How to Calculate Lifting Power at Constant Speed | HSC Physics

A hoist raises a \(20\ \text{kg}\) crate vertically at a constant speed of \(0.50\ \text{m s}^{-1}\). The crate isn’t accelerating, but the motor is certainly not doing nothing. It is increasing the crate’s gravitational potential energy every second.

So how much power does the motor need?

For a vertical lift at constant speed, the useful mechanical power delivered to the load is:

\[ \boxed{P=mgv} \]

If you’re given the lifting height and time instead of the speed, use the equivalent equation:

\[ \boxed{P=\frac{mgh}{t}} \]

Both equations calculate the rate at which gravitational potential energy is transferred to the load. The important part is knowing which one to use, and whether the question wants useful lifting power or the motor’s actual input power.

01Understanding why lifting power is mgv

Start with the forces acting on the load.

When a motor lifts a load vertically, two main forces act on it:

• The lifting force, acting upwards through the cable.
• The gravitational force, or weight, acting downwards.

The weight is:

\[ F_g=mg \]

where:

• \(F_g\) is the weight in newtons (N).
• \(m\) is the mass in kilograms (kg).
• \(g\) is gravitational acceleration, approximately \(9.8\ \text{m s}^{-2}\).

Why does constant speed matter?

If the load travels upwards at a constant speed, its acceleration is zero.

Newton’s second law tells us:

\[ F_{\text{net}}=ma \]

Since \(a=0\):

\[ F_{\text{net}}=0 \]

The upward lifting force must therefore balance the downward weight, assuming no additional resistance acts on the load.

\[ F_{\text{lift}}=mg \]

This is the force the lifting mechanism must apply to the load.

Connecting force, distance, and energy

Suppose the motor raises the load through a vertical height \(h\).

Work done is force multiplied by displacement in the direction of the force:

\[ W=F_{\text{lift}}h \]

Substituting \(F_{\text{lift}}=mg\):

\[ W=mgh \]

This work increases the gravitational potential energy of the load.

But we haven’t calculated power yet.

Power measures how quickly work is done. Its definition is:

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

Substituting our expression for work:

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

Now notice that vertical speed is:

\[ v=\frac{h}{t} \]

Therefore:

\[ \boxed{P=mgv} \]

This is the equation to remember for a load being lifted vertically at constant speed.

The symbols mean:

Unless a question specifies otherwise, use \(g=9.8\ \text{m s}^{-2}\).

Power is measured in watts:

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

That unit is worth remembering. One watt means one joule of energy transferred every second.

02Work and power are not the same thing

Here’s a distinction that makes these questions much easier.

Imagine two identical motors lifting identical crates through the same vertical height.

One motor takes four seconds. The other takes two seconds.

Both motors transfer the same amount of energy because the mass and height are identical.

However, the second motor transfers that energy twice as quickly.

Consider some illustrative calculations.

A \(20\ \text{kg}\) crate is raised through \(2.0\ \text{m}\).

The work done in either lift is:

\[ W=mgh \]

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

\[ W=392\ \text{J} \]

Now compare the power.

The energy transferred is unchanged, but the power doubles when the lifting time halves.

This gives us a useful way to think about the equations:

• \(W=mgh\) calculates the amount of energy transferred.
• \(P=mgh/t\) calculates how quickly that energy is transferred.
• \(P=mgv\) calculates the same rate using lifting speed.

A useful boundary case: holding a load stationary

Suppose the motor lifts a crate and then stops it at a particular height.

The motor must still support the crate against gravity. However, the crate is no longer moving.

Since \(v=0\):

\[ P=mg(0)=0 \]

The mechanical power transferred to the stationary load is zero.

A real electric motor may still consume electrical power while holding the load because of electrical resistance and other losses. But it is no longer increasing the load’s gravitational potential energy.

This is another reason to remember that power depends on both force and velocity.

03How could you test the lifting-power calculation?

The relationship \(P=mgh/t\) can be tested using a simple motorised lifting apparatus.

The question is straightforward: does the rate of gravitational potential energy gained by the load match the useful mechanical power delivered to it?

Setting up the apparatus

A suitable arrangement contains:

1. A load attached to a light cord.
2. A motorised drum mounted above the load, with the cord wound around it.
3. A vertical metre ruler positioned beside the load’s path.
4. An electronic balance to measure the load’s mass.
5. A stopwatch or video recording to measure the lifting time.
6. A DC power supply, with an ammeter connected in series with the motor and a voltmeter connected across it.

