Kinetic Energy, Work, Power, and Potential Energy

8.01

W05D1

Today’s Reading Assignment: W05D1

Young and Freedman: 6.1-6.4

Math Review Module: Scalar Product

Kinetic Energy

• Scalar quantity (reference frame dependent)

• SI unit is joule:

• Change in kinetic energy:

210

2K mv= ≥

2 21J 1kg m /s≡ ⋅

ΔK =

12

mvf2 −

12

mv02 =

12

m(vx, f2 + vy, f

2 + vz, f2 ) −

12

m(vx,02 + vy,0

2 + vz,02 ) ≥0

Concept Question: Work and Kinetic Energy

Compared to the amount of energy required to accelerate a car from rest to 10 mph (miles per hour), the amount of energy required to accelerate the same car from 10 mph to 20 mph is

(1) the same

(2) twice as much

(3) three times as much

(4) four times as much

(5) unsure.

Work Done by a Constant Force for One Dimensional

Motion Definition:

The work W done by a constant force with an x-component, Fx, in displacing an object by Δx is equal to the x-component of the force times the displacement:

W =FxΔx

Concept Question: Pushing against a wall

The work done by the contact force of the wall on the person as the person moves away from the wall is

1. positive.

2. negative.

3. zero.

4. impossible to determine from information given in question and the figure.

Concept Question: Work and Walking

When a person walks, the force of friction between the floor and the person's feet accelerates the person forward. The work done by the friction force is

1. positive.

2. negative.

3. zero.

Pushing a Stalled Car

Table Problem: Work Done by Gravity Near the Surface of the

Earth Consider an object of mass near the surface of

the earth falling directly towards the center of the earth. The gravitational force between the object and the earth is nearly constant. Suppose the object starts from an initial point that is a distance y0 from the surface of the earth and moves to a final point a distance yf from the surface of the earth. How much work does the gravitational force do on the object as it falls?

Work done by Non-Constant Force: One Dimensional Motion

(Infinitesimal) work is a scalar

Add up these scalar quantities to get the total work as area under graph of Fx vs x :

ΔWi =(Fx)i Δxi

1 1

( )i N i N

i x i ii i

W W F x= =

= =

= Δ = Δ∑ ∑

As and 0iN x→ ∞ Δ →

010

lim ( )f

i

x xi N

x i i xN

i x xx

W F x F dx==

→ ∞= =Δ →

= Δ =∑ ∫

Concept Question: Work due to Variable Force

A particle starts from rest at x = 0 and moves to x = L under the action of a variable force F(x), which is shown in the figure. What is the particle's kinetic energy at x=L/2 and at x=L?

• (Fmax)(L/2), (Fmax)(L)

(2) (Fmax)(L/4), 0

(3) (Fmax)(L), 0

(4) (Fmax)(L/4), (Fmax)(L/2)

(5) (Fmax)(L/2), (Fmax)(L/4)

Table Problem: Work Done by the Spring Force

Connect one end of a spring of length l0 with spring constant k to an object resting on a smooth table and fix the other end of the spring to a wall. Stretch the spring until it has length l and release the object.

How much work does the spring do on the object as a function of x = l - l0, the distance the spring has been stretched or compressed?

Worked Example: Work Done by Several Forces

A block of mass m slides along a horizontal table with speed v0. At x = 0 it hits a spring with spring constant k and begins to experience a friction force. The coefficient of kinetic friction is given by . How far did the spring compress when the block first momentarily comes to rest?

 

Recall integration formula for acceleration with respect

to time The x-component of the acceleration of an objectis the derivative of the x-component of the velocity

Therefore the integral of x-component of the acceleration with respect to time, is the x-component of the velocity

a

xdt

t0

t f∫ =dvx

dtdt

t0

tf∫ = dvxvx,0

vx, f∫ =vx, f −vx,0

a

x≡

dvx

dt

Integration formula for acceleration with respect to

displacement The integral of x-component of the acceleration with respect to the displacement of an object, is given by

Multiply both sides by the mass of the object giving integration formula

a

xdx

x0

x f∫ =dvx

dtdx

x0

xf∫ = dvx

dxdtx0

xf∫ = vxdvxvx,0

vx, f∫

a

xdx

x0

x f∫ = d (1 / 2)vx2( )

vx,0

vx, f∫ =12(vx, f

2 −vx,02 )

ma

xdx

x0

x f∫ = d (1 / 2)vx2( )

vx,0

vx, f∫ =12

mvx, f2 −

12

mvx,02 =ΔK

Work-Kinetic Energy Theorem One Dimensional Motion

Substitute Newton’s Second Law (in one dimension)

in definition of work integral which then becomes

Apply integration formula to get work-kinetic energy theorem

W = Fx dx

x0

xf∫ = max dxx0

xf∫

Fx=max

W = Fx dx

x0

xf∫ = max dxx0

xf∫ =ΔK

Concept Question: Work-Energy

1. a 2. b 3. c 4. d 5. e

An object is dropped to the earth from a height of 10m. Which of the following sketches best represent the kinetic energy of the object as it approaches the earth (neglect friction)?

