How to Separate Electrical Input Power from Mechanical Output Power

A winch motor draws \(2.0\text{ A}\) from a \(12\text{ V}\) supply while lifting a \(0.80\text{ kg}\) load through \(1.5\text{ m}\) in \(4.0\text{ s}\). If you calculate \(P=VI\), you get \(24\text{ W}\). Does that mean the motor delivers \(24\text{ W}\) of mechanical power to the load?

No. The \(24\text{ W}\) is the electrical input power: the rate at which electrical energy enters the motor. The mechanical output power is the rate at which the motor transfers mechanical energy to the load. In this example, the load gains \(11.76\text{ J}\) of gravitational potential energy in \(4.0\text{ s}\), giving an average useful mechanical output power of just \(2.94\text{ W}\).

The key is to calculate power on opposite sides of the energy converter, rather than treating electrical and mechanical power as interchangeable.

01Identify the input and output

Think of an electric motor as an energy converter.

Electrical energy enters through its terminals. The motor converts some of that energy into mechanical energy, which can then move a load, turn a wheel, or lift an object.

However, not all the electrical energy becomes useful mechanical energy. Some is transferred into thermal energy, sound, and vibration. Friction in the motor and its transmission can also dissipate energy.

For a motor operating normally, distinguish these two quantities:

Here:

• \(V\) is the potential difference across the motor terminals, measured in volts (V).
• \(I\) is the current entering the motor, measured in amperes (A).
• \(F\) is the force exerted by the motor, measured in newtons (N).
• \(v\) is the velocity of the object receiving that force, measured in metres per second (\(\text{m s}^{-1}\)).
• \(\theta\) is the angle between the force and velocity.

For the electrical calculation, use the voltage across the motor itself, not automatically the voltage displayed on the power supply.

For the mechanical calculation, use the force supplied by the motor, not automatically the net force acting on the load.

If you need to revise either foundation, see Electrical Power in HSC Physics: Using P = VI and Equivalent Forms and Mechanical Power with P = Fv cos θ for HSC Physics.

02Measure electrical input and mechanical output separately

A useful way to investigate the difference is to connect a small DC motor to a winch that lifts a known mass.

The experiment needs two sets of measurements because the input and output occur in different parts of the system.

Measuring electrical input power

Start with a DC power supply connected to the motor.

Connect an ammeter in series with the motor so that the current flowing through it can be measured. Connect a voltmeter in parallel across the motor terminals so that it measures the potential difference across the motor.

Once the motor is running under load, record both readings.

Calculate the electrical input power using:

\[ P_{\mathrm{in}}=VI \]

For example, if the motor operates at \(12.0\text{ V}\) and draws \(2.0\text{ A}\):

\[ P_{\mathrm{in}}=12.0\times2.0 \]

\[ P_{\mathrm{in}}=24.0\text{ W} \]

The motor receives electrical energy at a rate of \(24.0\text{ J}\) each second.

Measuring mechanical output power

Now consider the mechanical side.

Attach a string to a drum connected to the motor shaft. The string supports a known mass, which the motor lifts vertically.

Measure:

• the mass being lifted, \(m\), in kilograms;
• the vertical displacement, \(\Delta h\), in metres;
• the time taken, \(\Delta t\), in seconds.

If the mass moves at constant speed, the upward tension equals its weight:

\[ F=mg \]

The mechanical output power is therefore:

\[ P_{\mathrm{out}}=mgv \]

Alternatively, calculate the gravitational potential energy gained by the mass:

\[ \Delta E_{\mathrm{GPE}}=mg\Delta h \]

Then divide by the time taken:

\[ P_{\mathrm{out,avg}} =\frac{mg\Delta h}{\Delta t} \]

These two methods give the same answer for a constant-speed lift.

The energy method is particularly useful when a question provides a height and a time rather than an instantaneous velocity.

What should the experiment show?

You might initially predict that electrical input power and mechanical output power should be equal because energy is conserved.

The prediction is reasonable, but it overlooks energy transfers into other forms.

Consider the following illustrative measurements.

Take \(g=9.8\text{ m s}^{-2}\).