As the motor turns the drum, the cord winds around it and the load moves vertically upwards.

The metre ruler measures the vertical displacement, while the stopwatch measures the time taken.

The electrical instruments also allow us to calculate the motor’s electrical input power.

What should the experiment show?

Suppose we use a \(2.0\ \text{kg}\) load and raise it through \(0.80\ \text{m}\).

The gravitational potential energy gained is:

\[ \Delta E_p=mgh \]

\[ \Delta E_p=2.0\times9.8\times0.80 \]

\[ \Delta E_p=15.68\ \text{J} \]

The energy transferred is the same regardless of how quickly the load is raised.

Now consider three illustrative trials.

These are calculated illustrative values, not reported experimental measurements.

The results show two important relationships:

• Halving the time while keeping the mass and height unchanged doubles the power.
• Doubling the mass while keeping the height and time unchanged also doubles the power.

The pattern supports the relationship:

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

What could go wrong with the experiment?

There is an important limitation in the original measurement method.

Timing only the beginning and end of the lift gives average speed. It doesn’t prove that the load travelled at a constant speed throughout.

To improve the experiment, record the load’s position at regular time intervals using video analysis or a position sensor.

If the load travels equal vertical distances in equal time intervals, its speed is approximately constant.

You should also account for measurement uncertainty, motor friction, and electrical energy lost as heat.

For a real motor, electrical input power will generally be greater than the useful power delivered to the load.

That brings us to another important part of HSC power questions.

04How efficiency changes the power calculation

So far, we’ve calculated the useful mechanical power transferred to the load.

But what if the question asks how much electrical power the motor needs?

These are different quantities.

A real motor isn’t perfectly efficient. Some of its input energy is transferred to the surroundings rather than increasing the load’s gravitational potential energy.

Efficiency tells us what fraction of the input power becomes useful output power.

\[ \boxed{\eta=\frac{P_{\text{useful}}}{P_{\text{input}}}} \]

where:

• \(\eta\) is efficiency, expressed as a decimal.
• \(P_{\text{useful}}\) is the useful mechanical output power.
• \(P_{\text{input}}\) is the power supplied to the motor.

Rearranging:

\[ \boxed{P_{\text{input}}= \frac{P_{\text{useful}}}{\eta}} \]

For a vertically lifted load:

\[ \boxed{P_{\text{input}}= \frac{mgv}{\eta}} \]

This assumes the motor is the source of the energy used for the lift, with the stated efficiency applying to the complete conversion from input power to useful lifting power.

The most common efficiency mistake

Suppose a motor has an efficiency of \(80\%\).

A tempting approach is to multiply the useful power by \(0.80\) to find the input power.

That’s backwards.

The motor only converts \(80\%\) of its input power into useful output. Therefore, the input must be larger than the output.

For example, if the useful lifting power is \(400\ \text{W}\):

\[ P_{\text{input}}=\frac{400}{0.80} \]

\[ P_{\text{input}}=500\ \text{W} \]

The remaining \(100\ \text{W}\) is wasted power.

A useful decision table:

If a question gives electrical voltage and current for a DC motor, you can also calculate its electrical input power using:

\[ P_{\text{electrical}}=VI \]

The same distinction between input power, useful output power, and energy losses appears in Ideal Transformer Current and Power for HSC Physics.

05Worked examples

06A subtle trap: Constant speed does not mean zero power

There’s one more idea worth understanding before attempting exam questions.

At constant speed, the net force on the load is zero. The net work done on the load is also zero because its kinetic energy doesn’t change.

A student might therefore argue:

“The load has zero acceleration, so the net work is zero. That means the motor supplies zero power.”

The problem is that the student has confused net work with work done by the motor.

Consider the two forces acting on the load.

• The motor applies an upward force.
• Gravity applies an equal downward force.

As the load moves upwards, the motor does positive work while gravity does negative work.

For a lift through height \(h\):

\[ W_{\text{motor}}=+mgh \]

\[ W_{\text{gravity}}=-mgh \]

Therefore:

\[ W_{\text{net}}=mgh-mgh=0 \]

The net work is zero, but the motor has still transferred \(mgh\) joules of energy into the load’s gravitational potential energy.

Its useful power is still:

\[ \boxed{P_{\text{motor}}=mgv} \]

The distinction is:

• Net work determines the change in kinetic energy.
• Work done by the lifting mechanism increases gravitational potential energy.
• Power measures how quickly that energy transfer occurs.