Concept QuestionTwo objects are pushed on a frictionless surface from a starting line to a finish line with equal constant forces. One object is four times as massive as the other. Both objects are initially at rest. Which of the following statements is true when the objects reach the finish line?

1. The kinetic energies of the two objects are equal.

2. Object of mass 4m has the greater kinetic energy.

3. Object of mass m has the greater kinetic energy.

4. Not information is given to decide.

Worked Example: Work-Energy Theorem for Inverse Square

Gravitational Force Consider a magnetic rail gun that shoots an

object of mass m radially away from the surface of the earth (mass me). When the object leaves the rail gun it is at a distance ri from the center of the earth moving with speed vi . What speed of the object as a function of distance from the center of the earth?

Power

• The average power of an applied force is the rate of doing work

• SI units of power: Watts

• Instantaneous power

applied,applied,

xx x

F xWP F v

t t

ΔΔ= = =

Δ Δ

2 31W 1J/s 1kg m /s≡ = ⋅

P = lim

Δt→ 0

ΔWΔt

=Fapplied,x limΔt→ 0

ΔxΔt

⎛

⎝⎜⎞

⎠⎟=Fapplied,x vx

Work and the Dot Product

Dot ProductA scalar quantity

Magnitude:

The dot product can be positive, zero, or negative

Two types of projections: the dot product is the parallel component of one vector with respect to the second vector times the magnitude of the second vector

cosθ⋅ =A B A Br rr r

(cos ) Aθ⋅ = =A B A B BP

r rr r r(cos ) Bθ⋅ = =A B A B A P

r r rr r

Dot Product of Unit Vectors in Cartesian Coordinates

ˆ ˆ ˆ ˆ| || | cos(0) 1

ˆ ˆ ˆ ˆ| || |cos( /2) 0π

⋅ = =

⋅ = =

i i i i

i j i j

i, j

i ⋅i =j ⋅j =k ⋅k =1

i ⋅j =i ⋅k = j⋅k =0

For unit vectors

We have

Generally:

Example: Dot Product in Cartesian Coordinates

rA =Axi + Ayj + Azk,

rB =Bxi + Byj + Bzk

ThenrA ⋅

rB =(Axi + Ayj + Azk) ⋅(Bxi + Byj + Bzk)

=Axi ⋅Bxi + Axi ⋅Byj + Axi ⋅Bzk

+Ayj⋅Bxi + Ayj⋅Byj + Ayj⋅Bzk

+Azk ⋅Bxi + Azk ⋅Byj + Azk ⋅Bzk

=AxBx + AyBy + AzBz

Kinetic Energy and Dot Product

Velocity

Kinetic Energy:

Change in kinetic energy:

rv =vxi + vyj + vzk

K =

12

m(rv⋅rv) =12

m(vx2 + vy

2 + vz2 ) ≥0

ΔK =12

mvf2 −

12

mv02 =

12

m(rv f ⋅rv f ) −12

m(rv0 ⋅rv0 )

12

m(vx, f2 + vy, f

2 + vz, f2 ) −

12

m(vx,02 + vy,0

2 + vz,02 )

Work Done by a Constant Force

Definition: Work

The work done by a constant force on an object is equal to the component of the force in the direction of the displacement times the magnitude of the displacement:

Note that the component of the force in the direction of the displacement can be positive, zero, or negative so the work may be positive, zero, or negative

Fr

cos cosW Fθ θ= ⋅Δ = Δ = Δ = ΔF r F r F r rP

r r rr r r r

Concept Question: Work and Gravity 1

A ball is given an initial horizontal velocity and allowed to fall under the influence of gravity near the surface of the earth, as shown below. The work done by the force of gravity on the ball is:

(1) positive (2) zero (3) negative

Worked Example: Work Done by a Constant Force in

Two DimensionsLet the force exerted on an object be

ˆ ˆx yF F= +F i j

r

ˆDisplacement: xΔ =Δr ir

cos

ˆ ˆ ˆ( ) ( )x y x

W F x

F F x F x

β= ⋅Δ = Δ

= + ⋅Δ = Δ

F r

i j i

r r

Fx=F cosβ

Fy =F sinβ

Table Problem: Work Constant Forces and Dot Product

a) Use the dot product definition of work to calculate the work done by the normal force, the gravitational force, and the friction force as the object displaces a distance s down the inclined plane.

b) For each force, is the work done by the force positive or negative?

c) What is the sum of the work done by the three forces? Is this positive or negative?

An object of mass m, starting from rest, slides down an inclined plane of length s. The plane is inclined by an angle of θ to the ground. The coefficient of kinetic friction is μ.

