Physics 1501: Lecture 11, Pg 1 Physics 1501: Lecture 11 l Announcements çHW 04 will be due the following Friday (same week). çMidterm 1: Monday Oct. 3.

Physics 1501: Lecture 11, Pg 1

Physics 1501: Lecture 11Physics 1501: Lecture 11 Announcements

HW 04 will be due the following Friday (same week).

Midterm 1: Monday Oct. 3Wednesday

» Replacement (Sandipan Banerjee)

» Office hours Friday instead of Wednesday

TopicsWork & EnergyScalar Product


Chap.6: Work & EnergyChap.6: Work & Energy

One of the most important concepts in physics.Alternative approach to mechanics.

Many applications beyond mechanics.Thermodynamics (movement of heat).Quantum mechanics...

Very useful tools.You will learn new (sometimes much easier) ways to

solve problems.


Definition of Work:Definition of Work:

Ingredients: Ingredients: Force ( FF ), displacement ( rr )

Work, W, of a constant force FF

acting through a displacement rr

is:

W = FF · rr = F rr cos = Fr rr

FF

rr

displace

ment

Fr

“Dot Product”


Review: Scalar Product ( or Dot Product)Review: Scalar Product ( or Dot Product)

Definition:a a · bb = ab cos

= a[b cos ] = aba

= b[a cos ] = bab

Some properties:a a ·bb = b b ·aaq(a a ·bb) = (qbb) · a a = bb · (qaa) (q is a scalar)aa · (b b + cc) = (a a ·bb) + (aa ·cc) (cc is a vector)

The dot product of perpendicular vectors is 0 !!

aa

ab bb

aa

bb

ba


Review: Examples of dot productsReview: Examples of dot products

Suppose Then

aa = 1 i i + 2 j j + 3 k k

bb = 4 i i - 5 j j + 6 k k

aa · bb = 1x4 + 2x(-5) + 3x6 = 12

aa · aa = 1x1 + 2x2 + 3x3 = 14

bb · bb = 4x4 + (-5)x(-5) + 6x6 = 77

i i · ii = j j · j j = k k · k k = 1

i i · jj = j j · k k = k k · i i = 0x

y

z

ii

jj

kk


Review: Properties of dot productsReview: Properties of dot products

Magnitude:a2 = |a|2 = a · a

= (ax i i + ay j j ) · (ax i i + ay j j )= ax

2( i i · i i ) + ay 2( j j · j j ) + 2ax ay ( i i · j j )

= ax 2 + ay

2

Pythagorian Theorem !!

aa

ax

ay

ii

j j


Review: Properties of dot productsReview: Properties of dot products

Components:aa = ax i i + ay j j + az k k = (ax , ay , az ) = (aa · i i , aa · j j , aa · k k )

Derivatives:

Apply to velocity

So if v is constant (like for UCM):


Back to the definition of Work:Back to the definition of Work:

Work, W, of a force FF acting

through a displacement rr is:

W = FF · rrFF

rr


Lecture 11, Lecture 11, ACT 1ACT 1WorkWork

A box is pulled up a rough ( > 0) incline by a rope-pulley-weight arrangement as shown below.

How many forces are doing work on the box ?

(a)(a) 2

(b)(b) 3

(c)(c) 4


Work: 1-D Example Work: 1-D Example (constant force)(constant force)

A force FF = 10N pushes a box across a frictionless floor for a distance x x = 5m.

xx

FF

Work done byby F F onon box : WF = FF ·xx = F x (since FF is parallel to xx)

WF = (10 N)x(5m) = 50 N-m.


Units:Units:

N-m (Joule) Dyne-cm (erg)

= 10-7 J

BTU = 1054 J

calorie = 4.184 J

foot-lb = 1.356 J

eV = 1.6x10-19 J

cgs othermks

Force x Distance = Work

Newton x

[M][L] / [T]2

Meter = Joule

[L] [M][L]2 / [T]2


Work and Varying ForcesWork and Varying Forces

Consider a varying force,

W = Fxx

As x 0, x dx

Fx

xx

Area = Fxx


SpringsSprings

A very common problem with a variable force is a spring.

In this spring, the force gets greater as the spring is further compressed.

Hook’s Law,

FS = - k x

x is the amount the spring is stretched or compressed from it resting position.

Fx

Active Figure


Lecture 11,Lecture 11, ACT 2ACT 2HookHook’’s Laws Law

Remember Hook’s Law,

Fx = -k x

What are the units for the constant k ?

A) B) C) D)


Lecture 11,Lecture 11, ACT 3ACT 3HookHook’’s Laws Law

0.2 kg

9 cm

8 cm

What is k for this spring ??

