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AP* PHYSICS B

Work and Energy

Teacher Packet

AP* is a trademark of the College Entrance Examination Board. The College Entrance Examination Board was not involved in the production of this material.

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Work & Energy

Objective To review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on work and energy. Standards Work and energy are addressed in the topic outline of the College Board AP* Physics Course Description Guide as described below. AP Physics Exam Connections Topics relating to work and energy are tested every year on the multiple choice and in most years on the free response portion of the exam. The list below identifies free response questions that have been previously asked over work and energy, often in conjunction with momentum. These questions are available from the College Board and can be downloaded free of charge from AP Central. http://apcentral.collegeboard.com. Free Response Questions

2005 Question 2 2006 Form B Question 2 2002 Question 1 2005 Form B Question 2 2001 Question 2 2002 Form B Question 1 1999 Question 1 (ex a)

I. Newtonian Mechanics C. Work, energy, and power

1. Work and work-energy theorem 2. Conservative forces and potential energy 3. Conservation of energy 4. Power

F. Oscillations and Gravitation 1. Simple harmonic motion (dynamics and energy relationships) 2. Mass on a spring 3. Pendulum and other oscillations

http://apcentral.collegeboard.com/



Work, Energy, & Power

What I Absolutely Have to Know to Survive the AP* Exam

Work is the defined quantity from which the entire theory of energy begins. It is the scalar product of the force acting on an object and the displacement through which it acts. Power is the rate at which work is done. Objects are said to have energy if they have the ability to do work, either due to the fact that they are moving (kinetic energy) or due to their position in a force field (potential energy). Conservative forces are forces that do work that is path independent. Conserved quantities are quantities that do not change with time. If the mechanical energy of a system is conserved, then the sum of the system’s kinetic and potential energies at any given time will always add to the same number. Work done by a nonconservative force generally cannot be recovered as usable energy.

Key Formulas and Relationships Work: cosW θ= ⋅ =F d Fd unit = Joule = 1 N·m

Power: WPt

= unit = Watt = 1 Joule/second P = ⋅F v

Kinetic Energy: 212

K m= v unit = Joule

Gravitational Potential Energy: unit = Joule gU mg= h

Potential Energy stored in spring: 212sU k= x unit = Joule

Conservation of Energy: ' 'K U K U+ = +



Important Concepts

• Work is done by a force parallel to the displacement of the object: perpendicular forces (centripetal forces for example) do no work.

• If the force is at an angle to the displacement, you must resolve it into components. • Work is a scalar quantity but can be negative • Work-Energy Theorem: Work causes a change in energy: it is the method by which

energy is transferred (W = ΔE).

θ

F Fy

Fx

d

Only the x component of the force does work.

• Work is positive when the force and displacement are in the same direction (object gains energy): work is negative when the force and displacement are in opposite directions (object looses energy.)

• Energy is defined as the ability to do work. It is also a scalar and cannot be negative. • Kinetic energy – energy through motion. • Potential energy – energy through position (position in a gravitational or electric field

or position next to a stretched or compressed spring). • Conservative force – a force where the work done in moving an object between two

positions is the same regardless of the path taken o Three conservative forces you should know: Gravitational, Elastic (springs),

Electric o Example: the work done against gravity in lifting a box on top of a shelf is the

same regardless of whether you lift it straight up or push it up a ramp. • When only conservative forces act on an object, then the total energy (kinetic plus

potential energy) remains constant. • Non-conservative force – a force where the work done in moving an object between

two points does depend on the path taken o Examples: friction, drag o Work done by a non-conservative force is equal to the change in energy of the

object (remember the work-energy theorem)



A

B

2

1

More work is needed to push an object against friction along path 2 than path 1. Friction is a non-conservative force.

F (N)

x (m)

Work = area under Force vs. Displacement Graph

• Power is the rate at which work is done and is measured in Watts.

• Power is the rate at which work is done and is measured in Watts.