The electrical input power is:

\[ P_{\mathrm{in}}=VI \]

\[ P_{\mathrm{in}}=12.0\times2.0 =24.0\text{ W} \]

The gravitational potential energy gained by the mass is:

\[ \Delta E_{\mathrm{GPE}}=mg\Delta h \]

\[ =0.80\times9.8\times1.5 \]

\[ =11.76\text{ J} \]

Therefore, the average useful mechanical output power is:

\[ P_{\mathrm{out,avg}} =\frac{11.76}{4.0} \]

\[ =2.94\text{ W} \]

The electrical input is \(24.0\text{ W}\), while the useful mechanical output is \(2.94\text{ W}\).

The difference is not a violation of conservation of energy. Assuming the motor and lifting mechanism have no net change in stored kinetic energy over the measured interval, the remaining energy is transferred into other forms, including thermal energy, sound, and frictional losses.

This is the central observation: electrical input power and mechanical output power describe different stages of an energy transfer.

A measurement problem worth catching

There is an important detail in the apparatus.

Suppose the power supply displays \(12.0\text{ V}\), but the potential difference measured directly across the motor terminals is only \(10.5\text{ V}\) while the motor is operating.

If the current is \(2.0\text{ A}\), which voltage should you use?

For the electrical power entering the motor, use \(10.5\text{ V}\):

\[ P_{\mathrm{motor,in}}=10.5\times2.0 \]

\[ =21.0\text{ W} \]

Using the supply voltage would give:

\[ P_{\mathrm{supply}}=12.0\times2.0 =24.0\text{ W} \]

The difference is \(3.0\text{ W}\), which is dissipated in the connecting leads and other components between the measurement boundaries.

The first measurement design would incorrectly attribute this additional power to the motor.

The improved design places the voltmeter directly across the motor terminals. This makes the electrical input measurement consistent with the chosen system boundary.

The same principle applies to HSC exam questions: identify exactly where the electrical energy enters the device before choosing the voltage.

03Separate power from energy

Another common source of mistakes is confusing the amount of energy transferred with the rate of energy transfer.

Energy measures how much has been transferred. Power measures how quickly that transfer occurs.

The relationship is:

\[ P_{\mathrm{avg}}=\frac{\Delta E}{\Delta t} \]

where:

• \(P_{\mathrm{avg}}\) is average power in watts;
• \(\Delta E\) is energy transferred in joules;
• \(\Delta t\) is the time interval in seconds.

Remember:

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

For the winch example, the motor receives \(96\text{ J}\) of electrical energy over four seconds:

\[ E_{\mathrm{in}}=P_{\mathrm{in}}\Delta t \]

\[ =24.0\times4.0 =96.0\text{ J} \]

The load gains only \(11.76\text{ J}\) of gravitational potential energy.

The input and output energies describe the total amounts transferred over the interval. The corresponding powers describe their rates of transfer.

This distinction becomes particularly important when a motor’s electrical readings change during operation.

If the power is constant, simply multiply power by time.

If the power changes, calculate the energy transferred during each interval and add the results.

For example:

The average electrical input power is:

\[ P_{\mathrm{avg}}=\frac{84}{5} =16.8\text{ W} \]

It is not \(16\text{ W}\), which would be the simple arithmetic average of the two power readings.

The intervals have different durations, so their contributions must be weighted by time.

04Calculate efficiency using the correct input and output

Once you have separated electrical input power from mechanical output power, you can calculate efficiency.

Efficiency measures the proportion of the input energy that becomes useful output energy.

For a motor operating in its normal motoring mode:

\[ \eta= \frac{P_{\mathrm{useful,out}}} {P_{\mathrm{in}}}\times100\% \]

The same relationship can be expressed using energy:

\[ \eta= \frac{E_{\mathrm{useful,out}}} {E_{\mathrm{in}}}\times100\% \]

Both methods give the same result when the input and output quantities refer to the same time interval.

However, you must identify what the question means by useful mechanical output.

These are different system boundaries.

For example, if a motor drives a gearbox and lifting drum, calculating the gravitational potential energy gained by the load measures the useful efficiency of the complete lifting arrangement. It does not isolate the motor’s shaft efficiency because the gearbox and drum can also dissipate energy.

To measure the motor’s shaft efficiency separately, you would need to measure its mechanical shaft output rather than simply measuring the energy gained by the load.

