Transcript document

Chapter 7
Work and Kinetic Energy
Adapted from Hyde-Wright, ODU
©James Walker, Physics, 2nd Ed. Prentice Hall
Energy
• How are we here?





Nuclear energy in the sun,
is converted into light, which is absorbed by chlorophyll in plants,
making chemical energy from food,
which undergoes many transformations, until
our muscles convert the chemical energy into motion (kinetic energy).
• Energy can be defined (somewhat vaguely at this point) as the
ability to create change.
• The book E=mc2 documents how it took over a century after
Newton for the concept of energy to take hold.
Equations of
constant acceleration
v y (t )  v0 y  a y t
1
y (t )  y0  v0 y t  a y t 2
2
The parabola of y(t) is the curve whose slope v(t) is a straight line.
v y  v0 y
Solve for t as a function of v :
t
ay
Substitute into y (t ) :
2
 v y  v0 y  1  v y  v0 y 
y  y 0  v0 y 
  ay 

a
y

 2  a y 
1
a y(y-y 0 )  v0 y v y  v0 y  v y  v0 y v y  v0 y
2
1
a y(y-y 0 )  v0 y v y  v0 y  v y  v0 y v y  v0 y
2
1
a y(y-y 0 ) 
v y  v0 y v y  v0 y
2
1
1
a y(y-y 0 )  v 2y  v02 y [Eq. 2 - 12]
2
2


 



 




Motion in one Direction
Constant Acceleration
1 2 1 2
a y(y f -yi )  v fy  viy
2
2
1 2 1 2
ma y(y f -yi )  mv fy  mviy
2
2
1
1
FNet (y f -yi )  mv2fy  mviy2
2
2
• Net (constant) force, Fnet, times displacement, y-y0, equals change in kinetic
energy.
1 2
• Kinetic Energy:
K  mv
2
• Force times Displacement equals Work:
W  F ( y f  yi )
Constant acceleration motion in two or
more dimensions (e.g. projectile motion)
• Just add the equations for each coordinate
1 2 1 2
mv fy  mviy
2
2
1
1
 FNet , x(x f -xi )  mv 2fx  mvix2
2
2
FNet , y(y f -yi ) 



1
1
m v 2fx  v 2fy  m vix2  viy2
2
2
1  2 1 
 mv f  mvi2
2
2
FNet , x(x f -xi )  FNet , y(y f -yi ) 