• Work done by a variable force is equal to the area under a graph of force vs. displacement.

• Work done by a variable force is equal to the area under a graph of force vs. displacement.



Free Response Question 1 (15 pts) 1. A block of mass m is attached to an ideal spring of spring constant k, the other end of which is fixed. The block is on a level, frictionless surface as shown in the diagram. At time t0, the block is set into simple harmonic motion of period T by an external force pushing it to the right, giving the block initial velocity v0. Express all answers in terms of the given quantities and fundamental constants.

m

v0

A. Determine the amplitude of the block’s motion.

(4 points max) Kinitial = Us,max

2 2

0 m

1 1

2 2mv kx= ax

2

0max

mvx

k=

1 point For applying conservation of energy 1 point For equating the initial kinetic energy of the block with the maximum elastic potential energy 1 point For correctly substituting the definitions of kinetic and elastic potential energy 1 point For the correct answer in terms of the given quantities and fundamental constants

B. On the axis below, plot the kinetic energy of the block as a function of time for two periods. Label the vertical axis appropriately.



2T T 3

2T 2T

K (J)

t (s)

(4 points max) sinusoidal graph, positive, with max KE at 0, T/2, T, 3T/2, and 2T

1 point for drawing any sinusoidal graph 1 point for a graph that never shows negative kinetic energy (graph is always positive) 1 point for showing the maximum

kinetic energy is 2

0

1

2mv and the

minimum kinetic energy is zero 1 point for correctly showing that the maximum kinetic energy occurs at t = 0, T/2, T, 3T/2, and 2T and that the minimum kinetic energy occurs at t = T/4, 3T/4, 5T/4, and 7T/4.



C. The block is stopped and a second identical block is glued on top of the first. The

How does the amplitude of the motion of the two blocks together compare to

reater than __________ Less than ________ Equal ________

ii. Justify your answer.

(3 points max)

reater than __x

blocks are returned to simple harmonic motion with the same initial velocity as before. i.the amplitude found in part a)?

G

G __

the increased mass increases the kinetic

nd

sing the equation from part A, as the mass

energy when the block is set in motion, meaning the spring will be compressed astretch more uincreases so does the amplitude

2

0mvmaxx

k=

1 point For correctly indicating the

point for any reasonable justification

amplitude increases when the mass increases 1



uestion 2 (20 pts) f mass 650 kg is moving with a velocity of 15.0 m/s at the top of the

. Calculate the maximum velocity of the roller coaster.

(4 points max)

QA roller coaster car ofirst hill of a roller coaster track as shown in the diagram. The cart rolls without friction down the hill and through a vertical circular loop of radius 12 m.

15.0 m/s

55 m

12 m

30 m

A

( ) ( ) ( )

21 1mv mgh+ = 2

0 0 max

2

max 0 0

2

2max

max

2 2

12

2

12 15 9.8 55

2

36 ms

mv

v v gh

m mv ms s

v

= +

= +

=

⎡ ⎤⎢ ⎥⎣ ⎦

⎡ ⎤⎢ ⎥⎣ ⎦

1 point for a correct application of

point for recognizing that the

point for correctly adding the y

point for the correct answer

icant

conservation of energy 1maximum kinetic energy occurs when the roller coaster is at the bottom of the first hill (zero potential energy) 1initial kinetic and potential energto find the total energy 1including correct units and reasonable number of signifdigits



he roller coaster travels without friction through the circular loop.

i. On the diagram below, draw and label all of the forces acting on the roller coaster

(3 points max)

1 point for a correct force vector for the Normal force

t force vector for e weight force

raneous vectors

T

when it is upside down at the top of the loop.

1 point for a correcth 1 point for no ext

FgFN


p

ii. Calculate the magnitude of the normal force exerted on the roller coaster when it is upside down at the top of the loop.