This is why you shouldn’t automatically use the net work done on an object when a question asks for the work or power supplied by a particular force.

07Questions and solutions

Try each question before reading its solution. The difficulty increases as you progress, with the final question combining power, efficiency, and energy transfers in a more unusual lifting system.

Question 1

A \(12\ \text{kg}\) load is lifted vertically at a constant speed of \(0.75\ \text{m s}^{-1}\).

Calculate the useful mechanical power required to lift the load.

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

Solution 1

The required useful mechanical power is \(88.2\ \text{W}\).

Because the load moves at constant speed, its acceleration is zero. The lifting force therefore equals its weight.

Use:

\[ P=mgv \]

Substitute:

\[ P=12\times9.8\times0.75 \]

\[ \boxed{P=88.2\ \text{W}} \]

The lifting mechanism must transfer \(88.2\ \text{J}\) of gravitational potential energy to the load every second.

Question 2

An electric hoist raises a \(28\ \text{kg}\) load vertically through \(2.4\ \text{m}\) in \(3.0\ \text{s}\).

The load moves at constant speed, and the hoist has an overall efficiency of \(70\%\).

Calculate:

a. The useful mechanical output power.

b. The electrical input power.

c. The total energy wasted during the lift.

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

Solution 2

The useful output power is \(219.5\ \text{W}\), the electrical input power is \(313.6\ \text{W}\), and the wasted energy is \(282.2\ \text{J}\).

Step 1: Calculate the useful work done.

The load gains gravitational potential energy:

\[ W=mgh \]

\[ W=28\times9.8\times2.4 \]

\[ W=658.56\ \text{J} \]

Step 2: Calculate useful output power.

\[ P_{\text{useful}}=\frac{W}{t} \]

\[ P_{\text{useful}}= \frac{658.56}{3.0} \]

\[ \boxed{P_{\text{useful}}=219.52\ \text{W}} \]

Step 3: Calculate electrical input power.

Convert the efficiency:

\[ \eta=0.70 \]

Rearrange the efficiency equation:

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

Substitute:

\[ P_{\text{input}}= \frac{219.52}{0.70} \]

\[ \boxed{P_{\text{input}}=313.6\ \text{W}} \]

Step 4: Calculate wasted energy.

First calculate the total electrical energy supplied:

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

\[ E_{\text{input}}=313.6\times3.0 \]

\[ E_{\text{input}}=940.8\ \text{J} \]

The useful energy transferred is \(658.56\ \text{J}\).

Therefore:

\[ E_{\text{wasted}}= E_{\text{input}}-E_{\text{useful}} \]

\[ E_{\text{wasted}}=940.8-658.56 \]

\[ \boxed{E_{\text{wasted}}=282.24\ \text{J}} \]

The important point is that efficiency changes the required input power, not the gravitational potential energy gained by the load.

Question 3

A \(24\ \text{kg}\) load is lifted vertically through \(0.90\ \text{m}\) in \(3.0\ \text{s}\) using an ideal movable-pulley system.

The pulley arrangement has two rope segments supporting the load. Assume the rope and pulleys have negligible mass, and there is no friction.

Calculate:

a. The constant upward speed of the load.

b. The force required at the free end of the rope.

c. The speed at which the free end must be pulled.

d. The mechanical power supplied at the free end.

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

Solution 3

The load rises at \(0.30\ \text{m s}^{-1}\). The pulling force is \(117.6\ \text{N}\), the free end moves at \(0.60\ \text{m s}^{-1}\), and the mechanical power supplied is \(70.56\ \text{W}\).

This question tests an important feature of pulley systems: reducing the required force increases the distance and speed through which the rope must move.

Step 1: Calculate the load’s vertical speed.

\[ v_{\text{load}}=\frac{h}{t} \]

\[ v_{\text{load}}=\frac{0.90}{3.0} \]

\[ \boxed{v_{\text{load}}=0.30\ \text{m s}^{-1}} \]

Step 2: Calculate the force required at the free end.

The weight of the load is:

\[ F_g=mg \]

\[ F_g=24\times9.8 \]

\[ F_g=235.2\ \text{N} \]

Two rope segments support the movable pulley.

In an ideal arrangement, each segment provides an equal upward tension.

Therefore:

\[ 2T=235.2 \]

\[ T=\frac{235.2}{2} \]

\[ \boxed{T=117.6\ \text{N}} \]

The pulling force at the free end equals the rope tension.

