energy over the complete test.

Solution 2

Both pumps consume 20,000 J of electrical energy, but Pump B delivers more useful mechanical energy: 16,000 J compared with Pump A’s 14,000 J.

Part a: Electrical energy consumed

Use:

\[ E_{\mathrm{input}}=P_{\mathrm{input}}t \]

For Pump A:

\[ E_A=500\times40 \]

\[ \boxed{E_A=20\,000\ \mathrm{J}} \]

For Pump B:

\[ E_B=800\times25 \]

\[ \boxed{E_B=20\,000\ \mathrm{J}} \]

Despite the different power ratings, both pumps consume the same electrical energy.

Part b: Useful mechanical energy

Use:

\[ E_{\mathrm{useful}}=\eta E_{\mathrm{input}} \]

For Pump A:

\[ E_{\mathrm{useful,A}} =0.70\times20\,000 \]

\[ \boxed{E_{\mathrm{useful,A}}=14\,000\ \mathrm{J}} \]

For Pump B:

\[ E_{\mathrm{useful,B}} =0.80\times20\,000 \]

\[ \boxed{E_{\mathrm{useful,B}}=16\,000\ \mathrm{J}} \]

Part c: Compare useful energy

Pump B delivers 2,000 J more useful mechanical energy.

It converts a greater proportion of its electrical input into useful output.

Part d: Explain the difference between power and energy

Pump B draws 800 J of electrical energy per second, compared with Pump A’s 500 J per second.

However, Pump B operates for only 25 seconds, whereas Pump A operates for 40 seconds.

The higher input power is balanced by the shorter operating time.

Therefore, both pumps consume the same total electrical energy.

The important lesson is that power alone cannot establish total energy consumption. You must consider the entire operating period.

Question 3

An electric winch lifts a 40 kg crate vertically through 3.0 m in 10 seconds.

During the lifting process, the motor operates at an efficiency of 80%.

After reaching the required height, the crate remains stationary for another 20 seconds. During this holding period:

• The motor remains electrically energised.
• The motor’s shaft is stationary.
• The motor draws a constant electrical input power of 150 W.
• No useful mechanical energy is transferred to the stationary crate.

Calculate:

a) The electrical energy consumed during lifting.

b) The electrical energy consumed while holding the crate.

c) The total electrical energy consumed during the complete 30-second operation.

d) The total useful mechanical energy transferred to the crate.

e) The overall efficiency for the complete operation.

Explain why the motor’s 80% lifting efficiency cannot be applied to the entire 30-second period.

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

Solution 3

The complete operation consumes 4,470 J of electrical energy and delivers 1,176 J of useful mechanical energy. Its overall efficiency is approximately 26.3%.

The important detail is that the motor continues consuming electricity while the crate is stationary.

Part a: Electrical energy consumed during lifting

First, calculate the useful energy transferred to the crate:

\[ E_{\mathrm{useful}}=mgh \]

\[ =40\times9.8\times3.0 \]

\[ E_{\mathrm{useful}}=1176\ \mathrm{J} \]

During lifting, the efficiency is 80%:

\[ E_{\mathrm{input,lift}} =\frac{1176}{0.80} \]

\[ \boxed{E_{\mathrm{input,lift}}=1470\ \mathrm{J}} \]

Part b: Electrical energy consumed while holding

During the holding period, the electrical input power is 150 W.

The holding time is 20 seconds.

Therefore:

\[ E_{\mathrm{input,hold}}=Pt \]

\[ =150\times20 \]

\[ \boxed{E_{\mathrm{input,hold}}=3000\ \mathrm{J}} \]

Part c: Total electrical energy

Add the energy consumed during both stages:

\[ E_{\mathrm{input,total}} =1470+3000 \]

\[ \boxed{E_{\mathrm{input,total}}=4470\ \mathrm{J}} \]

Part d: Total useful mechanical energy

The crate gains gravitational potential energy during the lifting stage.

While stationary, it gains no additional gravitational potential energy.

Therefore:

\[ \boxed{E_{\mathrm{useful,total}}=1176\ \mathrm{J}} \]

Part e: Overall efficiency

Efficiency must be calculated using the total useful energy and total input energy over the same complete interval.

\[ \eta=\frac{E_{\mathrm{useful,total}}} {E_{\mathrm{input,total}}} \]

\[ \eta=\frac{1176}{4470} \]

\[ \eta=0.2631 \]

As a percentage:

\[ \boxed{\eta=26.3\%} \]

Why isn’t the overall efficiency 80%?

The 80% efficiency applies only while the motor is lifting the crate.

During the holding period, the motor consumes another 3,000 J of electrical energy without transferring useful mechanical energy to the stationary crate.

The mechanical power delivered to the crate is zero because its velocity is zero:

\[ P=Fv \]

\[ P=F\times0=0 \]

The motor still draws electrical power, but that energy is transferred through non-useful pathways under the stated assumptions.

This is a useful reminder that a motor can consume electrical energy even when it is doing no useful mechanical work on its load.

Question 4

An electric motor drives a conveyor belt during a 60-second operating cycle.