A) 50 N/m B) 100 N/m C) 200 N/m D) 400 N/m


What is the Work done by the Spring...What is the Work done by the Spring... The work done by the spring Ws during a displacement

from x1 to x2 is the area under the F(x) vs x plot between x1 and x2.

Ws = - 1/2 [ ( kx2) (x2) - (kx1) (x1) ]

x2 x1

F(x)

x

Ws

kx1

kx2

-kx

Ws

Active Figure


Work & Kinetic Energy:Work & Kinetic Energy:

A force FF = 10N pushes a box across a frictionlessfloor for a distance x x = 5m. The speed of the box is v1 before the push, and v2 after the push.

xx

FFv1 v2

ii

m


Work & Kinetic Energy...Work & Kinetic Energy...

Since the force FF is constant, acceleration aa will be constant. We have shown that for constant a:

W = (F) · d = ma · dFor constant a, a = (v-v0)/t

also, d = vt = (1/2) (v+v0)t

xx

FFv1 v2

aa

ii

m


Work & Kinetic Energy...Work & Kinetic Energy...

Altogether,

W = (F) · d = ma ·d = m ((v-v0)/t) · 1/2 (v+v0)t

W = (1/2) m (v2 - v02)

Define Kinetic Energy K: K = 1/2mv2

K2 - K1 = WF

WF = K (Work kinetic-energy theorem)(Work kinetic-energy theorem)

xx

FFv1 v2

aa

ii

m


Work Kinetic-Energy Theorem:Work Kinetic-Energy Theorem:

{NetNet WorkWork done on object}=

{changechange in kinetic energy kinetic energy of object}

We’ll prove this for a variable force later.


xx

vo

m

to

F

ExampleExampleWork Kinetic-Energy TheoremWork Kinetic-Energy Theorem

Km = 1/2 m v2 - 1/2 m vo2

= 0 - 1/2 m vo2

Km = Wspring

Wspring = -1/2 k x2 - xo2 )

= - 1/2 k x2

(assuming x0=0)so : 1/2 m vo

2 = 1/2 k x2

and : x = ( m vo

2 /k ) 1/2

How much will the spring compress to bring the object to a stop if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ?

spring compressed

spring at an equilibrium position

V=0

t m


Lecture 11, Lecture 11, ACT 4ACT 4Kinetic EnergyKinetic Energy

To practice your pitching you use two baseballs. The first time you throw a slow curve and clock the speed at 50 mph (~25 m/s). The second time you go with high heat and the radar gun clocks the pitch at 100 mph. What is the ratio of the kinetic energy of the fast ball versus the curve ball ?

(a)(a) 1/41/4 (b)(b) 1/2 (c)(c) 1 (d) 2 (e) 4


Lecture 11, Lecture 11, ACT 5ACT 5Kinetic EnergyKinetic Energy

To practice your pitching you use two baseballs. The first time you use a softball: you throw a slow curve and clock the speed at 50 mph (~25 m/s). The second time you use a baseball (about half the mass of a softball) and go with high heat: the radar gun clocks the pitch at 100 mph. What is the ratio of the kinetic energy of the fast ball versus the curve ball ?

(a)(a) 1/4 (b)(b) 1/2 (c)(c) 1 (d) 2 (e) 4


Lecture 11, Lecture 11, ACT 6ACT 6Work & EnergyWork & Energy

Two blocks having mass m1 and m2 where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. > 0) which slows them down to a stop.Which one will go farther before stopping ?

(a)(a) m1 (b)(b) m2 (c)(c) they will go the same distance

m1

m2


Lecture 11, Lecture 11, ACT 7ACT 7Work & EnergyWork & Energy

You like to drive home fast, slam on your brakes at the bottom of the driveway, and screech to a stop laying rubber all the way. It’s particularly fun when your mother is in the car with you. You practice this trick driving at 20 mph and with some groceries in your car with the same mass as your mama. You find that you only travel half way up the driveway. Thus when your mom joins you in the car, you try it driving twice as fast. How far will you go this time ?(a)(a) The same distance. Not so exciting.The same distance. Not so exciting.

b) 2 times as far (only 7/10 of the way up the driveway)

(c)(c) twice as far, right to the door. Whoopee!twice as far, right to the door. Whoopee!

(d)(d) four times as far. Crashes into house. Sorry Ma.four times as far. Crashes into house. Sorry Ma.

Physics 1501: Lecture 11, Pg 1 Physics 1501: Lecture 11 l Announcements çHW 04 will be due the following Friday (same week). çMidterm 1: Monday Oct. 3.

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