Step 3: Calculate the free-end speed.

As the load rises by a certain distance, both supporting rope segments become shorter by that distance.

Consequently, the free end must move twice as far as the load.

Therefore:

\[ v_{\text{rope}}=2v_{\text{load}} \]

\[ v_{\text{rope}}=2\times0.30 \]

\[ \boxed{v_{\text{rope}}=0.60\ \text{m s}^{-1}} \]

Step 4: Calculate mechanical input power.

The power supplied at the free end is:

\[ P=Fv \]

Substitute the pulling force and rope speed:

\[ P=117.6\times0.60 \]

\[ \boxed{P=70.56\ \text{W}} \]

Notice something interesting.

The pulling force is half the weight, but the rope moves twice as fast as the load.

The power is unchanged:

\[ P=mgv_{\text{load}} \]

\[ P=24\times9.8\times0.30 \]

\[ P=70.56\ \text{W} \]

An ideal pulley reduces the force needed, but it doesn’t reduce the total mechanical work or power required.

Question 4

A motor raises a \(15\ \text{kg}\) load vertically through \(2.0\ \text{m}\) in \(5.0\ \text{s}\) at constant speed.

A student makes the following claim:

“The net work done on the load is zero because it moves at constant speed. Therefore, the motor supplies zero power.”

Evaluate this claim.

Calculate:

a. The work done by the motor on the load.

b. The work done by gravity.

c. The net work done on the load.

d. The useful mechanical power supplied by the motor.

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

Solution 4

The student’s claim is incorrect. The net work is zero, but the motor still does \(294\ \text{J}\) of positive work and supplies \(58.8\ \text{W}\) of useful mechanical power.

Step 1: Calculate the lifting force.

Since the load moves at constant speed, the upward force balances its weight.

\[ F_{\text{lift}}=mg \]

\[ F_{\text{lift}}=15\times9.8 \]

\[ F_{\text{lift}}=147\ \text{N} \]

Step 2: Calculate the work done by the motor.

The motor’s force and the displacement are both upwards.

Therefore, the work is positive:

\[ W_{\text{motor}}=F_{\text{lift}}h \]

\[ W_{\text{motor}}=147\times2.0 \]

\[ \boxed{W_{\text{motor}}=+294\ \text{J}} \]

Step 3: Calculate the work done by gravity.

Gravity acts downwards while the load moves upwards.

The angle between the force and displacement is \(180^\circ\).

Therefore:

\[ W_{\text{gravity}}=mgh\cos180^\circ \]

Since \(\cos180^\circ=-1\):

\[ W_{\text{gravity}}=-mgh \]

\[ W_{\text{gravity}}=-15\times9.8\times2.0 \]

\[ \boxed{W_{\text{gravity}}=-294\ \text{J}} \]

Step 4: Calculate net work.

The total work is the sum of the work done by all forces:

\[ W_{\text{net}}= W_{\text{motor}}+W_{\text{gravity}} \]

\[ W_{\text{net}}=294-294 \]

\[ \boxed{W_{\text{net}}=0\ \text{J}} \]

This agrees with the work-energy theorem.

The load’s kinetic energy doesn’t change because its speed remains constant.

Step 5: Calculate the motor’s power.

The question asks for power supplied by the motor, not net power on the load.

Use the motor’s positive work:

\[ P_{\text{motor}}= \frac{W_{\text{motor}}}{t} \]

\[ P_{\text{motor}}=\frac{294}{5.0} \]

\[ \boxed{P_{\text{motor}}=58.8\ \text{W}} \]

The motor supplies \(58.8\ \text{J}\) of useful energy every second.

The key is to distinguish the work done by an individual force from the net work done on an object.

Question 5

An elevator car containing passengers has a combined mass of \(360\ \text{kg}\).

It rises vertically at a constant speed of \(1.2\ \text{m s}^{-1}\). A counterweight of mass \(300\ \text{kg}\) is connected to the elevator by a cable passing over a motor-driven pulley.

As the elevator rises, the counterweight descends at the same speed.

Assume:

• The elevator and counterweight move at constant speed.
• The cable and pulley have negligible mass.
• There is no friction in the guides or cable.
• The motor-drive has an efficiency of \(80\%\).
• This efficiency is defined as the motor’s net mechanical output power to the connected elevator-counterweight system divided by its electrical input power.