Work done by net force = change in Kinetic energy
Kinetic Energy
An object in motion has kinetic energy:
2
1
K  mv
2
m = mass
v = speed (magnitude of velocity)
v2 = vx2+ vy2
(Pythagoras)
The unit of kinetic energy is Joules (J).
Kinetic energy is a scalar (magnitude only)
Kinetic energy is non-negative (zero or positive)
Work done by a force
• For a constant force, F, the work done ON a mass m while
the mass moves through a displacement Dr = r2r1 is
(switch from rr0 to r2r1 notation)
 W = Fx(x2- x1) + Fy (y2 –y1) = work done by force F
 Work = [x-component of Force]  [x-component of displacement]
+ [y-component of Force]  [y-component of displacement] …
• Total work is the work done by the Net force.
 WTotal = Fnet,x(x2- x1) + Fnet,y (y2 –y1)
• Total work = sum of the work done by each force
 Fnet = F1 + F2 + F2 +…
 WTotal = W1 + W2 + W3 +…
Work,
in terms of magnitude & direction of force and displacement
• W = Fx(x2- x1) + Fy (y2 –y1) = work done
by force F
• Pick a coordinate system with the x-axis
along the direction of the force F
(remember, because of initial velocity and/or
other forces present, this force F does not
have to be parallel to the displacement)
• The displacement Dr makes an angle q
with respect to the x-axis.
 Fx = F
 Define d = | Dr |
 x2  x1 = d·cosq
• W = (F) (d) cosq
Dr= r2  r1
r2
r1
F
Dr= r2  r1
Fy = 0
y
q
x
F
• Work is a scalar
 Work has only magnitude, no direction.
 The value of W is independent of how we draw the
coordinate system (unlike the components of r or F)
• The SI unit of work is the Joule (J)
1 Joule  1 Newton·meter
1 J = 1 Nm = 1 kg m2/s2
• Note that if F is in the same direction as d, then q
= 0, and
W= Fd
• If 90º<q<270º, then
W = F d cosq is negative.
Negative Work and Total Work
Work can be positive, negative or zero depending on
the angle between the force and the displacement.
If there is more than one force, each force can do work.
The total work is calculated from the total (or net) force:
Wtotal = (Ftotal cosq)d = Ftotald cosq
Walker Problem 9, pg. 194
A 55-kg packing crate is pulled with constant speed
across a rough floor with a rope that is at an angle of
40.0° above the horizontal. If the tension in the rope
is 125 N, how much work is done on the crate to
move it 5.0 m?
a) How much work is done by the tension in the rope?
b) How much work is done by the floor (friction
force)?
c) How much work is done by the net force?
a) Wtension = (125 N) (5.0m) cos40.0º=625 J
b) Fnet = 0 (constant speed)
• 0 = Tx + Ffriction
• Wfriction = (125N)(cos40.0)(5.0m) =625J
c) WTotal = 0
Work-Energy Theorem
The net (total) work done on an object by the
total force acting on it is equal to the change
in the kinetic energy of the object:
Wtotal = DKE = KEfinal - KEinitial
Forms of Energy
• Kinetic Energy is one form of energy.
• Work is a transformation from Kinetic Energy
to another form of energy (potential energy,
dissipation into heat, see Chap 8…)
• 1 Calorie (nutrition) = 1kcal = 5.18·103 Joule
Typical Values of Work in our lives
Activity
Work or Energy (Joules)
Annual US energy use
8·1019
Mt St Helens (1981)
1018
Burning one gallon of gas
108
Human diet, one day
107
Melting an ice cube
104
100 watt light bulb for 1 hour
0.1 kiloWatt-Hr = 3.6·105
1 Human heart beat (pumping blood)
0.5
Turning page in book
103
Flea hop
103
One blue photon, absorbed on retina
5 ·1019
Breaking a bond in DNA
1020
Walker Problem 22, pg. 195
A 65-kg bicyclist rides his 10.0-kg bicycle with a speed of 12 m/s.
(a) How much work must be done by the brakes to bring the bike and rider to a
stop?
(b) How far does the bicycle travel if it takes 4.0 s to come to rest?
(c) What is the magnitude of the braking force?
b) Find acceleration first
Friction = only horizontal force
v = v0+at
Wnet = Kf-Ki <0
Wnet = 0 – (1/2) m vi2
0 = v0 + at, a= v0/t
a = (12m/s)/(4s) = 3 m/s2
Wnet =  (1/2) (10kg+65kg) (12m/s)2
x= x0+v0 t+(1/2)at2
Wnet =  5400 kg m 2 /s 2 = 5400 J
x= 0 + (12m/s)(4s) + (0.5)(-3m/s2)(4s)2
x= 48 m-24 m = 24 m
c) Braking force
Wnet = Fnet d = FFriction d
FFriction = Wnet /d = (5400 Nm)/(24m) = 225 N
a)
Force vs. Displacement Graph
The work done by a force can be found from the area
between the force curve and the x-axis (remember, area
below the x-axis is negative):
Work done by a variable force
Walker Problem 27, pg. 195
An object is acted on by the force
shown in the Figure. What is the
final position of the object if its
initial position is x = 0.50 m and
the work done on it is equal to (a)
0.12 J or (b) –0.29 J? (c) If the
force acts on a 0.13-kg object and
the initial velocity is 3 m/s, find
the object’s speed at x = 0.85 m.
a)
Area from 0.5m to 0.75 m = (0.4N)(0.25m) = 0.10 J
W=0.12 J = 0.10J + 0.02 J = 0.10 J + (0.2N)(x-0.75m)
0.02 J = (0.2 N)(x-0.75m)
(x-0.75m) = (0.02 J) / (0.2 N) = 0.1m
x=0.85m
b) Area from 0.5 m to 0.25 m = (0.8N)(0.25m)= 0.20J
W = 0.29 J = 0.20J + (0.6N)(x-0.25m)
0.09J = (0.6N)(x-0.25m)
(x-0.25m) = 0.15m
c) Area from 0.25m to 0.00 m = (0.6N)(0.25m) = 0.15J
Work done by a
variable force
Walker Problem 27, pg. 195
(c) If the force acts on a 0.13-kg
object and the initial velocity is 3
m/s, find the object’s speed at x =
0.85 m.
c)
Work = (0.4 N)(0.75m-0.5m) + (0.2 N)(0.85m-0.75m)
W = 0.10 J + 0.02 J = 0.12 J
W = K f – Ki
Kf = Ki + W
Kf = (0.5) m[vi2] + W = (0.5) (0.13kg)(3m/s) 2 + 0.12J
Kf = 0.585 J + 0.120J = 0.705 J
Kf = (0.5) m[vf2]
vf2 = 2 Kf / m = 2 (0.705 kg m2/s2)/ (0.13 kg) =10.85 m2/s2
Vf =3.29 m/s
Work Done by a
Hooke’s Law Force (Spring force)
If the force depends on the displacement, then it will
not be constant. Example: spring F = kx
F
Work done on mass m by a
spring as the mass stretches (x>0)
or compresses (x<0) the spring a
distance x from equilibrium:
F= kx
x
Area of triangle = (one half)
(base times height)
W = (1/2)(x)(-kx)
W = -½ kx2
Work & Spring-Force
• Text describes “Work done by a force
to stretch a spring”
• If a spring is stretched a distance x
(positive or negative) by some other
force (e.g. your hand pushing) such
that the mass on the end of the spring
travels at constant velocity (a=0) then
the net force = 0 and the pushing
force =  (spring force) = (kx) = kx
• Work by external force = (1/2) x (kx)
• W = ½ kx2
Fig 7-11
§7-3
• Fig 7-13 (motion of a car) does not explain that the backwards force labeled
Ffriction is actually kinetic friction in the bearings of the axle etc. and rolling
resistance of the tires as they flex.
• The figure also fails to explain that the forward force F is actually static
friction between the tires and road (which is the action-reaction partner to the
force of the tires pushing backwards on the road).
Fig. 7-13
Power
Average Power is defined as the time rate of
doing work :
W Fd
P

