Chapter 6Work and Energy

04/19/23 1Phys 201, Spring 2011

Work (dot product) Work by gravity, by springKinetic energy, power Work-kinetic energy theoremC.M. system

Definition of Work: Constant Force

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

Work, W, of a constant force FF

acting through a displacement Δrr :

Vector

“Dot Product”

€

W =r F ⋅Δr r = FΔr cosθ = FrΔr

Work is a scalar.

Units are Newton meter: Joule = N x m

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Work and Force DirectionWork and Force DirectionWork is done in lifting the box (you do +work; gravity -)No work is done on the bucket when held still, or to move horizontally

More on “dot product” (or scalar product)

Definition:

aa.bb = ab cos θ

= a[b cos θ] = aba

= b[a cos θ] = bab

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

The dot product of perpendicular vectors is 0 !!

θ

aa

ab bb

θ

aa

bb

ba

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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 = 12aa . aa = 1x1 + 2x2 + 3x3 = 14bb . 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

04/19/23 5Phys 201, Spring 2011

More 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 uniform circular motion):

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Which of the statements below is correct?

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What about multiple forces?

Suppose FFNET = FF1 + FF2 and the

displacement is Δrr.

The work done by each force is:

W1 = FF1 . Δr r W2 = FF2

. Δrr

WTOT = W1 + W2

= FF1 . Δr r + FF2

. Δrr

= (FF1 + FF2 ) . Δrr

WTOT = FFTOT . Δrr It’s the totaltotal force that matters!!

FFNET

ΔrrFF1

FF2

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Question 1Question 1You are towing a car up a hill with constant velocity. The work done on the car by the normal force is:1. positive2. negative3. zero

W

T

FN V

Normal force is perpendicular to displacement cos θ = 0

correct

04/19/23 Phys 201, Spring 2011 10

You are towing a car up a hill with constant velocity. The work done on the car by the gravitational force is:1. positive2. negative3. zero

There is a non-zero component of gravitational force pointing opposite the direction of motion.

W

T

FN V

correct

Question 2Question 2

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You are towing a car up a hill with constant velocity. The work done on the car by the tension force is:1. positive2. negative3. zero

T is pointing in the direction of motion - therefore, work done by this force is positive.

W

T

FN V

correct

Question 3Question 3

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You are towing a car up a hill with constant velocity. The total work done on the car by all forces is:1. positive2. negative3. zero

Constant velocity implies that there is no net force acting on the car, so there is no work being done overall

W

T

FN V

correct

Question 4Question 4

Lifting an object:

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A 3000–kg truck is to be loaded onto a ship by a crane that exerts an upward force of 31 kN on the truck. This force, which is strong enough to overcome the gravitational force and keep the truck moving upward, is applied over a distance of 2.0 m. Find (a) the work done on the truck by the crane, (b) the work done on the truck by gravity, (c) the net work done on the truck.

(a) the work done by the crane: Wc = Fc d = 31 kN x 2 m = 62 kJ

(b) the work done by gravity: Wg = Fg d = (-31 kN) x 2 m = -62 kJ (c) the net work done on the truck: W = Wc + Wg = 0

Work by a variable force:

Total work, W, of a force F actingthrough a displacement Δx = dx î is an integration:

Work = Area under F(x) curve

Fx

x

Δxi

Fi

€

W =r F i ⋅Δ

r x ii

∑

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Work done by a spring force

Force exerted by a spring extended a distance x is:

Fx = -kx (Hooke’s law)

Force done by spring on an object when the displacement of spring is changed from xi to xf is:

04/19/23 15Phys 201, Spring 2011

A person pushes against a spring, the opposite end of which is attached to a fixed wall. The spring compresses. Is the work done by the spring on the person positive, negative or zero?

A. positiveB. negativeC. zero

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EnergyEnergyEnergy is that quality of a substance or objectwhich “causes something to happen”; or “capability of exerting forces”;or “ability to do work”…

The vagueness of the definition is due to the factthat energy can result in many effects.

It is convertible into other forms without loss (i.e it is conserved)

Electrical EnergyChemical EnergyMechanical EnergyNuclear Energy

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Kinetic Energy Kinetic Energy The “energy of motion”.

Work done on the object increases its energy, -- by how much? (i.e. how to calculate the

value?)

