When Can You Use P = Fv for a Moving Vehicle?

A car travels along a level road at \(20\ \text{m s}^{-1}\). Its driven wheels provide a forward tractive force of \(700\ \text{N}\), while air resistance and rolling resistance combine to produce a backward force of \(700\ \text{N}\).

The car travels at constant speed. Has its engine stopped supplying power?

No. The tractive force supplies \(14\,000\ \text{W}\) of mechanical power at the wheels. Resistance removes mechanical energy at exactly the same rate. The net power is zero, but the engine is still doing work.

This is the key to using \(P=Fv\) for a moving vehicle: you must identify which force you are calculating the power of.

You can use

\[
\boxed{P=Fv}
\]

when the force acts parallel to the vehicle’s instantaneous direction of motion, with \(F\) representing the magnitude of that force and \(v\) representing the vehicle’s speed at the same instant.

For a vehicle, this commonly calculates the mechanical power delivered at the wheels by the tractive force. It does not automatically give the engine’s shaft power, fuel input power, or the net power changing the vehicle’s kinetic energy.

01When is \(P=Fv\) valid for a vehicle?

Start by looking at the direction of the force.

A moving vehicle experiences several forces:

• Tractive force from the road acting on the driven tyres, forwards.
• Air resistance acting backwards.
• Rolling resistance acting backwards.
• Gravity, acting vertically downwards.

On a level road, the forward tractive force and backward resistive forces are parallel or antiparallel to the vehicle’s velocity.

The general equation for mechanical power supplied by a force is

\[
P=Fv\cos\theta
\]

where:

• \(P\) is mechanical power in watts (\(\text{W}\)).
• \(F\) is the magnitude of the force in newtons (\(\text{N}\)).
• \(v\) is speed in metres per second (\(\text{m s}^{-1}\)).
• \(\theta\) is the angle between the force and the velocity.

For a force acting directly forwards, \(\theta=0^\circ\).

Since \(\cos 0^\circ=1\),

\[
P=Fv
\]

For a force acting directly backwards, \(\theta=180^\circ\).

Since \(\cos 180^\circ=-1\),

\[
P=-Fv
\]

The negative sign means the force removes mechanical energy from the vehicle.

This gives us three useful expressions:

Notice that these expressions calculate three different quantities.

If the force is not parallel to the velocity, you must retain the cosine term. You cannot simply multiply the full force magnitude by speed.

For a more detailed explanation of the general equation, see Mechanical Power with \(P=Fv\cos\theta\) for HSC Physics.

02The experiment: measuring power in a moving vehicle

Suppose we want to measure how much mechanical power a vehicle delivers through its wheels.

A tempting first approach is to measure the vehicle’s mass and acceleration, then calculate its net force using Newton’s second law.

We could then calculate:

\[
P_{\text{net}}=F_{\text{net}}v
\]

But would this tell us how much power the engine supplies?

Not necessarily.

Imagine measuring a car travelling at a constant \(20\ \text{m s}^{-1}\). Its acceleration is zero, so

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

The calculated net power is also zero.

Yet the engine is still supplying power to overcome air resistance and rolling resistance.

The original measurement design has a problem: it measures the rate at which the vehicle’s kinetic energy changes, rather than isolating the power delivered by the engine.

We need a better experimental arrangement.

Using a rolling-road dynamometer

A rolling-road dynamometer allows a vehicle to drive its wheels while remaining stationary relative to the room.

The basic arrangement works as follows:

1. The vehicle’s driven tyres rest on a pair of large rollers.
2. As the engine turns the wheels, the tyres rotate the rollers.
3. The rollers are connected to a braking mechanism that resists their rotation.
4. A load cell measures the force acting on a reaction arm attached to the braking mechanism.
5. An encoder measures the rollers’ angular speed.

The braking mechanism absorbs the mechanical energy transferred through the wheels.

The load cell measures a force, but the braking mechanism absorbs energy through rotational motion.

We therefore first calculate the braking torque:

\[
\tau=F_sL
\]

where:

• \(\tau\) is torque in newton metres (\(\text{N m}\)).
• \(F_s\) is the force measured by the load cell in newtons.
• \(L\) is the perpendicular distance from the shaft to the load cell’s line of action, in metres.

The power absorbed by the rotating rollers is

\[
P=\tau\omega
\]

where \(\omega\) is angular velocity in radians per second.

But we can connect this directly to \(P=Fv\).

