Power

Work measures energy transfer through a force acting over a displacement. Power measures how rapidly that work is done.

Two machines can perform the same amount of work while operating at very different power levels. A motor that lifts a load in two seconds transfers the same gravitational energy as a slower motor that lifts the same load through the same height in twenty seconds, but the first motor transfers that energy ten times faster.

The average power delivered during a time interval is

(1)

The instantaneous power is the limiting rate

(2)

For a force acting on a moving particle,

(3)

This dot-product form is one of the most useful relations in mechanics. It connects force, motion, work, and energy at a single instant.

1 Average power

Suppose a force does work ΔW during the interval from t₁ to t₂. The elapsed time is

(4)

The average power is

(5)

Average power says nothing about how uniformly the work was performed during the interval. A machine can deliver a large power for part of the interval and a smaller power during another part while having the same average.

If the work done over the interval is known,

(6)

This relation is exact for the average value over the stated interval.

2 Instantaneous power

To describe the rate of work at a particular instant, shrink the time interval:

(7)

Thus

(8)

If the power varies with time, the work done from t_i to t_f is

(9)

This relation is analogous to the variable-force work integral

(10)

On a graph of P versus t, the signed area under the curve equals work.

3 Derivation of P = F ⋅ v

Differential work is

(11)

Divide by dt:

(12)

Since

(13)

we obtain

(14)

If 𝜃 is the instantaneous angle between the force and velocity,

(15)

This is the instantaneous mechanical power delivered by the force to the particle.

4 The sign of power

The sign of P = F ⋅ v has a direct mechanical interpretation.

P > 0	the force is transferring energy to the particle,	(16)
P = 0	the force is doing no instantaneous work,	(17)
P < 0	the force is removing kinetic energy from the particle.	(18)

For a constant-mass particle,

(19)

Thus positive net power means kinetic energy is increasing, negative net power means kinetic energy is decreasing, and zero net power means the kinetic energy is instantaneously unchanged.

5 Power and kinetic energy

M03-02 established the differential work-energy relation

(20)

Therefore

(21)

For constant mass,

(22)

Differentiating,

	= m(v2)	(23)
	= mv	(24)

for one-dimensional motion in which v denotes the signed speed along the coordinate direction.

Hence

(25)

In vector form, the more general identity is

(26)

6 Units of power

The SI unit of power is the watt:

(27)

Since

(28)

we also have

(29)

Using base SI units,

(30)

Common multiples include

1 kW	= 103 W,	(31)
1 MW	= 106 W.	(32)

Mechanical horsepower is approximately

(33)

7 Power is not energy

A watt is a unit of power, not energy.

A kilowatt-hour is a unit of energy because it is power multiplied by time:

1 kWh	= (1000 W)(3600 s)	(34)
	= 3.6 × 106 J.	(35)

Thus

(36)

The distinction is fundamental:

(37)

when the power is constant, and more generally

(38)

8 Example 1: lifting a load

A motor lifts a 50 kg crate vertically upward through 8.0 m in 5.0 s at constant speed.

At constant speed, the upward motor force equals the Weight:

(39)

The work done by the motor is

W	= mgh	(40)
	= (50)(9.81)(8.0)	(41)
	= 3924 J.	(42)

The average power is

Pavg	=	(43)
	=	(44)
	= 785 W.	(45)

Thus

(46)

Because the lifting speed is constant, the instantaneous mechanical power is also constant.

9 Example 2: pulling at an angle

A force of magnitude 120 N pulls a cart moving at 4.0 m∕s. The force makes an angle of 30^∘ above the direction of motion.

The instantaneous power delivered by the force is

P	= Fv cos 𝜃	(47)
	= (120)(4.0) cos 30∘	(48)
	= 416 W.	(49)

Only the component of force parallel to the velocity transfers mechanical energy to the cart at that instant.

10 Example 3: braking power

A braking force of magnitude 5000 N acts opposite the velocity of a CAR moving at 20 m∕s.

The angle between force and velocity is 180^∘, so

P	= Fv cos 180∘	(50)
	= −(5000)(20)	(51)
	= −1.0 × 105 W.	(52)

Therefore

(53)

The negative sign means the braking force is removing mechanical energy from the car at a rate of 100 kJ∕s at that instant.

11 Constant power and changing speed

Suppose a constant net power P₀ is delivered to a particle of constant mass m moving in one dimension.