Its electrical input power changes as follows:

• During the first 10 seconds, while accelerating the conveyor, the motor draws 900 W.
• During the following 50 seconds, the motor draws 500 W.

Over the complete cycle, the motor transfers 24 kJ of useful mechanical energy to the driven system.

Calculate:

a) The electrical energy consumed during the first 10 seconds.

b) The electrical energy consumed during the following 50 seconds.

c) The total electrical energy consumed.

d) The average electrical input power over the 60-second cycle.

e) The overall efficiency of the motor during the complete cycle.

Explain why using 900 W as the motor’s power for the entire 60 seconds would give an incorrect result.

Solution 4

The motor consumes 34,000 J of electrical energy, has an average input power of approximately 567 W, and operates at an overall efficiency of approximately 70.6%.

The key is to calculate the energy separately for each interval because the input power changes.

Part a: Energy consumed during the first 10 seconds

Use:

\[ E_1=P_1t_1 \]

\[ E_1=900\times10 \]

\[ \boxed{E_1=9000\ \mathrm{J}} \]

Part b: Energy consumed during the following 50 seconds

\[ E_2=P_2t_2 \]

\[ E_2=500\times50 \]

\[ \boxed{E_2=25\,000\ \mathrm{J}} \]

Part c: Total electrical energy

Add the energy from both intervals:

\[ E_{\mathrm{input,total}}=E_1+E_2 \]

\[ =9000+25\,000 \]

\[ \boxed{E_{\mathrm{input,total}}=34\,000\ \mathrm{J}} \]

Part d: Average electrical input power

Average power is the total energy divided by the total time:

\[ P_{\mathrm{average}} =\frac{E_{\mathrm{input,total}}}{t_{\mathrm{total}}} \]

\[ =\frac{34\,000}{60} \]

\[ \boxed{P_{\mathrm{average}}=566.7\ \mathrm{W}} \]

Part e: Overall efficiency

The useful mechanical energy is:

\[ E_{\mathrm{useful}}=24\ \mathrm{kJ} =24\,000\ \mathrm{J} \]

Calculate efficiency:

\[ \eta=\frac{E_{\mathrm{useful}}} {E_{\mathrm{input,total}}} \]

\[ =\frac{24\,000}{34\,000} \]

\[ \eta=0.7059 \]

Therefore:

\[ \boxed{\eta=70.6\%} \]

Why can’t we simply multiply 900 W by 60 seconds?

Because 900 W is only the electrical input power during the first 10 seconds.

For the remaining 50 seconds, the motor draws only 500 W.

Using 900 W for the entire cycle would overestimate the electrical energy consumed.

The correct method is to calculate the energy transferred during each interval and then add the results.

Question 5

An idealised electric hoist raises a 20 kg crate from the floor to a platform 4.0 m above the ground.

Two motors, A and B, are tested on the same lifting task.

The lift consists of two consecutive stages, each involving a vertical displacement of 2.0 m.

During each stage, the motor operates at constant useful shaft power and constant efficiency.

The motors have the following operating characteristics:

Assume:

• The crate moves vertically upwards.
• Each stage is treated as a steady lift.
• The transition between stages takes negligible time and involves negligible energy transfer.
• The quoted efficiencies apply throughout their respective stages.
• The crate starts and finishes at rest, and any transition-related changes in stored energy are neglected.

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

Calculate:

a) The useful mechanical energy required during each stage.

b) The time taken by each motor to complete each stage and the total lifting time.

c) The electrical energy consumed by each motor during each stage and the total electrical energy consumed.

d) The overall efficiency of each motor for the complete lift.

e) Which motor is more powerful, which motor finishes the task faster, and which motor consumes less electrical energy?

Finally, explain why comparing the motors’ maximum useful output powers, or taking the arithmetic mean of their two stage efficiencies, would give an incomplete or incorrect comparison.

Solution 5

Motor B has the greater peak useful output power, at 392 W, but Motor A completes the lift faster, taking 4.0 seconds compared with Motor B’s 5.0 seconds.

Despite taking longer, Motor B consumes less electrical energy:

• Motor A: 1,274 J.
• Motor B: approximately 996 J.

Motor B also has the higher overall efficiency, at 78.75% compared with Motor A’s 61.5%.

This result demonstrates why peak power, operating time, and efficiency must be considered together.

Part a: Useful mechanical energy during each stage

Each stage involves lifting the same 20 kg crate through 2.0 m.

Therefore, the useful mechanical energy required for each stage is:

\[ E_{\mathrm{useful}}=mgh \]

\[ =20\times9.8\times2.0 \]

\[ \boxed{E_{\mathrm{useful}}=392\ \mathrm{J}} \]

The complete lift requires:

\[ E_{\mathrm{useful,total}}=392+392 \]

\[ \boxed{E_{\mathrm{useful,total}}=784\ \mathrm{J}} \]

Both motors must deliver 784 J of useful mechanical energy.