 Fv
t
t
The unit of power is the Watt (W).
1W = 1Joule/sec
Power
Example
Power (Watt)
Total US Electrical Generation
1012 W = 1TW
Large nuclear power plant
109 W = 1GW = 1000 MW
SUV on highway (12 mpg)
1.5·105
175 hp engine
(175hp)(746W/hp)= 1.3·105
100 W light bulb
100
Human metabolism (daily average)
80
Solar energy reaching 1 m2 of
earth’s surface at Norfolk on a clear
day
250
Walker Problem 39, pg. 196
In order to keep a leaking ship from sinking, it is necessary to
pump 10.0 lb of water each second from below deck up a
height of 2.00 m and over the side. What is the minimum
horsepower motor that can be used to save the ship?
2.2lb = (1kg)(9.81 m/s2) = 9.81 N
Force applied to water =
10 lb = (10 lb)(9.81 N)/(2.2 lb) =44.6 N
Velocity of water = (2.00 m) / (1.0sec) = 2.00 m/s
Power = F· v = (44.6 N) (2.0 m/s) = 89.2 W
P = (89.2 W) / (746 W/hp) = 0.060 hp
Work = Change in Kinetic Energy
Fd = ½ mvf2  ½ mvi2
• A car of mass m travels with speed v.
 A braking force F brings the car to rest in distance d.
• The car’s speed is now 2v.
 The same braking force F is applied to bring the car to rest.
 What is the new stopping distance?
•
•
•
•
d/2
d
2d
4d