For a roller with radius \(r\), its surface speed is

\[
v=\omega r
\]

The equivalent tangential force at the roller surface is

\[
F=\frac{\tau}{r}
\]

Substituting both expressions:

\[
\begin{aligned}
P&=Fv\\
&=\frac{\tau}{r}\times\omega r\\
&=\tau\omega
\end{aligned}
\]

The two methods give the same power.

This is why \(P=Fv\) is useful for vehicle testing: it connects a longitudinal force to the speed at which mechanical energy is transferred.

What should the measurements show?

Consider an idealised dynamometer with:

• Roller radius \(r=0.20\ \text{m}\).
• Brake reaction arm length \(L=0.50\ \text{m}\).
• Load cell reading \(F_s=120\ \text{N}\).

These are illustrative values, not measurements from an actual vehicle.

First, calculate the braking torque:

\[
\tau=F_sL
\]

\[
\tau=120\times0.50=60\ \text{N m}
\]

The equivalent tangential force at the roller surface is

\[
F=\frac{60}{0.20}=300\ \text{N}
\]

Now consider two different roller speeds.

For Run A:

\[
P=Fv=300\times8
\]

\[
P=2400\ \text{W}=2.4\ \text{kW}
\]

For Run B:

\[
P=300\times12
\]

\[
P=3600\ \text{W}=3.6\ \text{kW}
\]

The force remained unchanged, but power increased by 50% because the roller surface speed increased by 50%.

The observations agree with the relationship \(P=Fv\).

There is an important experimental limitation. A real dynamometer must account for losses in the rollers, bearings, tyres, and other components. The simplified calculation assumes negligible losses or that appropriate corrections have been applied.

It also measures mechanical power delivered at the wheels, not the chemical energy supplied by the fuel.

03Why you must not automatically use the net force

This is one of the most important distinctions in vehicle power questions.

Consider a car travelling at constant speed along a level road.

Its tractive force is \(900\ \text{N}\) forwards, while its total resistive force is \(900\ \text{N}\) backwards.

The net force is

\[
F_{\text{net}}=900-900=0
\]

Therefore,

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

But the power supplied by the tractive force is not zero.

If the car travels at \(20\ \text{m s}^{-1}\),

\[
P_{\text{drive}}=900\times20
\]

\[
P_{\text{drive}}=18\,000\ \text{W}
\]

or \(18\ \text{kW}\).

The resistive forces remove the same amount of mechanical power:

\[
P_{\text{res}}=-900\times20
\]

\[
P_{\text{res}}=-18\ \text{kW}
\]

The complete power balance is

\[
P_{\text{net}}=18-18=0\ \text{kW}
\]

The car’s kinetic energy stays constant because the energy supplied by the engine is transferred into other forms, mainly thermal energy in the tyres, road, and surrounding air.

The correct force depends on what the question asks:

• Engine or wheel power: use the tractive force.
• Power dissipated by resistance: use the resistive force, with a negative sign.
• Rate of increase of kinetic energy: use the net force.

For straight-line motion,

\[
P_{\text{net}}=F_{\text{net}}v
\]

Since \(F_{\text{net}}=ma\),

\[
P_{\text{net}}=mav
\]

This is the rate at which the vehicle’s kinetic energy changes. It is not automatically the engine’s power.

04Power is a rate, not an amount of energy

A common mistake is to treat power and energy as interchangeable.

They are related, but they measure different things.

Power measures how quickly energy is transferred.

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

where:

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

One watt means one joule of energy transferred every second:

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

The equation \(P=Fv\) gives the instantaneous mechanical power associated with a force.

It does not directly give the total energy transferred over an interval.

If power remains constant, you can calculate energy using

\[
E=P\Delta t
\]

For example, a vehicle delivering \(12\ \text{kW}\) of constant mechanical power for \(30\ \text{s}\) transfers

\[
E=12\,000\times30
\]

\[
E=360\,000\ \text{J}
\]

or \(360\ \text{kJ}\).

What happens if the vehicle speeds up while the tractive force stays constant?

Its power increases.

Suppose the tractive force is \(600\ \text{N}\).

At \(10\ \text{m s}^{-1}\):

\[
P=600\times10=6.0\ \text{kW}
\]

At \(20\ \text{m s}^{-1}\):

\[
P=600\times20=12.0\ \text{kW}
\]

The force is identical, but the faster vehicle transfers mechanical energy at twice the rate.

If speed or force changes during a question, instantaneous power changes too. You cannot automatically multiply the initial power by the entire time interval.

Instead, calculate the work done or use the average power over that interval.