A useful convention for HSC questions

Unless the question specifies otherwise:

1. Identify the electrical input to the device.
2. Identify the useful mechanical output.
3. Make sure both refer to the same system and time interval.
4. Calculate the ratio of useful output to total input.
5. Express efficiency as a percentage.

For ordinary steady-state operation, with all energy inputs accounted for, efficiency cannot exceed \(100\%\).

If your calculation produces more than \(100\%\), check your system boundary and whether stored energy is being released.

We will return to that last point in Question 5.

05A contrasting case: a motor that draws power but produces no mechanical output

Imagine connecting a DC motor to a power supply and then preventing its shaft from rotating.

The motor still draws electrical current. It can also produce a substantial turning force, called torque.

But is it delivering mechanical power?

No. Its shaft is stationary.

Mechanical power depends on both force and velocity:

\[ P=Fv\cos\theta \]

When the shaft is stationary, its tangential velocity is zero. Therefore:

\[ P_{\mathrm{mech,out}}=0 \]

The motor can have a non-zero torque while producing zero mechanical output power.

At a steady stall, the electrical energy supplied is predominantly dissipated as thermal energy in the motor windings and other electrical losses.

This exposes an important misconception: having a force or torque does not automatically mean that mechanical power is being produced.

There is another related trap.

Suppose a motor lifts a load at constant speed. The net force on the load is zero, so the net power acting on the load is also zero.

Does that mean the motor produces zero mechanical power?

Again, no.

The motor supplies positive mechanical power, while gravity supplies negative mechanical power. Their contributions balance, so the load’s kinetic energy remains constant.

The motor is still transferring energy into the load-Earth system by increasing its gravitational potential energy.

When calculating the motor’s mechanical output, use the motor’s force rather than the net force.

06Worked examples

07Questions and solutions

Question 1

An electric winch operates at a terminal voltage of \(12\text{ V}\) and draws a current of \(1.5\text{ A}\).

It lifts a \(2.0\text{ kg}\) mass vertically through \(0.60\text{ m}\) in \(4.0\text{ s}\).

Assume the mass starts and finishes at rest, the winch has no net change in stored kinetic energy, and the gravitational field strength is \(9.8\text{ m s}^{-2}\).

Calculate:

a. the electrical input power;

b. the average useful mechanical output power;

c. the efficiency of the lifting system.

Solution 1

The electrical input power is \(18.0\text{ W}\), the useful mechanical output power is \(2.94\text{ W}\), and the lifting system efficiency is \(16.3\%\).

Part a: Electrical input power

Use:

\[ P_{\mathrm{in}}=VI \]

Substitute:

\[ P_{\mathrm{in}}=12\times1.5 \]

\[ P_{\mathrm{in}}=18.0\text{ W} \]

Part b: Mechanical output power

The useful mechanical energy gained by the mass is its increase in gravitational potential energy:

\[ \Delta E_{\mathrm{GPE}}=mg\Delta h \]

\[ =2.0\times9.8\times0.60 \]

\[ =11.76\text{ J} \]

Divide by the time taken:

\[ P_{\mathrm{out,avg}}=\frac{11.76}{4.0} \]

\[ P_{\mathrm{out,avg}}=2.94\text{ W} \]

Part c: Efficiency

\[ \eta= \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} \times100\% \]

\[ =\frac{2.94}{18.0}\times100\% \]

\[ =16.3\% \]

The system converts approximately \(16.3\%\) of its electrical input energy into useful gravitational potential energy for the load.

The remaining energy is dissipated through other processes, assuming there is no net change in stored energy.

Question 2

An electric cart is travelling uphill along a straight ramp.

At a particular instant:

• the cart has a total mass of \(15\text{ kg}\);
• the ramp is inclined at \(10^\circ\) to the horizontal;
• the cart is travelling uphill at \(2.0\text{ m s}^{-1}\);
• its uphill acceleration is \(0.40\text{ m s}^{-2}\);
• the total resistive force is \(8.0\text{ N}\), acting down the slope.

The electric motor has a terminal voltage of \(24\text{ V}\) and draws \(4.0\text{ A}\).

