7/23/2019 Lecture6_2101

1/16

Different energies: Kinetic/translation

Gravitational potential

Heat energy

scalar which defines a state of a

system at a given time.

Strain or elastic energy

Kinetic Energy is associated with the state of motion

21

2KE mv= Units of Joules: 1 J = kgm2

/s2

KE depends on speed (v) not

KE doesnt depend on which way something is moving

v ( herev2 =

v

v =

v2

)

or even if its changing direction

KE is ALWAYS a positive scalar

7/23/2019 Lecture6_2101

2/16

How much is Kinetic Energy

Kinetic Energy is associated with the state of motion

KE depends on speed not

v ( here v2 =

v2 )

or even if its changing direction

1) Electron (e-) moving in Copper

me= 9.1110-31 kg and v ~ 1106 m/s KE = 710-19 J ( ~ 4 eV)

2) Bullet traveling at 950 m/s (3100 ft/s).

m = 4.2 g and v ~ 950 m/s (3100 ft/s) KE = 2000 J

3) Football Linebacker

m = 240 lbs and v ~ 18 m h 7 m/s KE = 2800 J4) Aircraft Carrier Nimitz

m = 91,400 tons and v ~ 1 knot KE = 10 MJ

7/23/2019 Lecture6_2101

3/16

~

WorkWork

Somehow force is related to KE energy

If we transfer energy via a force, this is work.

Doing work is the act of transferring energy.

Work (W) is said to be done on an object by a force.

.

Energy transferred from an object is negative work.

7/23/2019 Lecture6_2101

4/16

How to find an alternate form of Newtons 2ndLaw that relates osition and velocit .??

Work and Kinetic EnergyWork and Kinetic Energy

Start in 1-D (e.g. Bead along wire ), we know x

Fx,net

= max

= mdv

dt

Fx net

dx

= m

dv

dx

= mv

dv

1 2

1 2

Fx,net

dt

= 2

dtx,net

=2

=2

1

Wnet

Fx,net

dx = KE2KE

1= KE

Fx,net

dx = d 12mv

2(

Wnet = KE

7/23/2019 Lecture6_2101

5/16

WorkWork--Kinetic Energy TheoremKinetic Energy Theorem

Work-Kinetic

Energy Theorem

energy of an object the particle=

If is not a function of , thennet

F x

Wnet

Fnet

d = KE

component of the force along the objects line of motion.component of the force along the objects line of motion.

7/23/2019 Lecture6_2101

6/16

QuestionQuestion

ua tat ve y, ow muc wor am o ng on t e ow ng a

I walk across the room with a constant velocity?

.

2. None

7/23/2019 Lecture6_2101

7/16

Weight lifting: apply a FORCE up and DISPLACE the

barbell up

both the force and displacement are in the +y

direction so work is positive

On the downward motion the FORCE is still up and

the force is in the +y but the displacement is in y

direction so work is negative

External Force acts on box

.

Rank: Work done on box by F

7/23/2019 Lecture6_2101

8/16

QuestionQuestion

Two forces act on the box shown in the drawing, causing it to

move across the floor. The two force vectors are drawn to

. 12. F

23. Theyre both zero (F1 =F2=0)

4. Theyre the same, but not zero(F1 =F20)

7/23/2019 Lecture6_2101

9/16

ExampleExample

Durin a storm a crate is slidin across

a slick, oily parking lot through a

displacement while a

steady wind pushes against the crate

d = 3.0m( )i

with a force

F = 2.0N( )i + 6.0N( )j

(a) How much work does the force from the wind do on the crate during the

displacement? Negative Work

net net

Energy transferred from object

KE ecreases

,

the kinetic energy at the end of ?

d

W = KE= KE KE KE = W + KEor

ne ne

E lE l

7/23/2019 Lecture6_2101

10/16

ExampleExample

uestion 7-4: The fi ure shows the values of a force F directed alon an x axis

that will act on a particle at the corresponding values of x. If the particlebegins at rest a x=0, what is the particles coordinate when it has

net net f iW F dx KE KE KE = =

Area under F-x plot is W.

(b) the minimum speed

Q estion: Th fi h th l f

7/23/2019 Lecture6_2101

11/16

Question: The figure shows the values of a

force F, directed along an x axis, that

will act on a particle at thepos. work -> KE increase

corresponding values of x. If the

particle begins at rest a x=0, what isthe particles coordinate when it has

Greatest KE here

(a) the greatest speed

(b) the minimum speed ->

zero KE here

.

Greatest speed at x=3 m Minimum speed (zero) at x=6 m

W kW k

7/23/2019 Lecture6_2101

12/16

i net net iW W F d KE KE KE = = =

WorkWork

all

Wi

= Wnet Work is sum total of all transfer of energy (pos and neg)

2

( ) where

r

W F r dr F d r r d = =

all

No work is done on an object by a force unless there is a

component of the force along the objects line of motion

1r

Sum of the area under net F-r plot yields net work

Wnet

= KE If speed goes up then W is positive -- energy transferred to object

If speed goes down then W is negative -- energy transferred from object

spee s constant t en

KE= KE KEChan e in kinetic ener is chan e in s eed i.e. v2

What about centripetal force?

S i l CS i l C

7/23/2019 Lecture6_2101

13/16

Work done by Constant Gravitational ForceSpecial Case:Special Case:

Wg

Fg

d

x =

Fg

d

where Fg

= mg( ) j( ) and g = 9.81m /s 2

If an object is displaced upward ( y positive), then the work done

by the gravitational force on the object is negative.

If an object is displaced downward (y negative), then the workdone by the gravitational force on the object is positive.

What is the change in KE due to Gravitational Force

,

Wnet = Wg

=

Fg

d( )= KE= KEf

KEi

( )If an object is displaced upward ( y positive), the change in

Kinetic Energy is negative (it slows down).

If an object is displaced downward (y negative), the change inKinetic Energy is positive (it speeds up).

Wh t k i d d t lift l bj t?

7/23/2019 Lecture6_2101

14/16

What work is needed to lift or lower an object?

In order to lift an object, we must apply an external

force to counteract the gravitational force.

Wnet

Wg

+ Wext

= KE

If (i.e. ), thenKE= 0 vf = vi Wg = Wext

If an object is displaced upward ( y positive), then the.

If an object is displaced downward (y negative), then the.

D hill SkiExampleExample

7/23/2019 Lecture6_2101

15/16

Downhill SkierExampleExample

A 58-kg skier coasts down a 25 slope where thekinetic frictional force is fk=70N. If the starting

s eed is v =3.6 m/s then what is the s eed after a

displacement of 57 m?

Along the direction of motion (+x) the forces are:

= mgs n k

= 240N 70N

=170N

The work this force does is:

This gives a change in kinetic energy that is:

W = 1 mv2 1 mv22 2

1 2

=

1 2

=

1 2 v =2(10100J)

2 2 0 2 .

=10100J

58kg

=19m /s

W k d t F i tiSpecial Case:Special Case:

7/23/2019 Lecture6_2101

16/16

Work due to Friction:Special Case:Special Case:

WORK due to friction is ALWAYS NEGATIVE- Energy is transferred OUT

- Kinetic energy decreases or KE < 0 (slow down)

Where did the ener o? THERMAL/Sound