(6 points max)

N g cF F F F= + =∑

2

N

mvF mg

r= −

( ) ( ) ( )

,

2423, 475 650 9.8 30

232, 375

top total g topK E U

mJ kg ms

K J

K

= −

= −

=

( )

21

2

2 2 232, 375

650

26.74

K mv

K Jv

m k

mv

g

s

=

= =

=

( ) ( )( ) ( )

2

2

2

650 26.74650 9.8

1232, 000

N

N

N

mvF mg

r

mkg s mF kgsm

F N

= −

= −

=

1 point for recognizing that the sum of the forces provide a centripetal force on the car 1 point for recognizing that the sum of the forces is the normal force minus the weight force 1 point for correctly solving for the normal force and substituting the expression for the centripetal force 1 point for using conservation of energy to find the velocity 1 point for the correctly solving for the velocity 1 point for the correct answer including correct units and reasonable number of significant digits

iii. The safety engineer determines that the acceleration of the riders is too great while they are passing through the loop. Describe a way the engineer can modify the ride to safely reduce the acceleration of the passengers as they go through the loop. Justify your answer.




(2 points max) increase the radius, increase height of loop, decrease initial velocity, introduce drag or friction increasing the height of the loop would require the roller coaster to gain more potential energy and thus it would have less kinetic energy and would be traveling slower

1 point for any reasonable modification that would reduce the acceleration of the riders through the loop 1 point for an appropriate justification

C. At the end of the ride, the roller coaster car returns to its initial height of 55 m before being brought to a stop by friction. Determine the amount of work that must be done in stopping the roller coaster at this height.

(5 points max) top end fE E W= +

W = ΔK

( ) ( )221 1

650 152 273,125

73,125

mK mv kg s

K J

K W J

= =

=

Δ = = −

Alternate solution

( ) ( ) ( )2650 9.8 55

350, 350

g

g

g

U mgh

mU kgs

U J

=

=

=

m

423, 475 350, 250

73,125

73,125

K J J

K J

K W J

= −

=

Δ = = −

1 point for any indication of conservation of energy 1 point for indicating that the work done against friction is equal to the change in energy as the roller coaster comes to a stop 1 point for indicating that at the beginning, the roller coaster has both potential and kinetic energy 1 point for indicating that at the end, the roller coaster has only potential energy 1 point for the correct answer including correct units and reasonable number of significant digits



bject

Question 3 (15 pts) An 80 kg object is to be pulled to the top of a 6.0 m ramp by a rope, the other end of which is pulled up by a 2400 W electric winch. The ramp forms a 30° angle with the horizontal. The coefficient of kinetic friction between the ramp and the object is 0.6 and the coefficient of static friction is 0.8. The object moves up the ramp at a constant

elocity. v

6 m

30 °

A. On the diagram of the o below, draw vectorsshowing all the forces on the object. Label each one.

(4 points max)

1 point for a correct force vector for the Normal force 1 point for a correct force vector for the tension 1 point for a correct force vector for the frictional force 1 point for a correct force vector for the weight force

1 point deducted for any extraneous vectors with a maximum of four points.

Fg

FT

FN

Ff



k

B. Determine the maximum constant speed at which the winch can pull the object up the ramp.

(7 points max)

, , 0x g x f k TF F F F= + − =∑

, ,T g x fF F F= +

sin cosT kF mg mgθ μ θ= +

( ) ( ) ( )

( ) ( ) ( )

2

2

80 9.8 sin 30

.6 80 9.8 cos 30

799

T

T

mF kgs

mkgs

F N

= °

+ °

=

P = Fv

2400

799

3.0 ms

P Fv

JP svF N

v

=

= =

=

1 point for any statement indicating that the net force is zero since the speed is constant 1 point for correctly summing the forces to find the tension 1 point for correctly substituting expressions for the x-component of weight and the force of friction 1 point for using the coefficient of kinetic friction and not static friction 1 point for a statement that the maximum constant speed at which the winch can pull the object up the ramp is power or the rate at which work is done 1 point for a correct application of P = Fv to find the speed 1 point for the correct answer including correct units and reasonable number of significant digits


d

C. Determine the total amount of work that the winch must do in pulling the object up the ramp.

(4 points max)