Calculate:

a. the force supplied by the motor;

b. the instantaneous mechanical output power;

c. the motor’s mechanical conversion efficiency.

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

Solution 2

The motor supplies approximately \(39.5\text{ N}\) of driving force and \(79.1\text{ W}\) of mechanical output power. Its efficiency is approximately \(82.4\%\).

Part a: Motor force

Take uphill as the positive direction.

The forces along the slope are:

• motor force uphill;
• gravitational component \(mg\sin\theta\) downhill;
• resistive force downhill.

Apply Newton’s second law:

\[ F_{\mathrm{motor}}-mg\sin\theta-F_r=ma \]

Rearrange:

\[ F_{\mathrm{motor}} =ma+mg\sin\theta+F_r \]

Substitute:

\[ =15(0.40)+15(9.8)\sin10^\circ+8.0 \]

\[ =6.0+25.5+8.0 \]

\[ F_{\mathrm{motor}}\approx39.5\text{ N} \]

Part b: Mechanical output power

The motor force and cart velocity point in the same direction.

Therefore:

\[ P_{\mathrm{out}}=F_{\mathrm{motor}}v \]

\[ =39.5\times2.0 \]

\[ P_{\mathrm{out}}\approx79.1\text{ W} \]

Part c: Efficiency

First calculate electrical input power:

\[ P_{\mathrm{in}}=VI \]

\[ =24\times4.0 \]

\[ =96\text{ W} \]

Then calculate efficiency:

\[ \eta=\frac{79.1}{96}\times100\% \]

\[ \eta\approx82.4\% \]

The motor converts approximately \(82.4\%\) of its electrical input power into mechanical power delivered to the cart.

The tempting mistake is to calculate the net force and multiply it by velocity. That would give the rate of change of the cart’s kinetic energy, not the mechanical output power of the motor.

The motor must also supply energy to increase gravitational potential energy and overcome resistance.

Question 3

A student investigates an electric winch and obtains the following measurements:

• power supply terminal voltage: \(18.0\text{ V}\);
• voltage directly across the motor terminals: \(15.0\text{ V}\);
• current through the motor: \(2.0\text{ A}\);
• mass being lifted: \(2.5\text{ kg}\);
• constant upward speed of the mass: \(0.50\text{ m s}^{-1}\).

The motor drives the lifting drum directly, and the measured load power represents the useful mechanical output of the complete motor-and-winch assembly.

Assume the connecting leads account for the difference between the supply voltage and motor terminal voltage.

Calculate:

a. the electrical power entering the motor;

b. the useful mechanical output power;

c. the efficiency measured from the motor terminals to the load;

d. the efficiency measured from the supply terminals to the load.

Explain why the two efficiencies differ.

Solution 3

The electrical power entering the motor is \(30.0\text{ W}\), the useful mechanical output is \(12.25\text{ W}\), the motor-and-winch efficiency is \(40.8\%\), and the supply-to-load efficiency is \(34.0\%\).

The difference arises because the second efficiency calculation also includes energy dissipated in the connecting leads.

Part a: Electrical input power

For the motor itself, use the voltage across its terminals:

\[ P_{\mathrm{motor,in}}=VI \]

\[ =15.0\times2.0 \]

\[ =30.0\text{ W} \]

Do not use \(18.0\text{ V}\) here. That voltage belongs to the supply boundary.

Part b: Useful mechanical output power

The load moves upward at constant speed, so the lifting force equals its weight:

\[ F=mg \]

\[ =2.5\times9.8 \]

\[ =24.5\text{ N} \]

The mechanical output power is:

\[ P_{\mathrm{out}}=Fv \]

\[ =24.5\times0.50 \]

\[ =12.25\text{ W} \]

Part c: Motor-and-winch efficiency

\[ \eta_{\mathrm{motor+winch}} =\frac{12.25}{30.0}\times100\% \]

\[ =40.8\% \]

This measures the useful mechanical output relative to the electrical power entering the motor.

Part d: Supply-to-load efficiency

The electrical power leaving the supply is:

\[ P_{\mathrm{supply}}=VI \]

\[ =18.0\times2.0 \]

\[ =36.0\text{ W} \]

Therefore:

\[ \eta_{\mathrm{supply\ to\ load}} =\frac{12.25}{36.0}\times100\% \]

\[ =34.0\% \]

The difference in power is:

\[ P_{\mathrm{leads}}=36.0-30.0 \]

\[ =6.0\text{ W} \]

This power is dissipated in the connecting leads.

