11/1/20151 Norah Ali Al Moneef king Saud university.

042023 1Norah Ali Al Moneef king Saud university

2

bull W = F Δx cos bull Work = the product of force and displacement

times the cosine of the angle between the force and the direction of the displacement

bull How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m

bull W = FΔx cos bull W = 15 N (12 m) (cos 20) = 169J

20o

15 N

W = F Δx units Nm or joule j

042023 Norah Ali Al Moneef king Saud university

+ve work Force acts in the same direction as the displacement

object gains energy

ve work Force acts in the opposite direction as the displacement

object loses energy

Sign of work

cos 90 = 0

No work is done on the load

When force F is at right angles to displacement s (F s) perp

F Δ Δxx cos = 0

No work is done ifFF = 0 orΔΔx x = 0 or = 90o


bull A 1500 kg car moves down the freeway at 30 msKE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ

bull A 2 kg fish jumps out of the water with a speed of 1 ms KE = frac12(2 kg)(1 ms)2= 1 kgm2s2 = 1 J

bullCalculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cmwork = force x distancethe child must exert an upward force equal to its weightthe distance moved upwards equals (12 x 20cm) = 24mwork = 300 N x 24 m= 720 J

example


6-2 Kinetic EnergySame units as work

Remember the Eq of motion

Multiply both sides by m 1

2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi2

2ax

KE 1

2mv2

When work is done on a system and the only change in the system is in its speed the work done by the net force equals the change in kinetic energy of the system

The speed of the system increases if the work done on it is positiveThe speed of the system decreases if the net work is negative

W = F xW = K Work-Energy Theorem

KE f KE i Wnet


6

Kinetic Energy ndash the energy of motion

bull KE = frac12 m v 2

bull units kg (ms) 2 = (kgms2) mbull = Nm = joule

bull Work Energy Theorem W = KEbull Work = the change in kinetic energy of a

systemNote v = speed NOT velocity The direction of motion has no relevance to kinetic energy042023 Norah Ali Al Moneef

king Saud university

Factors Affecting Kinetic Energybull The kinetic energy of an object depends on both its mass and its velocity bull Kinetic energy increases as mass increases bull For example think about rolling a bowling ball and a golf ball down a bowling

lane at the same velocity as shown in Figure 2 bull The bowling ball has more mass than the golf ball bull Therefore you use more energy to roll the bowling ball than to roll the golf ball bull The bowling ball is more likely to knock down the pins because it has more

kinetic energy than the golf ball


bull Kinetic energy also increases when velocity increases bull For example suppose you have two identical bowling

balls and you roll one ball so it moves at a greater velocity than the other

bull You must throw the faster ball harder to give it the greater velocity

bull In other words you transfer more energy to it bull Therefore the faster ball has more kinetic energy

Figure 2Kinetic Energy Kinetic energy increases as mass and velocity increase Predicting In each example which object will transfer more energy to the pins Why


Calculating Kinetic Energybull There is a mathematical relationship between

kinetic energy mass and velocity

Example A 6 kg bowling ball moves at 4 msa) How much kinetic energy does the bowling ball haveb) How fast must a 25 kg tennis ball move in order to have the same kinetic energy as the bowling ball K = frac12 mb vbsup2

K = frac12 (6 kg) (4 ms)sup2 KE = 48 J

KE = frac12 mt vtsup2v = radic2 Km = 620 ms


bull Do changes in velocity and mass have the same effect on kinetic energy

bull Nomdashchanging the velocity of an object will have a greater effect on its kinetic energy than changing its mass

bull This is because velocity is squared in the kinetic energy equation

bull For instance doubling the mass of an object will double its kinetic energy

bull But doubling its velocity will quadruple its kinetic energy


Calculate the speed of a car of mass 1200kg if its kinetic energy is 15 000JK = frac12 m v2

15 000J = frac12 x 1200kg x v2

15 000 = 600 x v2

15 000 divide 600 = v2

25 = v2

v = 25speed = 50 ms-1

example


exampleCalculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms-1 if its brakes exert a constant force of 3 kN

kE of car = frac12 m v2

= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J

The work done by the brakes will be equal to this kinetic energyW = F Δx180 000 J = 3 kN x Δx180 000 = 3000 x ΔxΔx = 180 000 3000

braking distance = 60 m


bull An object does not have to be moving to have energy bull Some objects have stored energy as a result of their positions or

shapesbull When you lift a book up to your desk from the floor or compress a

spring to wind a toy you transfer energy to it bull The energy you transfer is stored or held in readiness bull It might be used later when the book falls to the floor

bull Stored energy that results from the position or shape of an object is called potential energy

bull This type of energy has the potential to do work


Gravitational Potential Energybull Potential energy related to an objectrsquos height is called

gravitational potential energy bull The gravitational potential energy of an object is equal to the

work done to lift it bull Remember that Work = Force times Distance bull The force you use to lift the object is equal to its weight bull The distance you move the object is its height bull You can calculate an objectrsquos gravitational potential energy

using this formulabull Gravitational Potential Energy = Weight x Heightchange in Gravitational potential energy