05Which power does the engine actually produce?

There is another distinction worth understanding before using vehicle power equations in efficiency questions.

A vehicle has several stages of energy transfer.

The fuel supplies chemical energy to the engine. The engine converts some of this energy into mechanical energy at its output shaft. The transmission then transfers mechanical energy to the wheels.

These stages do not have identical power outputs.

We can distinguish three quantities:

The efficiency of a device is defined as

\[
\eta=\frac{P_{\text{useful output}}}
{P_{\text{input}}}\times100\%
\]

The important detail is that you must identify the input and output of the particular device or system being considered.

For a drivetrain:

\[
\eta_{\text{drivetrain}}
=
\frac{P_{\text{wheel}}}
{P_{\text{shaft}}}
\times100\%
\]

For the engine:

\[
\eta_{\text{engine}}
=
\frac{P_{\text{shaft}}}
{P_{\text{fuel}}}
\times100\%
\]

For the complete fuel-to-wheel system:

\[
\eta_{\text{overall}}
=
\frac{P_{\text{wheel}}}
{P_{\text{fuel}}}
\times100\%
\]

If the stages are in sequence, their efficiencies multiply when expressed as decimals.

This means a vehicle’s wheel power is generally lower than its engine shaft power, which is itself lower than its fuel input power.

Do not use \(P=Fv\) to calculate fuel input power unless you have also accounted for the relevant efficiency.

For more practice identifying useful output power and input power, see How to Find Useful Output Power from Efficiency and Input Power.

06Worked examples

07A quick decision process for vehicle power questions

Before substituting into \(P=Fv\), work through these checks.

1. What power is the question asking for?

Is it asking for power supplied by the engine, power delivered at the wheels, power dissipated by resistance, or net power?

2. Which force belongs in the equation?

Choose the tractive force for wheel power. Choose the net force for the rate of change of kinetic energy.

3. Is the force parallel to the velocity?

If yes, use \(P=Fv\). If not, use \(P=Fv\cos\theta\).

4. Are force and speed measured at the same instant?

For instantaneous power, they must correspond to the same instant.

5. Are you calculating power or energy?

Power is the rate of energy transfer. To calculate energy over a time interval, use work done or average power multiplied by time.

6. Is efficiency involved?

Identify the input and useful output before applying the efficiency equation.

There are also two useful boundary cases:

• A vehicle at rest has zero translational mechanical power from a force because \(v=0\), even if the engine is producing torque and consuming fuel.
• A vehicle braking while moving has negative mechanical power associated with the braking force. In regenerative braking, some of that mechanical energy can be converted into electrical energy.

08Questions and solutions

The following questions progress from straightforward wheel-power calculations to situations where force balance, changing speed, and efficiency must be considered together.

Question 1

A car travels along a level road at a constant speed of \(15\ \text{m s}^{-1}\). Its wheels provide a forward tractive force of \(500\ \text{N}\).

Calculate:

a) The mechanical power delivered at the wheels.

b) The mechanical energy transferred by the tractive force over \(120\ \text{s}\), assuming the speed and tractive force remain constant.

Solution 1

The mechanical power delivered at the wheels is \(7.5\ \text{kW}\), and the energy transferred over \(120\ \text{s}\) is \(900\ \text{kJ}\).

Part a: Calculate power.

The tractive force is parallel to the velocity, so

\[
P=Fv
\]

Substitute:

\[
P=500\times15
\]

\[
P=7500\ \text{W}
\]

Therefore,

\[
P=7.5\ \text{kW}
\]

Part b: Calculate energy.

Since power is constant,

\[
E=P\Delta t
\]

\[
E=7500\times120
\]

\[
E=900\,000\ \text{J}
\]

Therefore,

\[
E=900\ \text{kJ}
\]

The vehicle receives \(900\ \text{kJ}\) of mechanical energy from its tractive force. Because it travels at constant speed, this energy is transferred into other forms through resistive forces rather than increasing its kinetic energy.

Question 2

A \(1400\ \text{kg}\) car travels along a level road at an instantaneous speed of \(20\ \text{m s}^{-1}\). It is accelerating forwards at \(0.75\ \text{m s}^{-2}\).

The combined air resistance and rolling resistance is \(650\ \text{N}\), acting backwards.

Calculate:

a) The net force acting on the car.

b) The forward tractive force.

c) The mechanical power supplied by the tractive force.

d) The net power acting on the car.

Explain why the answers to parts (c) and (d) are different.