Concept Question: Work and inverse square gravity

A comet is speeding along a hyperbolic orbit toward the Sun. While the comet is moving away from the Sun, the work done by the Sun on the comet is:

(1) positive (2) zero (3) negative

Work Done Along an Arbitrary Path

ΔWi =rFi ⋅Δ

rri

Work done by force for small displacement

Work done by force along path from A to B

WAB

= limN→ ∞Δrri → 0

rFi ⋅Δ

rri

i=1

i=N

∑ ≡rF ⋅d

rr

A

B

∫

Work-Energy Theorem in Three-Dimensions

As you will show in the problem set, the one dimensional work-kinetic energy theorem generalizes to three dimensions

WAB

=rF ⋅d

rr

A

B

∫ = mra ⋅d

rr

A

B

∫ = md

rv

dt⋅d

rr

A

B

∫ = mdrv ⋅

drr

dtA

B

∫ = mdrv ⋅

rv

A

B

∫

=1

2mv

B2 −

1

2mv

A2 = K

B− K

A

= ΔK

Work: Path Dependent Line Integral

Work done by force along path from A to B

In order to calculate the line integral, in principle, requires a knowledge of the path. However we will consider an important class of forces in which the work line integral is independent of the path and only depends on the starting and end points

WAB

= limN→ ∞Δrri → 0

rFi ⋅Δ

rri

i=1

i=N

∑ ≡rF ⋅d

rr

A

B

∫

Conservative Forces

Definition: Conservative Force If the work done by a force in moving an object from point A to point B is independent of the path (1 or 2),

then the force is called a conservative force which we denote by . Then the work done only depends on the location of the points A and B.

c (path independent)B

c

A

W d≡ ⋅∫F rr r

cFr

Example: Gravitational Force Consider the motion of an object under the influence of a gravitational force near the surface of the earth

The work done by gravity depends only on the change in the vertical position

g gW F y mg y= Δ =− Δ

Potential Energy Difference

c

B

c

A

U d WΔ ≡− ⋅ =−∫F rr r

Definition: Potential Energy Difference between the points A and B associated with a conservative force is the negative of the work done by the conservative force in moving the body along any path connecting the points A and B.

cFr

Potential Energy Differnece: Constant Gravity

Force:

Work:

Potential Energy:

Choice of Zero Point: Choose and choose . Then

Potential Energy:

rF =m

rg=F y j =−mgj

W =FyΔy=−mgΔy

ΔU =−W =mgΔy=mg yf −y0( )

U ( y0) ≡0

U ( y) =mgy

y0=0

Worked Example: Change in Potential Energy for Inverse Square Gravitational Force

Consider an object of mass m1 moving towards the sun (mass m2). Initially the object is at a distance r0 from the center of the sun. The object moves to a final distance rf from the center of the sun. For the object-sun system, what is the change in potential during this motion?

Worked Example Solution: Inverse Square Gravity

Force:

Work done:

Potential Energy Change:

Zero Point:

Potential Energy Function

1 2

1 2, 2

ˆm m

Gm m

r=−F r

r

00 0

1 2 1 21 22

0

1 1f f fr r r

r fr r

Gm m Gm mW d dr Gm m

r r r r

⎛ ⎞⎛ ⎞= ⋅ = − = = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫F r

r r

U (r0=∞) =0

U (r) = −

Gm1m2

r

Table Problem: Change in Potential Energy Spring Force Connect one end of a spring of length l0 with

spring constant k to an object resting on a smooth table and fix the other end of the spring to a wall. Stretch the spring until it has length l and release the object. Consider the object-spring as the system. When the spring returns to its equilibrium length what is the change in potential energy of the system?

Potential Energy Difference: Spring Force

Force:

Work done:

Potential Energy Change:

Zero Point:

Potential Energy

xˆ ˆF kx= =−F i i

r

W = −kx( )dx=−12x=x0

x=xf

∫ k xf2 −x0

2( )

ΔU =−W =

12

k xf2 −x0

2( )

U (x =0) =0

U (x) =

12

kx2

Work-Energy Theorem: Conservative Forces

The work done by the force in moving an object from A to B is equal to the change in kinetic energy

When the only forces acting on the object are conservative forces

then the change in potential energy is

Therefore

W ≡

rF ⋅d

rr

A

B

∫ =12

mvB2 −

12

mvA2 ≡ΔK

ΔU =−W

−ΔU =ΔK

rF =

rFc

Table Problem : Asteroid about Sun

An asteroid of mass m is in a non-circular closed orbit about the sun. Initially it is a distance ri from the sun, with speed vi. What is the change in the kinetic energy of the asteroid when it is a distance is rf, from the sun?

Next Reading Assignment: W05D2

Young and Freedman: 7.1-7.5,12.3

Experiment 3: Energy Transformations

44