= mass x gravitational field strength x change in height

ΔU = m g Δh 042023 14Norah Ali Al Moneef


Gravitational potential energyhgmU g

-Potential energy only depends on y (height) and not on x (lateral distance)

-MUST pick a point where potential energy is considered zero

)( oog hhmgUUU

)( oog hhmgUUU


Calculate the change in Gravitational potential energy (gpe) when a mass of 200 g is lifted upwards by 30 cm (g = 98 Nkg-1) ΔU = m g Δh= 200 g x 98 Nkg-1 x 30 cm= 0200 kg x 98 Nkg-1 x 030 mchange in gpe = 059 J

example


Q Calculate the gravitational potential energy in the following systems

a a car with a mass of 1200 kg at the top of a 42 m high hill(1200 kg)( 98mss)(42 m) = 49 x 105

b a 65 kg climber on top of Mt Everest (8800 m high)

(65 kg) (98mss) (8800 m) = 56 x 106 J

c a 052 kg bird flying at an altitude of 550 m (52 kg) (98mss)(550) = 28 x 103 J

example


Conservation of Energybull Law of conservation of energy

- energy cannot be created or destroyedclosed system

- all energy remains in the system- nothing can enter or leave

open system- energy present at the beginning of the system may not be at present at the end


Conservation of Energybull When we say that something is conserved it means that

it remains constant it doesnrsquot mean that the quantity can not change form during that time but it will always have the same amount

bull Conservation of Mechanical EnergyMEi = MEf

bull initial mechanical energy = final mechanical energy

If the only force acting on an object is the force of gravityKo + Uo = K + U

frac12 mvosup2 + mgho = frac12 mvsup2 + mgh042023 19Norah Ali Al Moneef


bull At a construction site a 150 kg brick is dropped from rest and hit the ground at a speed of 260 ms Assuming air resistance can be ignored calculate the gravitational potential energy of the brick before it was dropped

K+U = Ko+Uo

frac12 mv2+mgh=frac12 mv2o+ mgho

frac12 mv2+0 = =o+ mgho

U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example


exampleA child of mass 40 kg climbs up a wall of height 20 m and then steps off Assuming no significant air resistance calculate the maximum(a) gravitational potential energy (gpe) of the child(b) speed of the child

g = 98 Nkg-1

(a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wallΔ U = mgΔh= 40 x 98 x 20max gpe = 784 J

(b) max speed occurs when the child reaches the groundfrac12 m v2 = m g Δh frac12 m v2 = 784 J v2 = (2 x 784) 40v2 = 392v = 392max speed = 63 ms-1


A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distancem = 100 kg vi = 0

Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg

A How much work is done by each force on the cartWg = 0 WN = 0

Wp = Fp Δx cos = (2500 N)(100 m)cos 0Wp = 2500 J

Wk = Fk Δx cos = (500 N)(100 m)cos 180= -500 J= -500 J

example


B How much kinetic energy has the cart gainedWnet = ∆KEWp + Wk = KEf - KEi

2500 J + -500 J = KEf - 0KEf = 2000J

C What is the carts final speed

KE = 12 m v2

v = radic((2KE)(m))v = radic((2(2000 J))(100 kg)) v = 20 ms


Nonconservative Forcesbull A nonconservative force does not satisfy the conditions of

conservative forcesbull Nonconservative forces acting in a system cause a change

in the mechanical energy of the systembull The work done against friction is greater along

the brown path than along the blue pathbull Because the work done depends on the path

friction is a nonconservative force

dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00


On a frozen pond a person kicks a 100 kg sled giving it an initial speed of 22 ms How far does it travel if the coefficient of kinetic friction between the sled and the ice is 10

m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10

W = ∆KEFΔx= Kf - Ki

microk = Fk FN

Fk = microk (mg)microk (-mg) Δx = 12 m vf

2 - 12 m vi2

microk (-mg) Δx = - 12 m vi2

Δx = (- 12 m vi2) microk (-mg)

Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example


A 20-g projectile strikes a mud bank penetrating a distance of 6 cm before stopping Find the stopping force F if the entrance velocity is 80 ms

W =Δ KF Δ x cosθ = K ndash K0

F Δ x cosθ = frac12 mvsup2 - frac12 mvosup2 F (006 m) cos 1800 = 0 - 0 5 (002 kg)(80 ms)2

F (006 m)(-1) = -64 J F = 1067 NWork to stop bullet = change in KE for bullet

Example

W = frac12 mvsup2 - frac12 mvosup2 = 0


A bus slams on brakes to avoid an accident The tread marks of the tires are 80 m long If μk = 07 what was the speed before applying brakes

Work = F Δx (cos θ)

f = μkN = μk mg

Work = - μk mg Δx

ΔK = 12 mv2 - frac12 mvo2

W = ΔK

vo = (2μk g Δ x) frac12

example


bull Power is the rate at which work is donebull Power is the rate of energy transfer by any methodbull The SI unit of power is the watt Wbull 1 watt = 1 Joulesbull 1000 watts = 1 kW Power

Work

time

Work = force middot distancePower = forcemiddotdistancetimePower = forcemiddotvelocityP = Fmiddotv

P KEt

Fxt042023 28Norah Ali Al Moneef


Calculate the power of a car engine that exerts a force of 40 kN over a distance of 20 m for 10 secondsW = FΔx= 40 kN x 20 m = 40 000 x 20 m= 800 000 J

example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W

bull In 1935 R Goddard the America rocket pioneer launched several A-series sounding rockets Given that the engine had a constant thrust of 890 N how much power did it transfer to the rocket while traveling at its maximum speed of 1139 ms

bull P = Fv = 890N x 3139 ms = 279 kW042023 29Norah Ali Al Moneef


30

examplebull Together two students exert a force of 825 N

in pushing a car a distance of 35 mbull A How much work do they do on the carbull W = F Δx = 825N (35m) = bull B If the force was doubled how much work

would they do pushing the car the same distance

bull W = 2F Δx = 2(825N)(35m) =


31

A rock climber wears a 75 kg backpack while scaling a cliff After 300 min the climber is 82 m above the starting point

a How much work does the climber do on the backpack

W = Fd = m g Δx= 75kg(98ms2)(82m)

bHow much power does the climber expend in this effort

P = W = 603 J = 034 watt

t 1800 s

example


Example A 200 kg curtain needs to be raised 8m in as close to 5 s as possible You need to decide among three motors to buy for this each motor cost a little more the bigger the power rating The power rating for the three motors are listed as 10 kw 35 kw and 55 kw Which motor is the best for the job

bull Given m = 200 kg Δx = 8 m ∆t = 5sbull Unknown Power and work

W = FmiddotΔxW = mmiddotgmiddoth

W = (200 kg)middot(981 mssup2)middot(8 m) = 15696 JoulesP = W∆t

P = 15696 J 5 s= 3139 watts042023 32Norah Ali Al Moneef


exampleCalculate the power of a car that maintains a constant speed of 30 ms-1 against air resistance forces of 2 kN

As the car is travelling at a constant speed the carrsquos engine must be exerting a force equal to the opposing air resistance forces

P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW


What power is consumed in lifting a 70-kg robber 16 m in 050 s

A 100-kg cheetah moves from rest to 30 ms in 4 s What is the powerthe work is equal to the change in kinetic energy


Summarybull If the force is constant and parallel to the displacement work is force times distance

bull If the force is not parallel to the displacement

bull The total work is the work done by the net force

bull SI unit of work the joule J

bull Total work is equal to the change in kinetic energy

where

bull Power is the rate at which work is donebull SI unit of power the watt W


Slide 1

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Slide 3

Slide 4

6-2 Kinetic Energy


Factors Affecting Kinetic Energy

Slide 8

Calculating Kinetic Energy

Slide 10

Slide 11

example

6-3 Potential energy and conservative forces

Gravitational Potential Energy

Slide 15

Slide 16

Slide 17

Conservation of Energy

Slide 19

Slide 20

Slide 21

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distance

B How much kinetic energy has the cart gained

Nonconservative Forces


Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35

2

bull W = F Δx cos bull Work = the product of force and displacement

times the cosine of the angle between the force and the direction of the displacement

bull How much work is done pulling with a 15 N force applied at 20o over a distance of 12 m

