Power, steady speed & hills | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Average power and instantaneous power answer different questions

Definition

What is meant by power?

Model answer: The rate at which work is done or energy is transferred.

P_{\mathrm{avg}}=\frac{W}{t}

Symbols

$P_{\mathrm{avg}}$ = average power over an interval (W)
$W$ = work done in that interval (J)
$t$ = time taken (s)

P=Fv

Symbols

$P$ = instantaneous power (W)
$F$ = force in the direction of motion (N)
$v$ = instantaneous speed (m s⁻¹)

Instantaneous means “at that exact moment”. When the force acts along the motion, $W=Fd$ and $v=d/t$ , so $P=W/t=Fd/t=Fv$ .

$1\text{ W}=1\text{ J s}^{-1}$ .
$1\text{ kW}=1000\text{ W}$ ; convert before using $P=Fv$ .
If force is not along the motion, use its component along the motion. The syllabus does not require scalar-product notation.

Common mistake

Power is not work or energy. It is their rate. Two engines can do the same work but have different powers if they take different times.

At constant power, driving force changes with speed

F_{\mathrm{drive}}=\frac{P}{v}

Swipe left or right to see the whole diagram →

For a constant-power engine, driving force is P divided by v. Then use Newton’s law along the slope.

Constant-power hill lab

g = 10 m s⁻²

Engine power60 kWCurrent speed20 m s⁻¹Car mass1200 kgHill angle5 °Resistance300 N

1. Driving force = 60000 ÷ 20 = 3000 N

2. Weight component = 1045.87 N

3. Resultant = 3000 − 1345.87 = 1654.13 N

4. Acceleration = 1.38 m s⁻²

Motion state: speeding up uphill

At this speed, exactly 26.92 kW would give zero acceleration.

At a lower speed, the same power gives a larger driving force.
At a higher speed, the same power gives a smaller driving force.
Constant power therefore does not usually produce constant acceleration.

Examiner note

A current paper first used

P/v

to obtain the cyclist's pedalling force, then used Newton's law. For steady uphill motion it set the acceleration to zero while retaining resistance and the downslope component of weight. (9709/42/O/N/24 Q3)

Steady or maximum speed means zero resultant force, not zero force

For a car moving uphill at angle $\theta$ against resistance $R$ , take uphill as positive.

\frac{P}{v}-R-mg\sin\theta=ma

At steady or maximum speed, $a=0$ , so

\frac{P}{v}=R+mg\sin\theta

Here “maximum speed” means the limiting steady speed under the stated constant conditions: the driving force has fallen until it exactly balances the opposition.

Worked example

Resistance depends on speed

If resistance is $kv^2$ on a level road and a car moves at steady speed, then $P/v=kv^2$ , so $P=kv^3$ . Do not treat resistance as constant when its given model changes with speed.

(9709/42/M/J/24 Q4)

Key idea

Once the steady-speed equation is formed, the calculator equation solver is the fast normal method. Keep the positive speed that fits the model and substitute it back to check the force balance.

Your turnIndependent check [6 marks]

A 1000 kg car travels uphill where $\sin\theta=0.05$ . Its engine works at 80 kW, its current speed is 20 m s⁻¹ and the resistance is 500 N. Find the driving force, the instantaneous acceleration and the power required for steady motion at the same speed.

Show worked answer

F_{\mathrm{drive}}=\frac{80000}{20}

F_{\mathrm{drive}}=4000\text{ N}

mg\sin\theta=1000(10)(0.05)

mg\sin\theta=500\text{ N}

4000-500-500=1000a

a=3\text{ m s}^{-2}

P_{\mathrm{steady}}=(500+500)(20)

P_{\mathrm{steady}}=20000\text{ W}

P_{\mathrm{steady}}=20\text{ kW}

Answer

4000\text{ N}

3\text{ m s}^{-2}

20\text{ kW}

Average power and instantaneous power answer different questions

Definition

What is meant by power?

Model answer: The rate at which work is done or energy is transferred.

P_{\mathrm{avg}}=\frac{W}{t}

Symbols

$P_{\mathrm{avg}}$ = average power over an interval (W)
$W$ = work done in that interval (J)
$t$ = time taken (s)

P=Fv

Symbols

$P$ = instantaneous power (W)
$F$ = force in the direction of motion (N)
$v$ = instantaneous speed (m s⁻¹)

Instantaneous means “at that exact moment”. When the force acts along the motion, $W=Fd$ and $v=d/t$ , so $P=W/t=Fd/t=Fv$ .

$1\text{ W}=1\text{ J s}^{-1}$ .
$1\text{ kW}=1000\text{ W}$ ; convert before using $P=Fv$ .
If force is not along the motion, use its component along the motion. The syllabus does not require scalar-product notation.

Common mistake

Power is not work or energy. It is their rate. Two engines can do the same work but have different powers if they take different times.

At constant power, driving force changes with speed

F_{\mathrm{drive}}=\frac{P}{v}

Swipe left or right to see the whole diagram →

For a constant-power engine, driving force is P divided by v. Then use Newton’s law along the slope.

Constant-power hill lab

g = 10 m s⁻²

Engine power60 kWCurrent speed20 m s⁻¹Car mass1200 kgHill angle5 °Resistance300 N

1. Driving force = 60000 ÷ 20 = 3000 N

2. Weight component = 1045.87 N

3. Resultant = 3000 − 1345.87 = 1654.13 N

4. Acceleration = 1.38 m s⁻²

Motion state: speeding up uphill

At this speed, exactly 26.92 kW would give zero acceleration.

At a lower speed, the same power gives a larger driving force.
At a higher speed, the same power gives a smaller driving force.
Constant power therefore does not usually produce constant acceleration.

Examiner note

A current paper first used

P/v

Steady or maximum speed means zero resultant force, not zero force

For a car moving uphill at angle $\theta$ against resistance $R$ , take uphill as positive.

\frac{P}{v}-R-mg\sin\theta=ma

At steady or maximum speed, $a=0$ , so

\frac{P}{v}=R+mg\sin\theta

Here “maximum speed” means the limiting steady speed under the stated constant conditions: the driving force has fallen until it exactly balances the opposition.

Worked example

Resistance depends on speed

If resistance is $kv^2$ on a level road and a car moves at steady speed, then $P/v=kv^2$ , so $P=kv^3$ . Do not treat resistance as constant when its given model changes with speed.

(9709/42/M/J/24 Q4)

Key idea

Once the steady-speed equation is formed, the calculator equation solver is the fast normal method. Keep the positive speed that fits the model and substitute it back to check the force balance.

Your turnIndependent check [6 marks]

Show worked answer

F_{\mathrm{drive}}=\frac{80000}{20}

F_{\mathrm{drive}}=4000\text{ N}

mg\sin\theta=1000(10)(0.05)

mg\sin\theta=500\text{ N}

4000-500-500=1000a

a=3\text{ m s}^{-2}

P_{\mathrm{steady}}=(500+500)(20)

P_{\mathrm{steady}}=20000\text{ W}

P_{\mathrm{steady}}=20\text{ kW}

Answer

4000\text{ N}

3\text{ m s}^{-2}

20\text{ kW}