W F= ⋅∑ ∑

( )6

12sin 30

md m= =

°

( ) (799 12W N=∑ )m

W = 9790 J

1 point for an indication that the net work is the net force times the distance 1 point for indicating the distance the block travels along the ramp 1 point for using the net force from part B 1 point for the correct answer including correct units and reasonable number of significant digits




Multiple Choice 1. A hydraulic car lift is capable of lifting a 400 kg car to a height of 2.5 m in .8 s. At what rate does the car do work in lifting the car? a) 500 W b) 1,250 W c) 5,000 W d) 8,000 W e) 12,500 W Power is the rate at which work is done

( ) ( ) ( )2400 10 2.512,500

.8

W F d mgdP

t t t

mkg msP W

s

⋅= = =

= =

E

F = 64 N

Ff = 14 N 60°

2. An object of mass 0.5 kg is initially at rest. A constant force of 64 N is exerted on the object at a 60° relative to the horizontal as shown in the diagram. An opposing frictional force of 14 N pulls the object in the opposite direction. After being displaced 0.5 m to the right, the speed of the object is most nearly a) 0 m/s b) 6.0 m/s c) 8.5 m/s d) 9.0 m/s e) 10.0 m/s



Work is equal to change in energy Work is done by forces parallel to displacement

Forces parallel to displacement are the horizontal component of F (Fx) and friction (Ff)

( )64 cos 60 14 18x x fF F F N N N= − = − =∑

( ) ( )

( )

2

18 .5 9

1

2

2 2 96

.5

W F d N m J

K mv

K J mvsm kg

= ⋅ = =

=

= = =

B

3. While testing a new polymer fishing line, a student ties a mass of 0.25 kg to the end of a 0.5 m long piece of fishing line and swings it in a horizontal circle such that the mass completes ten revolutions each second. While swinging the mass in this manner, the work done by the fishing line on the mass during one revolution is most nearly a) 0 J b) 0.125 J c) 25 J d) 50π3 J e) 100π3 J Work is done by forces parallel to displacement

The force of tension is centripetal (perpendicular to displacement) and thus does no work A

4. In which of the following situations would a force be exerted on an object and no work be done on the object? I. a centripetal force is exerted on a moving object II. a force is exerted in the opposite direction as the object is moving III. a force is exerted on an object that remains at rest a) I only b) I and II c) I and III d) II and III e) I, II, and III Work is done by forces parallel to displacement

In I the force is perpendicular to displacement and in III there is no displacement. C

Questions 5 – 6 An object of mass 0.8 kg is initially at rest on a horizontal frictionless surface. The object is acted upon by a horizontal force F, the magnitude of which varies as function of the displacement of the object d as shown in the graph.



F (N) 20

0 0 0.2

d (m)

5. Which of the following would be most likely to produce the force shown in the graph? a) a string tied to a falling mass b) a stretched spring c) a magnet placed in the direction of the displacement d) kinetic friction e) static friction The force is decreasing as the object is displaced

The graph can only be produced by a variable force that gets weaker as the object moves – a stretched spring that pulls the object as it relaxes is the only one capable of producing such a graph

B

6. The amount of work done by the force F in displacing the object 0.2 m is most nearly a) 0 J b) 1 J c) 2 J d) 4 J e) 16 J Work done by a variable force is equal to the area under the curve of F/d graph