Part b: Calculate the time taken by each motor

Rearrange the power equation:

\[ P_{\mathrm{useful}}=\frac{E_{\mathrm{useful}}}{t} \]

\[ t=\frac{E_{\mathrm{useful}}}{P_{\mathrm{useful}}} \]

For Motor A, first stage:

\[ t_{A1}=\frac{392}{196} \]

\[ t_{A1}=2.0\ \mathrm{s} \]

For Motor A, second stage:

\[ t_{A2}=\frac{392}{196} \]

\[ t_{A2}=2.0\ \mathrm{s} \]

Total time:

\[ t_A=2.0+2.0 \]

\[ \boxed{t_A=4.0\ \mathrm{s}} \]

For Motor B, first stage:

\[ t_{B1}=\frac{392}{392} \]

\[ t_{B1}=1.0\ \mathrm{s} \]

For Motor B, second stage:

\[ t_{B2}=\frac{392}{98} \]

\[ t_{B2}=4.0\ \mathrm{s} \]

Total time:

\[ t_B=1.0+4.0 \]

\[ \boxed{t_B=5.0\ \mathrm{s}} \]

Motor B has a higher peak useful output power, but it does not finish the task faster.

Its second stage operates at a much lower output power, which increases the total lifting time.

Part c: Calculate the electrical energy consumed

Use the efficiency relationship separately for each stage:

\[ E_{\mathrm{input}} =\frac{E_{\mathrm{useful}}}{\eta} \]

For Motor A, first stage:

\[ E_{A1}=\frac{392}{0.80} \]

\[ E_{A1}=490\ \mathrm{J} \]

For Motor A, second stage:

\[ E_{A2}=\frac{392}{0.50} \]

\[ E_{A2}=784\ \mathrm{J} \]

Total electrical energy:

\[ E_A=490+784 \]

\[ \boxed{E_A=1274\ \mathrm{J}} \]

Now calculate Motor B.

First stage:

\[ E_{B1}=\frac{392}{0.70} \]

\[ E_{B1}=560\ \mathrm{J} \]

Second stage:

\[ E_{B2}=\frac{392}{0.90} \]

\[ E_{B2}=435.6\ \mathrm{J} \]

Total electrical energy:

\[ E_B=560+435.6 \]

\[ \boxed{E_B=995.6\ \mathrm{J}} \]

Motor B consumes approximately 278.4 J less electrical energy than Motor A.

The difference comes from the motors’ efficiencies at their different operating stages.

Part d: Calculate the overall efficiency

This is where students often make a mistake.

You cannot simply take the arithmetic mean of the two stage efficiencies.

For example, Motor A has efficiencies of 80% and 50%.

A simple average would give:

\[ \frac{80+50}{2}=65\% \]

But this is not the overall efficiency.

The correct method is to divide the total useful energy by the total electrical input energy.

For Motor A:

\[ \eta_A=\frac{784}{1274} \]

\[ \boxed{\eta_A=61.5\%} \]

For Motor B:

\[ \eta_B=\frac{784}{995.6} \]

\[ \boxed{\eta_B=78.75\%} \]

Overall efficiency must be calculated using the total energy transferred over the entire operation, rather than by simply averaging the individual percentages.

Part e: Compare the motors

Motor B has twice the peak useful output power, but it takes longer to complete the lift and consumes less electrical energy.

Why?

Look at the second stage.

Motor A delivers 196 W at only 50% efficiency. Its electrical input power is:

\[ P_{\mathrm{input,A2}} =\frac{196}{0.50} =392\ \mathrm{W} \]

It operates in this stage for 2 seconds:

\[ E_{A2}=392\times2=784\ \mathrm{J} \]

Motor B delivers only 98 W during its second stage, but operates at 90% efficiency.

Its electrical input power is:

\[ P_{\mathrm{input,B2}} =\frac{98}{0.90} \approx108.9\ \mathrm{W} \]

It operates for 4 seconds:

\[ E_{B2}=108.9\times4 \approx435.6\ \mathrm{J} \]

Motor B takes twice as long during this stage, but its electrical input power is much lower.

Consequently, it consumes substantially less energy.

This is the important constraint that a comparison of maximum power ratings misses.

A motor’s peak output power tells you what it can deliver under a particular operating condition. It does not describe its complete power profile, efficiency, or total energy consumption.

07The key idea to remember

A more powerful electric motor does not automatically consume less electrical energy.

Power measures the rate of energy transfer, while energy measures the total amount transferred.

When comparing motors, keep three quantities separate:

• Useful mechanical work: How much energy does the motor need to deliver to complete the task?
• Efficiency: What proportion of the electrical input becomes useful mechanical output?
• Operating time and power: How quickly is the energy transferred, and does the actual input power change during operation?

For the same mechanical task, a more efficient motor generally requires less electrical energy. A more powerful motor may complete the task faster, but its power rating alone cannot tell you how much energy it will consume.

The next useful step in a motor calculation is to determine the mechanical power required by the load. For a vertical lift at constant speed, this is \(P_{\mathrm{useful}}=mgv\). Once you know that, efficiency connects the mechanical output to the electrical input, allowing you to calculate both the power drawn and the total energy consumed.