Solution 2

The tractive force is \(1700\ \text{N}\), supplying \(34.0\ \text{kW}\) at the wheels. The net power is only \(21.0\ \text{kW}\).

Part a: Calculate the net force.

Using Newton’s second law:

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

\[
F_{\text{net}}=1400\times0.75
\]

\[
F_{\text{net}}=1050\ \text{N}
\]

The net force is \(1050\ \text{N}\) forwards.

Part b: Calculate the tractive force.

Take forwards as positive.

\[
F_{\text{drive}}-650=1050
\]

Therefore,

\[
F_{\text{drive}}=1700\ \text{N}
\]

Part c: Calculate tractive power.

The tractive force acts in the direction of motion.

\[
P_{\text{drive}}=F_{\text{drive}}v
\]

\[
P_{\text{drive}}=1700\times20
\]

\[
P_{\text{drive}}=34\,000\ \text{W}
\]

Therefore,

\[
P_{\text{drive}}=34.0\ \text{kW}
\]

Part d: Calculate net power.

Use the net force:

\[
P_{\text{net}}=F_{\text{net}}v
\]

\[
P_{\text{net}}=1050\times20
\]

\[
P_{\text{net}}=21\,000\ \text{W}
\]

Therefore,

\[
P_{\text{net}}=21.0\ \text{kW}
\]

The difference is the power transferred out of the vehicle’s mechanical motion by resistance.

\[
P_{\text{res}}=-650\times20
\]

\[
P_{\text{res}}=-13.0\ \text{kW}
\]

The power balance is

\[
34.0-13.0=21.0\ \text{kW}
\]

The tempting mistake is to use \(F_{\text{net}}v\) as the engine power. That only calculates the net rate of change of kinetic energy. The engine must also supply the \(13.0\ \text{kW}\) needed to overcome resistance.

Question 3

A \(1250\ \text{kg}\) car climbs a straight hill at a constant speed of \(16\ \text{m s}^{-1}\).

The incline has \(\sin\theta=0.080\), and the total air resistance and rolling resistance is \(420\ \text{N}\), acting down the slope.

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

Calculate:

a) The component of gravitational force acting down the slope.

b) The tractive force required to maintain constant speed.

c) The mechanical power delivered at the wheels.

d) The rate at which gravitational potential energy increases.

e) The rate at which energy is dissipated by resistive forces.

Solution 3

The required tractive force is \(1400\ \text{N}\), and the wheel power is \(22.4\ \text{kW}\). Of this power, \(15.68\ \text{kW}\) increases gravitational potential energy, while \(6.72\ \text{kW}\) is dissipated by resistance.

Part a: Calculate the gravitational force component.

The component of weight parallel to the incline is

\[
F_g=mg\sin\theta
\]

Substitute:

\[
F_g=1250\times9.8\times0.080
\]

\[
F_g=980\ \text{N}
\]

Part b: Calculate the tractive force.

The car travels at constant speed, so its acceleration is zero.

Therefore, the net force along the slope is zero.

\[
F_{\text{drive}}-980-420=0
\]

\[
F_{\text{drive}}=1400\ \text{N}
\]

Part c: Calculate wheel power.

The tractive force acts uphill, in the direction of motion.

\[
P_{\text{wheel}}=F_{\text{drive}}v
\]

\[
P_{\text{wheel}}=1400\times16
\]

\[
P_{\text{wheel}}=22\,400\ \text{W}
\]

Therefore,

\[
P_{\text{wheel}}=22.4\ \text{kW}
\]

Part d: Calculate the rate of increase of gravitational potential energy.

The rate of gravitational potential energy gain is

\[
P_{\text{GPE}}=mg\sin\theta\times v
\]

\[
P_{\text{GPE}}=980\times16
\]

\[
P_{\text{GPE}}=15\,680\ \text{W}
\]

Therefore,

\[
P_{\text{GPE}}=15.68\ \text{kW}
\]

Part e: Calculate the rate of energy dissipation.

The resistive force acts opposite to the motion.

Its power is

\[
P_{\text{res}}=-F_{\text{res}}v
\]

\[
P_{\text{res}}=-420\times16
\]

\[
P_{\text{res}}=-6720\ \text{W}
\]

Therefore, the rate of energy dissipation is \(6.72\ \text{kW}\).

Check the power balance:

\[
22.4=15.68+6.72
\]

The engine supplies \(22.4\ \text{kW}\). All of this power is accounted for by the increase in gravitational potential energy and energy dissipated by resistance.