bull W = FΔx cos bull W = 15 N (12 m) (cos 20) = 169J

20o

15 N

W = F Δx units Nm or joule j



object gains energy


object loses energy

Sign of work

cos 90 = 0



F Δ Δxx cos = 0






example





2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi2

2ax

KE 1

2mv2




KE f KE i Wnet


6






























15 000J = frac12 x 1200kg x v2

15 000 = 600 x v2

15 000 divide 600 = v2

25 = v2

v = 25speed = 50 ms-1

example




= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J




















)( oog hhmgUUU

)( oog hhmgUUU



example





(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

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Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35


object gains energy


object loses energy

Sign of work

cos 90 = 0



F Δ Δxx cos = 0






example





2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi2

2ax

KE 1

2mv2




KE f KE i Wnet


6






























15 000J = frac12 x 1200kg x v2

15 000 = 600 x v2

15 000 divide 600 = v2

25 = v2

v = 25speed = 50 ms-1

example




= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J




















)( oog hhmgUUU

)( oog hhmgUUU



example





(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

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Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35




example





2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi2

2ax

KE 1

2mv2




KE f KE i Wnet


6






























15 000J = frac12 x 1200kg x v2

15 000 = 600 x v2

15 000 divide 600 = v2

25 = v2

v = 25speed = 50 ms-1

example




= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J




















)( oog hhmgUUU

)( oog hhmgUUU



example





(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

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Slide 19

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Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35




2mv f

2 1

2mvi

2 maxKE f KEi Fx

v f2

2vi2

2ax

KE 1

2mv2




KE f KE i Wnet


6






























15 000J = frac12 x 1200kg x v2

15 000 = 600 x v2

15 000 divide 600 = v2

25 = v2

v = 25speed = 50 ms-1

example




= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J




















)( oog hhmgUUU

)( oog hhmgUUU



example





(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35

6






























15 000J = frac12 x 1200kg x v2

15 000 = 600 x v2

15 000 divide 600 = v2

25 = v2

v = 25speed = 50 ms-1

example




= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J




















)( oog hhmgUUU

)( oog hhmgUUU



example





(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35
























15 000J = frac12 x 1200kg x v2

15 000 = 600 x v2

15 000 divide 600 = v2

25 = v2

v = 25speed = 50 ms-1

example




= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J




















)( oog hhmgUUU

)( oog hhmgUUU



example





(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35



= frac12 x 900kg x (20ms-1)2

= frac12 x 900 x 400= 450 x 400k = 180 000 J




















)( oog hhmgUUU

)( oog hhmgUUU



example





(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35

















)( oog hhmgUUU

)( oog hhmgUUU



example





(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35


example





(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35




(65 kg) (98mss) (8800 m) = 56 x 106 J


example















K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35














K+U = Ko+Uo



U0 = mgho = frac12 (150 kg) (260ms)2 = 507 J

example



g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35


g = 98 Nkg-1





Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35


Fp = 2500 N Fk = 500 N

Δx = 100 m FpFk

FN

Fg




example



2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35


2500 J + -500 J = KEf - 0KEf = 2000J


KE = 12 m v2








dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35






dNUKUK

dUKUK

WUKUK

k

k

friction

micro

f

00

00

00



m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35


m = 100 kgvi = 22 ms

vf = 0 ms

microk = 10


microk = Fk FN


2 - 12 m vi2



Δx = (- 12 (100 kg) (22 ms)2) (10 (-100 kg) (98 ms2))Δx = 247 m

example






Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35





Example





f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35



f = μkN = μk mg

Work = - μk mg Δx


W = ΔK


example



Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35


Work

time


P KEt




example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35


example

P = ΔW Δt= 800 000 J 10 spower = 80 000 W




30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35

30




bull W = 2F Δx = 2(825N)(35m) =


31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35

31



W = Fd = m g Δx= 75kg(98ms2)(82m)


P = W = 603 J = 034 watt

t 1800 s

example










P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35









P = F v= 2 kN x 30 ms-1 = 2 000 N x 30 ms-1

power = 60 kW










where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35









where



Slide 1

Slide 2

Slide 3

Slide 4

6-2 Kinetic Energy



Slide 8


Slide 10

Slide 11

example



Slide 15

Slide 16

Slide 17


Slide 19

Slide 20

Slide 21





Slide 26

Slide 27

6-9 power

Slide 29

example

Slide 31


Slide 33

Slide 34

Slide 35

11/1/20151 Norah Ali Al Moneef king Saud university.

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Transcript of 11/1/20151 Norah Ali Al Moneef king Saud university.