Calculate:

a. The rate at which the elevator car gains gravitational potential energy.

b. The rate at which the counterweight loses gravitational potential energy.

c. The net mechanical power the motor must supply to the connected system.

d. The electrical input power required by the motor-drive.

Explain why simply calculating \(P=mgv\) for the elevator car and dividing by the efficiency gives an incorrect answer for the motor’s electrical input power.

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

Solution 5

The elevator car gains gravitational potential energy at \(4233.6\ \text{W}\), while the counterweight releases gravitational potential energy at \(3528\ \text{W}\).

The motor must therefore supply only \(705.6\ \text{W}\) of net mechanical power. Its electrical input power is \(882\ \text{W}\).

The important detail is that the counterweight is also moving. Its decreasing gravitational potential energy helps lift the elevator car.

Step 1: Calculate the rate of gravitational potential energy gain of the elevator.

The elevator rises at a constant speed, so:

\[ P_{\text{car}}=m_{\text{car}}gv \]

Substituting:

\[ P_{\text{car}}=360\times9.8\times1.2 \]

\[ \boxed{P_{\text{car}}=4233.6\ \text{W}} \]

The elevator gains gravitational potential energy at a rate of approximately \(4.23\ \text{kW}\).

Step 2: Calculate the rate of gravitational potential energy loss of the counterweight.

The counterweight moves downwards, so its gravitational potential energy decreases.

The magnitude of its energy-loss rate is:

\[ P_{\text{counterweight}}=m_{\text{counterweight}}gv \]

Substituting:

\[ P_{\text{counterweight}}= 300\times9.8\times1.2 \]

\[ \boxed{P_{\text{counterweight}}=3528\ \text{W}} \]

The counterweight releases \(3528\ \text{J}\) of gravitational potential energy every second.

Step 3: Calculate the net mechanical power supplied by the motor.

This is the critical step.

The elevator gains gravitational potential energy, but the counterweight loses gravitational potential energy.

Therefore, the motor only needs to supply the difference.

\[ P_{\text{motor}}= P_{\text{car}}-P_{\text{counterweight}} \]

Substitute:

\[ P_{\text{motor}}=4233.6-3528 \]

\[ \boxed{P_{\text{motor}}=705.6\ \text{W}} \]

Because the system moves at constant speed, its kinetic energy remains unchanged.

With no friction or other losses in the lifting mechanism, this net increase in gravitational potential energy must come from the motor.

Step 4: Calculate electrical input power.

The motor-drive efficiency is:

\[ \eta= \frac{P_{\text{motor}}}{P_{\text{electrical}}} \]

Rearrange:

\[ P_{\text{electrical}}= \frac{P_{\text{motor}}}{\eta} \]

Substitute:

\[ P_{\text{electrical}}= \frac{705.6}{0.80} \]

\[ \boxed{P_{\text{electrical}}=882\ \text{W}} \]

Step 5: Explain why the tempting calculation is incorrect.

A student might calculate:

\[ P=360\times9.8\times1.2 \]

and then divide by \(0.80\).

This would give:

\[ P_{\text{electrical}}=5292\ \text{W} \]

But this calculation assumes the motor supplies all the energy required to raise the elevator.

It ignores the counterweight.

The counterweight is simultaneously losing gravitational potential energy, providing \(3528\ \text{W}\) to assist the lifting process.

The motor only needs to supply the remaining \(705.6\ \text{W}\) mechanically.

This is a useful reminder that the correct system boundary matters. When several objects exchange gravitational potential energy, calculate the net energy change before determining the motor’s required power.

If the counterweight were heavier than the elevator car, the gravitational energy released could even drive the system downwards, requiring the motor to act as a brake or absorb energy rather than supply positive mechanical power.

08Where to go next

For ordinary vertical lifting questions, the central relationship is:

\[ \boxed{P_{\text{useful}}=mgv} \]

If height and time are supplied, use:

\[ \boxed{P_{\text{useful}}=\frac{mgh}{t}} \]

If the question includes efficiency, distinguish useful output power from input power:

\[ \boxed{P_{\text{input}}= \frac{P_{\text{useful}}}{\eta}} \]

Always check what is moving, where the energy is coming from, and which power the question actually wants.

The next useful connection is electrical power in motors: comparing the electrical energy supplied to a motor with the mechanical energy it transfers to a moving load. That brings together \(P=VI\), \(P=Fv\), gravitational potential energy, and efficiency in one problem.