The net power is zero because the car’s kinetic energy is unchanged. Its gravitational potential energy is still increasing.

Question 4

A \(1200\ \text{kg}\) vehicle travels along a level test track at an initial speed of \(24\ \text{m s}^{-1}\).

Its brakes exert a constant force of \(3600\ \text{N}\) opposite to the direction of motion for \(4.0\ \text{s}\).

Ignore all other horizontal forces.

Calculate:

a) The instantaneous braking power at the beginning of the braking interval.

b) The vehicle’s speed after \(4.0\ \text{s}\).

c) The energy removed from the vehicle during the interval.

d) The average braking power over the \(4.0\ \text{s}\).

Explain why multiplying the initial braking power by \(4.0\ \text{s}\) gives an incorrect energy value.

Solution 4

The initial braking power is \(-86.4\ \text{kW}\), the final speed is \(12.0\ \text{m s}^{-1}\), and the brakes remove \(259.2\ \text{kJ}\) of mechanical energy. The average braking power is \(-64.8\ \text{kW}\).

Take forwards as positive.

Part a: Calculate initial braking power.

The braking force acts opposite to the velocity, so the power is negative.

\[
P=-Fv
\]

\[
P=-3600\times24
\]

\[
P=-86\,400\ \text{W}
\]

Therefore,

\[
P=-86.4\ \text{kW}
\]

The negative sign indicates that the brakes remove mechanical energy from the vehicle.

Part b: Calculate the final speed.

Using Newton’s second law:

\[
a=\frac{F}{m}
\]

\[
a=\frac{-3600}{1200}
\]

\[
a=-3.0\ \text{m s}^{-2}
\]

Now use the constant-acceleration equation:

\[
v=u+at
\]

Substitute:

\[
v=24+(-3.0)(4.0)
\]

\[
v=12.0\ \text{m s}^{-1}
\]

Part c: Calculate energy removed.

First, calculate the displacement.

\[
s=ut+\frac12at^2
\]

\[
s=(24)(4.0)+\frac12(-3.0)(4.0)^2
\]

\[
s=96-24
\]

\[
s=72\ \text{m}
\]

The work done by the braking force is

\[
W=Fs
\]

Since the force opposes displacement,

\[
W=-3600\times72
\]

\[
W=-259\,200\ \text{J}
\]

Therefore, the brakes remove

\[
259.2\ \text{kJ}
\]

of mechanical energy.

Part d: Calculate average braking power.

Average power is total work divided by time.

\[
P_{\text{avg}}=\frac{W}{\Delta t}
\]

\[
P_{\text{avg}}=\frac{-259\,200}{4.0}
\]

\[
P_{\text{avg}}=-64\,800\ \text{W}
\]

Therefore,

\[
P_{\text{avg}}=-64.8\ \text{kW}
\]

The tempting approach is to multiply the initial power by the full time:

\[
86.4\times4.0=345.6\ \text{kJ}
\]

But this assumes the braking power remains constant.

It does not. The braking force is constant, but the vehicle’s speed decreases.

At the end of the interval, the instantaneous braking power is

\[
P=-3600\times12
\]

\[
P=-43.2\ \text{kW}
\]

The braking power decreases in magnitude from \(86.4\ \text{kW}\) to \(43.2\ \text{kW}\).

This is why the average braking power is \(64.8\ \text{kW}\), rather than the initial \(86.4\ \text{kW}\).

Question 5

An electric vehicle descends a straight road at a constant speed of \(20\ \text{m s}^{-1}\).

The vehicle and its occupants have a combined mass of \(1600\ \text{kg}\). The road has an incline such that

\[
\sin\theta=0.050
\]

Air resistance and rolling resistance combine to produce an uphill force of \(500\ \text{N}\).

The vehicle’s regenerative braking system is the only other uphill force. It converts 70% of the mechanical power absorbed at the wheels into electrical power.

Assume:

• The vehicle travels at constant speed.
• The engine is switched off.
• No friction brakes are being used.
• All the stated forces act parallel to the slope.
• \(g=9.8\ \text{m s}^{-2}\).

Calculate:

a) The gravitational force component acting downhill.

b) The magnitude of the regenerative braking force.

c) The mechanical power absorbed by the regenerative braking system.

d) The electrical power produced.

e) The percentage of the gravitational power that becomes electrical power.

Explain why it would be incorrect to multiply the total gravitational power directly by 70%.

Solution 5

The regenerative braking force is \(284\ \text{N}\), the electrical output power is approximately \(3.98\ \text{kW}\), and approximately 25.4% of the gravitational power becomes electrical power.

