GRE Physics Companion: Power

GRE Physics Companion: Power

Power problems are usually short once the correct form is recognized.

The central relations are

(1)

and

(2)

For the net force,

(3)

1 High-value GRE facts

1. Average power is work divided by elapsed time.
2. instantaneous power is dW∕dt.
3. For a force acting on a particle,

   (4)

4. A force perpendicular to velocity delivers zero instantaneous power.
5. Negative power means that the force removes mechanical energy from the particle.
6. The area under a P versus t graph is work.
7. A kilowatt-hour is energy, not power.
8. For constant net power,

   (5)

9. For collinear motion at constant power,

   (6)

10. If drag satisfies F_D ∝ v², then the power required to overcome drag satisfies P _D ∝ v³.

Part I: Original GRE-style problems

Problem 1: average power

A machine does 18 kJ of work in 6.0 s. Its average power is

1. 0.33 kW
2. 3.0 kW
3. 6.0 kW
4. 18 kW
5. 108 kW

Problem 2: force and velocity

A 100 N force acts on a particle moving at 5.0 m∕s. The angle between force and velocity is 60^∘. The power delivered by the force is

1. 0 W
2. 100 W
3. 250 W
4. 500 W
5. 1000 W

Problem 3: perpendicular force

A particle moves in uniform circular motion under a central force. The instantaneous power delivered by the central force is

1. negative
2. zero
3. proportional to v
4. proportional to v²
5. impossible to determine

Problem 4: braking power

A braking force of magnitude 4000 N acts opposite a CAR moving at 15 m∕s. The instantaneous power of the braking force is

1. −60 kW
2. −15 kW
3. 0
4. 15 kW
5. 60 kW

Problem 5: lifting

A 60 kg person climbs a vertical height of 5.0 m in 4.0 s. Taking g = 9.8 m∕s², the average mechanical power against gravity is closest to

1. 0.37 kW
2. 0.74 kW
3. 1.5 kW
4. 2.9 kW
5. 7.4 kW

Problem 6: kilowatt-hour

One kilowatt-hour is equal to

1. 3.6 × 10³ J
2. 3.6 × 10⁴ J
3. 3.6 × 10⁵ J
4. 3.6 × 10⁶ J
5. 3.6 × 10⁹ J

Problem 7: constant power

A motor delivers constant mechanical power P to a vehicle moving in a straight line. Neglecting all resistive forces, the drive force at speed v is

1. Pv
2. P∕v
3. P∕v²
4. v∕P
5. independent of v

Problem 8: quadratic drag

A vehicle moves at steady speed through a regime in which the drag force is proportional to v². If its speed is doubled, the mechanical power required to overcome drag is multiplied by

1. 2
2. 4
3. 6
4. 8
5. 16

Problem 9: efficiency

A motor has efficiency η = 0.80 and must deliver 4.0 kW of useful mechanical output. The required input power is

1. 3.2 kW
2. 4.0 kW
3. 4.8 kW
4. 5.0 kW
5. 8.0 kW

Problem 10: power-time graph

The power delivered to a system increases linearly from 0 to 100 W during a 4.0 s interval. The work done during the interval is

1. 50 J
2. 100 J
3. 200 J
4. 400 J
5. 800 J

Problem 11: kinetic energy rate

A 2.0 kg particle moves in one dimension at 6.0 m∕s and has acceleration 3.0 m∕s² in the same direction. The net instantaneous power is

1. 12 W
2. 18 W
3. 24 W
4. 36 W
5. 72 W

Problem 12: constant net power from rest

A 4.0 kg particle starts from rest and receives constant net power 200 W for 5.0 s. Its final speed is closest to

1. 5.0 m∕s
2. 11.2 m∕s
3. 15.8 m∕s
4. 22.4 m∕s
5. 50 m∕s

Part II: Complete worked solutions

Solution 1

Average power is

(7)

Thus

Pavg	=	(8)
	= 3000 W	(9)
	= 3.0 kW.	(10)

Answer: (B).

Solution 2

Use

(11)

Therefore

P	= (100)(5.0) cos 60∘	(12)
	= 250 W.	(13)

Answer: (C).

Solution 3

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

(14)

Therefore

(15)

Answer: (B).

Solution 4

The force is opposite the velocity, so

(16)

Thus

P	= Fv cos 180∘	(17)
	= −(4000)(15)	(18)
	= −60 000 W	(19)
	= −60 kW.	(20)

Answer: (A).

Solution 5

The work done against gravity is

W	= mgh	(21)
	= (60)(9.8)(5.0)	(22)
	= 2940 J.	(23)

Average power is

Pavg	=	(24)
	= 735 W	(25)
	= 0.735 kW.	(26)

Answer: (B).

Solution 6

Use

1 kWh	= (1000 J∕s)(3600 s)	(27)
	= 3.6 × 106 J.	(28)

Answer: (D).

Solution 7

For force parallel to velocity,

(29)

Therefore

(30)

Answer: (B).

Solution 8

If

(31)

then

(32)

Doubling speed gives

(33)

Answer: (D).

Solution 9

Efficiency is

(34)

Thus

(35)

Answer: (D).

Solution 10

Work is the area under the power-time graph. The graph is a triangle:

(36)

Answer: (C).

Solution 11

Net power can be written as

(37)

The acceleration and velocity are parallel, so

Pnet	= mav	(38)
	= (2.0)(3.0)(6.0)	(39)
	= 36 W.	(40)

Answer: (D).

Solution 12

Constant net power changes kinetic energy according to

(41)

Starting from rest,

Kf	= (200)(5.0)	(42)
	= 1000 J.	(43)

Then

(44)

For m = 4.0 kg,

2vf2	= 1000,	(45)
vf2	= 500,	(46)
vf	= 22.4 m∕s.	(47)

Answer: (D).

2 GRE checklist

Before calculating, identify what kind of power is being asked for.

1. If total work and elapsed time are given, use P_avg = ΔW∕Δt.
2. If force and velocity are given, use P = F ⋅ v.
3. If kinetic energy is changing, use P_net = dK∕dt.
4. If a power-versus-time graph is given, find the signed area.
5. If efficiency is involved, distinguish input power from useful output power.
6. If constant power drives collinear motion, remember F = P∕v.
7. If resistance is proportional to v², remember that required power scales as v³.
8. Check whether the sign of power should be positive, zero, or negative.

References

References