mg

FN

x

FN

mg

fk fk

s

s

0EEW fNC If no net non-conservative forces

Then, conservation of mechanical energy holds

EEEW fNC 00

PEKE Crate on Incline Revisited

FBD

h

NCkf

Cg

N

WsmgWWmghsmgW

W

cos sin

0

The crate starts from rest, v0=0

0

0

0

000

cos

,0

EEsmgmghWEE

EEWWmghPEEKE

f

KNCf

fNC

g

Some energy, WNC is loss from the system

In this case it is due to the non-conservative friction force energy loss in the form of heat

Because of friction, the final speed is only 9.3 m/s as we found earlier

If the incline is frictionless, the final speed would be:

sm

221

00

3.12kg 100

J) 7510(22v

v0,0 since ,

m

W

WmghmPEKEEE

gf

gf

ff

Because of the loss of energy, due to friction, the final velocity is reduced. It seems that energy is not conserved

Conservation of Energy There is an overall principle of conservation of energy

Unlike the principle of conservation of mechanical energy, which can be broken, this principle can not

It says: ``The total energy of the universe is, has always been, and always will be constant. Energy can neither be created nor destroyed, only converted from one form to another.’’

So far, we have only been concerned with mechanical energy

There are other forms of energy: heat, electromagnetic, chemical, nuclear, rest mass (Em=mc2)

QEEWWEE

EEEEEEEE

EE

othersf

othersNC

NCmechmech

f

othersf

othersmechmechf

othersf

mechf

othersmech

totalf

total

0

0

00

00

0

Q (WNC) is the energy lost (or gained) by the mechanical system

The electrical utility industry does not produce energy, but merely converts energy Lake

River

Hydro-power plant

h

PE=mgh

KE=mv2/2

electricityLight,

heat

Power

Average power: t

WP

Units of J/s=Watt

(W)

Measures the rate at which work is being done

v cos cos

Ft

sFP

ExampleA car accelerates uniformly from rest to 27 m/s in 7.0 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 1.2x104 N, and (b) the weight of the car is 1.6x104 N.

Solution:

Given: v0=0, vf=27 m/s, t=7.0 s, (a) mg= 1.2x104 N, (b) mg= 1.6x104 N

Method: determine the acceleration

t

sF

t

WP

cos

We don’t know the displacement s

The car’s motor provides the force F to accelerate the car – F and s point in same direction

s

F

mm

t

mP

sasa

t

sma

t

FsP

f

fssf

s

522

027

s) 0.7(2

)vv(2

)vv(2vv

220

2

20

220

2

Need ass

Wx105.880.9

)x106.1(52

Wx104.680.9

)x102.1(5252

44

44

P

g

mgP (a)

(b)

Or from work-energy theorem

t

m

t

WP

mKEW

f

f

2

vv

2

221

Same as on previous slide

Example: Problem 6.77A particle starting from point A, is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height of 4.00 m above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A.

3.00 m4.00 m

A0v