The graph forms a triangle 1

2A bh=⎡ ⎤⎢ ⎥⎣ ⎦

so the work is

( ) ( )1.2 20

22W m N J= =

C


7. A helicopter ascends vertically upwards with constant velocity v1. When the helicopter is at height h, a passenger throws a baseball of mass m out of the window of the helicopter horizontally with velocity v2. Just before striking the ground, the kinetic energy of the baseball is a) K mgh=

b) 21

12

K mgh mv= +

c) 22

12

K mgh mv= +

d) ( )2 21 2

12

K m v v= +

e) ( )2 21 2

12

K mgh m v v= + +

Conservation of Energy – the total energy of the ball remains constant

The baseball begins with two sources of energy: its vertical velocity (kinetic) and its height (potential). When thrown, additional kinetic energy is given to the ball. As it falls to the ground, potential energy is converted to kinetic energy. Thus, the sum of the three give the total kinetic energy just before it reaches the ground. E

( )2 2

1 2

1

2

vertical horizontal gK K K U

K m v v mgh

= + +

= + +

8. A 30 kg sled is sliding on a frictionless sheet of ice at a velocity of 4 m/s. The sled encounters a rough patch of ice and begins to slow down. After traveling on the rough patch of ice for 3 m, the sled’s velocity is 2 m/s. Determine the coefficient of friction between the rough ice and the sled. a) 0.07 b) 0.12 c) 0.20 d) 0.27 e) 0.60




Work is equal to change in energy

Find change in energy, which is equal to the work done:

( ) ( )2 212 4 1

2f im mW K K K m s s= Δ = − = − = −⎡ ⎤

⎣ ⎦80J

Friction is the only force on the sled:

( ) ( ) ( )2

180.2

30 10 3

fW F d mg d

W Jmmg d kg m

s

μ

μ

= ⋅ = ⋅

= = =⋅

C

Questions 9 – 10 A simple pendulum is constructed from a string of length l and a bob of mass m as shown in the diagram. It is released from rest at point I, which is a vertical distance y from the equilibrium position. The bob has zero potential energy at point II. Friction and drag are negligible.

I.

II.

III.

y y

l

9. The speed of the pendulum bob when it is in position II is most nearly

a) 212

ml

b) 2gy

c) ( )2g l y−

d) ( )2 22g l y−

e) ( )2g l ym−



onservation of Energy At I, the bob has zero kinetic energy and at II the bob has Czero potential energy. Potential energy at I is converted to kinetic energy at II (energy is conserved).

,g I IIU K=

21

2

2

mgy mv

v gy

=

=

B

0. Which of the following graphs best shows the total energy of the pendulum bob as a

onservation of Energy Since there are no non-conservative forces acting on it, the E

1function of displacement?

I. II. III. I. II. III. I. II. III.

a) b) c)

d) e)

I. II. III. I. II. III.

C

total energy of the bob remains constant. Graph E shows this


11. A block of mass m is pulled across a rough horizontal surface by a force F at a constant speed v. The coefficient of kinetic friction between the block and the surface is μ. The rate at which the force does work on the block is

a) 2mgvμ

b) vmgμ

c) mgv

μ

d) mgvμ

e) mgvμ Power is the rate at which work is done

Since the block moves with constant velocity, the frictional force is equal to the force F = μmg. E P F v mgvμ= ⋅ =

12. An archer does 38 J of work in drawing a bowstring back. When he releases the bowstring, the arrow, which has a mass of 0.25 kg, reaches a maximum height of 12 m. Determine the velocity of the arrow when it is at its maximum height. Assume all of the energy from the bow was transferred to the arrow. a) 0 m/s b) 4.0 m/s c) 8.0 m/s d) 15 m/s e) 17 m/s Conservation of Energy Energy is conserved in this situation, meaning that the total

energy of the arrow must always equal 38 J. Find the potential energy of the arrow at its max height:

( ) ( ) ( )2.25 10 12 30gmU mgh kg m J

s= = =

Since there is 30J of potential energy, the arrow must have 8 J of kinetic energy. From this, we can find the velocity C

( )

21

2

2 2 88

.25

K mv

K J mv sm kg

=

= = =


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