Since

(54)

integration gives

(55)

Using

(56)

we obtain

(57)

Therefore

(58)

and

(59)

For motion in the same direction as the force,

(60)

so

(61)

Thus a system delivering constant mechanical power produces less force as speed increases.

12 Power required to overcome resistance

Suppose a vehicle moves at steady speed v while a resistive force F_R(v) acts opposite the motion.

Steady speed requires the driving force to balance the resistance:

(62)

The mechanical power required at the point of application is

(63)

This produces important speed scaling.

If resistance is approximately constant,

(64)

then

(65)

If drag is approximately proportional to v²,

(66)

then

(67)

Therefore doubling speed under quadratic drag multiplies the required drag power by eight.

13 Efficiency and input power

Real machines are not perfectly efficient. If a machine receives input power P_in and delivers useful output power P_out, define the efficiency

(68)

For an efficiency between zero and one,

(69)

Hence

(70)

and

(71)

For example, if a motor must deliver

(72)

with

(73)

then the required input power is

(74)

The missing 0.5 kW is transferred into other forms such as thermal energy, sound, or internal mechanical losses.

14 Power delivered by several forces

If several forces act on a particle,

(75)

The net power is

Pnet	= Fnet ⋅ v	(76)
	= ⋅ v	(77)
	= ∑ iFi ⋅ v.	(78)

Therefore

(79)

Individual forces may deliver positive, negative, or zero power at the same instant.

For example, a vehicle moving uphill at steady speed may receive positive power from its engine while gravity and drag each contribute negative power. The net power is zero because the kinetic energy remains constant.

15 Power in uniform circular motion

In uniform circular motion, the centripetal force is perpendicular to the velocity:

(80)

Thus

(81)

The force is nonzero and the acceleration is nonzero, but the force does not change the kinetic energy because it changes only the direction of velocity.

This is another example of why power must be interpreted through a dot product rather than merely as force multiplied by speed.

16 Power versus force

Large force does not necessarily mean large power.

From

(82)

power depends on force, speed, and direction.

A large force acting on an object that is nearly stationary can correspond to small mechanical power.

A smaller force acting on an object moving rapidly in the force direction can correspond to much greater power.

This distinction is important in engines, motors, actuators, pumps, and biological systems.

17 Common mistakes

1. Confusing power with energy.
2. Treating the kilowatt-hour as a unit of power.
3. Using P = Fv when the force is not parallel to the velocity.
4. Forgetting the sign of F ⋅ v.
5. Assuming that zero power means zero force.
6. Using average power when the problem asks for instantaneous power.
7. Assuming constant power implies constant force.
8. Forgetting that at constant power the available force decreases as 1∕v for collinear motion.
9. Ignoring efficiency when converting required output power into input power.
10. Forgetting that area under a power-versus-time graph represents work or energy transfer.

18 Practice exercises

1. A machine does 12 kJ of work in 3.0 s. Find its average power.
2. A 200 N horizontal force pushes an object moving at 5.0 m∕s in the force direction. Find the instantaneous power.
3. A 150 N force acts at 60^∘ to a velocity of 8.0 m∕s. Find the power delivered by the force.
4. A braking force of 3000 N acts opposite a car moving at 25 m∕s. Find the instantaneous power of the braking force.
5. A 75 kg person climbs vertically through 12 m in 18 s. Estimate the average mechanical power against gravity.
6. Show that 1 kWh = 3.6 × 10⁶ J.
7. A motor delivers constant net power P₀ to a mass m initially at rest. Derive v(t) = .
8. A vehicle experiences drag F_D = cv². Show that the power required to balance the drag is proportional to v³.
9. An engine delivers 40 kW of mechanical power while the vehicle moves at 20 m∕s. If the driving force is parallel to the velocity, find the force.
10. A motor must provide 5.0 kW of useful mechanical output at 75% efficiency. Find the input power.
11. A particle moves in uniform circular motion under a central force. Explain why the force can be large while its mechanical power is zero.
12. A power history is P(t) = P₀t∕T for 0 ≤ t ≤ T. Integrate the power to find the work done during the interval.

19 Summary

Average power is

(83)

Instantaneous power is

(84)

For a force acting on a moving particle,

(85)

For the net force,

(86)

Work is recovered from power by integration:

(87)

Under constant net power,

(88)

For collinear force and velocity,

(89)

The next article, M03-05, develops conservative forces and potential energy.

References

References