chapter 7 work and kinetic energy adapted from hyde-wright, odu ©james walker, physics, 2nd ed....
TRANSCRIPT
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
12]-2 [Eq.
2
1
2
10
2
10
2
10
2
10
2
1
:)( into Substitute
: offunction a as for Solve
line.straight a is slope whosecurve theis of parabola The2
1)(
)(
20
2
00
0000
0000
200
00
0
200
0
yyy
yyyyy
yyyyyyyy
yyyyyyyy
y
yyy
y
yyy
y
yy
yy
yyy
vv ) (y-ya
vvvv ) (y-ya
vvvvvv v) (y-ya
vvvvvv v) (y-ya
a
vva
a
vvvyy
ty
a
vvtvt
v(t)y(t)
tatvyty
tavtv
Motion in one DirectionMotion in one DirectionConstant Acceleration
2iy
2fyifNet
2iy
2fyify
2iy
2fyify
mv2
1mv
2
1 ) -y(yF
mv2
1mv
2
1 ) -y(yma
v2
1v
2
1 ) -y(ya
• Net (constant) force, Fnet, times displacement, y-y0, equals change in kinetic energy.
• Kinetic Energy:
• Force times Displacement equals Work:
2
2
1mvK
)( if yyFW
Constant acceleration motion in two or more dimensions (e.g. projectile motion)
• Just add the equations for each coordinate
2ix
2fxifxNet
2iy
2fyifyNet
mv2
1mv
2
1 ) -x(xF
mv2
1mv
2
1 ) -y(yF
,
,
2i
2f
2iy
2ix
2fy
2fxifyNetifxNet
vm2
1vm
2
1
vvm2
1vvm
2
1)-y(yF) -x(xF
,,
Work done by net force = change in Kinetic energy
Kinetic Energy
An object in motion has kinetic energy:
m = massv = speed (magnitude of velocity)v2 = vx
2+ 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)
221 mvK
Work done by a force
• For a constant force, F, the work done ON a mass m while the mass moves through a displacement r = r2r1is (switch from rr0to r2r1notation) 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 r makes an angle with respect to the x-axis. Fx = F Fy = 0
Define d = | r | x2 x1 = d·cos
• W = (F) (d) cos
F
r2
r1
r= r2 r1
F
r= r2 r1
x
y
|r|=
d
• 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 = 0, and
W= Fd
• If 90º<<270º, then W = F d cos 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 cos)d = Ftotald cos
Walker Problem 9, pg. 194A 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 Jb) 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 = KE = 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 livesActivity 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 103
Flea hop 103
One blue photon, absorbed on retina 5 ·1019
Breaking a bond in DNA 1020
Walker Problem 22, pg. 195A 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?
a) Friction = only horizontal forceWnet = Kf-Ki <0Wnet = 0 – (1/2) m vi
2
Wnet = (1/2) (10kg+65kg) (12m/s)2
Wnet = 5400 kg m 2 /s 2 = 5400 J
b) Find acceleration firstv = v0+at0 = v0 + at, a= v0/ta = (12m/s)/(4s) = 3 m/s2
x= x0+v0 t+(1/2)at2
x= 0 + (12m/s)(4s) + (0.5)(-3m/s2)(4s)2
x= 48 m-24 m = 24 mc) Braking force
Wnet = Fnet d = FFriction d
FFriction = Wnet /d = ( Nm)/(24m) = 225 N
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 forceWalker 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 JW=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=0.85mb) 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 forceWalker 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 JW = Kf – 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
W = -½ kx2
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:
Area of triangle = (one half) (base times height)
W = (1/2)(x)(-kx)
F
xF= kx
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 kx2Fig 7-11§7-3
Fig. 7-13
• 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).
PowerAverage Power is defined as the time rate of
doing work :
The unit of power is the Watt (W).
1W = 1Joule/sec
Fvt
Fd
t
WP
PowerExample 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. 196In 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 NForce 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/sPower = F· v = (44.6 N) (2.0 m/s) = 89.2 WP = (89.2 W) / (746 W/hp) = 0.060 hp
Work = Change in Kinetic Energy Fd = ½ mvf
2 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