F = maW = F d

W = (ma) d

V = V + 2 a d2 20

W = mV - m VW = mV - m V12

12

22

0

a = 2 d

V - V22

0

2 d

V - V22

0W = m d

K.E. = mV K.E. = mV 12

2

Phys 201, Spring 2011

Definition of Kinetic EnergyKinetic energy (K.E.) of a particle of mass m moving with speed v is defined as

K.E. = ½mv2

Kinetic energy is a useful concept because of theWork/Kinetic Energy theorem, which relates the work done on an object to the change in kinetic energy.

04/19/23 19Phys 201, Spring 2011

A falling object

What is the speed of an object that starts at rest and then falls a vertical distance H?

Work done by gravitational force WG = FΔr = mgH

Work/Kinetic Energy Theorem:

€

WG = mgH = 12 mv2

⇒ v = 2gH

Δrrmg g

H

j j

v0 = 0

v

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A skier of mass 50 kg is moving at speed 10 m/s at point P1 down a ski slope with negligible friction. What is the skier’s kinetic

energy when she is at point P2, 20 m below P1?

A. 2500 JB. 9800 JC. 12300 JD. 13100 JE. 15000 J

H

P1

P2

Work-kinetic energy theorem:KEf – KEi = Wg = mgHKEf = ½×50×102 + 50×9.8×20 = 2500 J + 9800 J

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Energy and Newton’s Laws

The importance of mechanical energy in classical mechanics is a consequence of Newton’s laws.

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The concept of energy turns out to be even more general than Newton’s Laws.

Problem: Work and Energy

Two blocks have masses 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 with μ>0, which slows them down to a stop.Which one will go farther before stopping?(A) m1 (B) m2 (C) They will go the same distance

m1

m2

Solution next page

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Problem: Work and Energy (Solution)

The net work done to stop a box is -fD = -μmgD.The work-kinetic energy theorem says that for any object, WNET = ΔK so WNET=Kf-Ki=0-Ki.

Since the boxes start out with the same kinetic energy, we haveμm1gD1=μm2gD2 and D1/D2=m2/m1.

Since m1>m2, we must have D2>D1.

m1

D1

m2

D2

04/19/23 24Phys 201, Spring 2011

Problem:

The magnitude of the single force acting on a particle of mass m is given by F = bx2, where b is a constant. The particle starts from rest at x=0. After it travels a distance L, determine its (a) kinetic energy and (b) speed.

€

W =r F ⋅dr x

0

L

∫ = bx2dx0

L

∫ =b3

L3work done by force:

So (a) kinetic energy is K=bL3/3

(b) Since ½mv2=bL3/3,

€

v =2bL3

3m04/19/23 25Phys 201, Spring 2011

Power: P: Work done by unit time.

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Units: Watt

Question

€

ddt

12 mv2

( ) = C

€

12 mv2 = Ct

A sports car accelerates from zero to 30 mph in 1.5 s. How long does it take for it to accelerate from zero to 60 mph, assuming the power of the engine to be independent of velocity and neglecting friction?

Integrating with respect to time, and noting that the initial velocity is zero, one gets . So getting to twice the speed takes 4 times as long, and the time to reach 60 mph is 4×1.5 = 6s.

Power is constant, so , a constant.

04/19/23 27Phys 201, Spring 2011

If a fighter jet doubles its speed, by what factor should the power from the engine change? (Assume that the drag force on the plane is proportional to the square of the plane’s speed.)

Magnitude of power is Fv. When the velocity v is doubled, the drag force goes up by a factor of 4, and Fv goes up by a factor of 8

04/19/23 28Phys 201, Spring 2011

Recall: Center of Mass

Definition of the center of mass:

rRcm =

mirri

i∑

mii∑

⇒ racm =

mirai

i∑

mii∑

.

Because , rFi =mi

rai

rFi

i∑ = mi

rai

i∑ = mi

i∑⎛⎝⎜

⎞⎠⎟

mirai

i∑

mii∑⎛⎝⎜

⎞⎠⎟

=MTOTraCM

04/19/23 29Phys 201, Spring 2011

Center-Of-Mass WorkFor systems of particles that are not all moving at the same velocity, there is a work-kinetic energy relation for the center of mass.

So

where is the translational kinetic energy

Net work done on collection of objects

change in translational kinetic energy of system=

Ktrans =12

mvCM2

04/19/23 30Phys 201, Spring 2011