Both efficiencies are valid because they use different system boundaries.

The mistake would be to calculate efficiency using the supply voltage while claiming to measure the motor’s efficiency alone.

Question 4

An electric winch raises a \(2.0\text{ kg}\) mass through \(1.5\text{ m}\) in \(5.0\text{ s}\).

The motor operates under two different electrical conditions:

Assume the readings are constant within each interval. The mass starts and finishes at rest, and the winch has no net change in stored kinetic energy.

Calculate:

a. the total electrical energy supplied;

b. the average electrical input power;

c. the average useful mechanical output power;

d. the overall efficiency.

A student argues that the average input power is simply the arithmetic mean of the two power readings.

Explain whether this is correct.

Solution 4

The total electrical energy supplied is \(84\text{ J}\), the average electrical input power is \(16.8\text{ W}\), the average useful mechanical output power is \(5.88\text{ W}\), and the overall efficiency is \(35.0\%\).

The student’s arithmetic average is incorrect because the two power readings apply for different lengths of time.

Part a: Total electrical energy

First calculate the electrical power during each interval.

For the first interval:

\[ P_1=12\times1.0=12\text{ W} \]

The energy supplied is:

\[ E_1=P_1t_1 \]

\[ =12\times2.0 \]

\[ =24\text{ J} \]

For the second interval:

\[ P_2=10\times2.0=20\text{ W} \]

The energy supplied is:

\[ E_2=P_2t_2 \]

\[ =20\times3.0 \]

\[ =60\text{ J} \]

Add the energy transfers:

\[ E_{\mathrm{in}}=E_1+E_2 \]

\[ =24+60 \]

\[ =84\text{ J} \]

Part b: Average electrical input power

The total time is:

\[ \Delta t=2.0+3.0=5.0\text{ s} \]

Therefore:

\[ P_{\mathrm{in,avg}} =\frac{E_{\mathrm{in}}}{\Delta t} \]

\[ =\frac{84}{5.0} \]

\[ =16.8\text{ W} \]

The arithmetic average of the two power readings would be:

\[ \frac{12+20}{2}=16\text{ W} \]

This is incorrect because the motor operates at \(12\text{ W}\) for two seconds and \(20\text{ W}\) for three seconds.

The correct average must account for the duration of each interval.

Part c: Average useful mechanical output power

The load gains gravitational potential energy:

\[ \Delta E_{\mathrm{GPE}}=mg\Delta h \]

\[ =2.0\times9.8\times1.5 \]

\[ =29.4\text{ J} \]

The average mechanical output power is:

\[ P_{\mathrm{out,avg}} =\frac{29.4}{5.0} \]

\[ =5.88\text{ W} \]

Part d: Overall efficiency

Use the total energy transferred over the complete interval:

\[ \eta= \frac{E_{\mathrm{useful,out}}} {E_{\mathrm{in}}}\times100\% \]

\[ =\frac{29.4}{84}\times100\% \]

\[ =35.0\% \]

The overall efficiency is \(35.0\%\).

Notice that calculating efficiency from the ratio of average powers would give the same answer:

\[ \eta=\frac{5.88}{16.8}\times100\% \]

\[ =35.0\% \]

The important condition is that both average powers refer to the same five-second interval.

Question 5

An electric winch raises a \(4.0\text{ kg}\) platform through a vertical height of \(2.8\text{ m}\) in \(2.0\text{ s}\).

During the lift:

• the motor terminal voltage remains at \(24\text{ V}\);
• the motor draws a constant current of \(2.0\text{ A}\);
• a flywheel attached to the motor shaft slows down, releasing \(20\text{ J}\) of stored rotational kinetic energy.

Assume:

• the platform starts and finishes at rest;
• the flywheel’s rotational kinetic energy decreases by exactly \(20\text{ J}\);
• there is no net change in other stored kinetic or potential energy within the winch;
• the remaining energy is dissipated through friction, heating, sound, and other losses;
• \(g=9.8\text{ m s}^{-2}\).