The important distinction is that the 70% efficiency applies only to the mechanical power entering the regenerative braking system, not to all the gravitational power supplied to the vehicle.

Part a: Calculate the gravitational force component.

Take downhill as the positive direction.

The component of gravity parallel to the slope is

\[
F_g=mg\sin\theta
\]

Substitute:

\[
F_g=1600\times9.8\times0.050
\]

\[
F_g=784\ \text{N}
\]

Gravity provides a downhill force of \(784\ \text{N}\).

Part b: Calculate the regenerative braking force.

The vehicle moves at constant speed, so its acceleration is zero.

Therefore, the net force along the slope must be zero.

Taking downhill as positive:

\[
F_g-F_{\text{res}}-F_{\text{regen}}=0
\]

Substitute:

\[
784-500-F_{\text{regen}}=0
\]

Rearrange:

\[
F_{\text{regen}}=784-500
\]

\[
F_{\text{regen}}=284\ \text{N}
\]

The regenerative braking system provides an uphill force of \(284\ \text{N}\).

Part c: Calculate mechanical power absorbed by regenerative braking.

The regenerative braking force acts opposite to the vehicle’s velocity.

Therefore, its power acting on the vehicle is negative:

\[
P_{\text{regen}}=-F_{\text{regen}}v
\]

\[
P_{\text{regen}}=-284\times20
\]

\[
P_{\text{regen}}=-5680\ \text{W}
\]

The regenerative system absorbs \(5.68\ \text{kW}\) of mechanical power.

Part d: Calculate electrical output power.

The regenerative braking system has an efficiency of 70%.

Convert this to a decimal:

\[
\eta=0.70
\]

Efficiency is useful output power divided by input power.

Therefore,

\[
P_{\text{electrical}}=0.70\times5.68
\]

\[
P_{\text{electrical}}=3.976\ \text{kW}
\]

The electrical output power is approximately

\[
P_{\text{electrical}}=3.98\ \text{kW}
\]

Part e: Calculate the percentage of gravitational power converted into electricity.

First, calculate the gravitational power supplied to the vehicle.

\[
P_g=F_gv
\]

\[
P_g=784\times20
\]

\[
P_g=15\,680\ \text{W}
\]

Therefore,

\[
P_g=15.68\ \text{kW}
\]

Now compare the electrical output power with the gravitational input power.

\[
\eta_{\text{overall}}
=
\frac{3.976}{15.68}\times100\%
\]

\[
\eta_{\text{overall}}=25.4\%
\]

Therefore, approximately 25.4% of the gravitational power becomes electrical power.

Why isn’t the answer 70%?

The 70% efficiency applies only to the mechanical power absorbed by the regenerative braking system.

The vehicle has two separate mechanisms removing mechanical energy:

• Air resistance and rolling resistance remove \(10.0\ \text{kW}\).
• Regenerative braking absorbs \(5.68\ \text{kW}\).

Only the second contribution enters the regenerative braking system.

The full power balance is

\[
15.68=10.00+5.68
\]

All values are in kilowatts.

The regenerative system then converts 70% of its \(5.68\ \text{kW}\) input into electricity.

The remaining 30% is lost through inefficiencies in the conversion process.

This question exposes two important mistakes.

First, zero net power does not mean that no energy is being transferred. Gravity supplies power continuously, even though the vehicle’s speed stays constant.

Second, efficiency must be calculated using the input and output of the correct system. Applying 70% to the entire gravitational power ignores the energy already dissipated by air resistance and rolling resistance.

08What to remember when using \(P=Fv\)

For a moving vehicle, the equation is straightforward once you identify the force.

\[
\boxed{P_{\text{wheel}}=F_{\text{drive}}v}
\]

Use it when calculating the mechanical power delivered at the wheels by a tractive force parallel to the vehicle’s motion.

Remember the distinctions:

• Tractive force gives positive mechanical power when the vehicle moves forwards.
• Resistive and braking forces give negative mechanical power.
• Net force gives net power, which describes the rate of change of kinetic energy.
• Power measures energy transferred per second, not the total energy transferred.
• Wheel power, engine shaft power, and fuel input power are different quantities.
• Efficiency is always useful output divided by input, using a clearly defined system boundary.

The next useful step is to apply the same reasoning to vehicles that change direction, travel around curves, or use regenerative braking. In each case, start with the force’s direction relative to the instantaneous velocity before choosing the power equation.