A student calculates the electrical input power and useful mechanical output power, then obtains an apparent efficiency greater than \(100\%\).

They conclude that the measurements must be impossible because energy cannot be created.

Assess this conclusion.

Calculate:

a. the electrical input energy during the lift;

b. the useful mechanical energy gained by the platform;

c. the apparent ratio of useful mechanical output energy to electrical input energy;

d. the actual energy-conversion efficiency when the released flywheel energy is included in the total energy supplied;

e. the energy dissipated through losses.

Solution 5

The student’s conclusion is incorrect. The apparent electrical-input-to-load ratio is approximately \(114.3\%\), but this is not the true energy-conversion efficiency because the flywheel releases an additional \(20\text{ J}\) of stored energy during the lift.

When this stored energy is included, the actual efficiency is approximately \(94.6\%\), and energy is conserved.

Part a: Electrical input energy

First calculate electrical input power:

\[ P_{\mathrm{in}}=VI \]

\[ =24\times2.0 \]

\[ =48\text{ W} \]

The lift takes \(2.0\text{ s}\), so:

\[ E_{\mathrm{electrical}}=P_{\mathrm{in}}\Delta t \]

\[ =48\times2.0 \]

\[ =96\text{ J} \]

Part b: Useful mechanical energy gained

The platform gains gravitational potential energy:

\[ \Delta E_{\mathrm{GPE}}=mg\Delta h \]

\[ =4.0\times9.8\times2.8 \]

\[ =109.76\text{ J} \]

Therefore, the useful mechanical energy output is \(109.76\text{ J}\).

Part c: Apparent electrical-input-to-load ratio

If we compare the useful mechanical output energy directly with the electrical input energy:

\[ \eta_{\mathrm{apparent}} =\frac{109.76}{96}\times100\% \]

\[ =114.3\% \]

This result looks impossible if electrical energy is assumed to be the only energy supplied.

However, that assumption is incorrect.

The flywheel is an additional energy store. Its rotational kinetic energy decreases by \(20\text{ J}\), meaning it supplies energy to the lifting process.

Part d: Actual energy-conversion efficiency

The total energy supplied is the sum of the electrical input and the energy released by the flywheel:

\[ E_{\mathrm{total,in}} =E_{\mathrm{electrical}}+E_{\mathrm{flywheel}} \]

\[ =96+20 \]

\[ =116\text{ J} \]

Now calculate the actual efficiency:

\[ \eta= \frac{E_{\mathrm{useful,out}}} {E_{\mathrm{total,in}}}\times100\% \]

\[ =\frac{109.76}{116}\times100\% \]

\[ =94.6\% \]

Part e: Energy dissipated through losses

The energy balance is:

\[ E_{\mathrm{total,in}} =E_{\mathrm{useful,out}}+E_{\mathrm{loss}} \]

Rearrange:

\[ E_{\mathrm{loss}} =116-109.76 \]

\[ =6.24\text{ J} \]

The energy accounting is therefore:

Everything balances.

The important distinction is that the motor-win​ch system is not relying solely on electrical energy during this interval. It is also drawing on energy stored in the rotating flywheel.

If the flywheel maintained constant rotational kinetic energy, this additional energy source would not exist. Under that steady-state condition, and with all energy inputs accounted for, the useful output could not exceed the total input.

This is why efficiency calculations must consider not only the input and output powers, but also changes in stored energy within the system.

08What to check in your next power question

When you encounter a motor or another energy-conversion device in HSC Physics, make three decisions before substituting numbers:

1. Where does the energy enter? For an ordinary DC motor, electrical input power is calculated using the voltage across its terminals and the current entering it.
2. Where does the useful energy leave? Identify whether the question wants shaft power, power delivered to a load, or an increase in gravitational potential energy.
3. Are you calculating an instantaneous rate or an average over time? Use \(Fv\cos\theta\) for instantaneous mechanical power, or useful energy transferred divided by the time interval for average power.

Once those decisions are clear, the efficiency calculation becomes straightforward.

The next useful skill is deciding whether a motor’s mechanical output should be calculated from an instantaneous force and velocity or from the total energy gained by its load. That distinction becomes especially important when the load accelerates, the motor changes operating conditions, or the system contains stored energy.