work and energy
DESCRIPTION
Work and Energy. Scalars are back. Review. Equations for Motion Along One Dimension. Review. Motion Equations for Constant Acceleration. 1. 2. 3. 4. Review. 3 Laws of Motion If in Equilibrium If not in equilibrium Change in Motion is Due to Force - PowerPoint PPT PresentationTRANSCRIPT
WORK AND ENERGYScalars are back
REVIEW Equations for Motion Along One Dimension
dtdx
txv
txv
t
ave
0lim
dtdv
tva
tva
t
ave
0lim
REVIEW Motion Equations for Constant Acceleration
•1.
•2.
•3.
•4.
atvv 0
221
00 attvxx
20vvvave
xavv 220
2
REVIEW 3 Laws of Motion If in Equilibrium
If not in equilibrium Change in Motion is Due to Force
Force causes a change in acceleration
0F
maF
SPRINGS AND OTHER PROBLEMS Force exerted by a
spring is dependent on amount of deformity of the spring
Amount of force applied changes continuously over time
What is the velocity of an object launched from the spring?
WORK Work done on an
object by all forces is equal to the change in kinetic energy of the object.
This definition is valid even if the force is not constant
WORK – CONSTANT FORCE When a force, F, is
doing work on an object, the object will move and be displaced.
The work done, by the force, F, is defined as
Where d is the objects displacement
FdW
WORK – CONSTANT FORCE We are only
interested in the component of the force that is parallel to the direction of motion dFW ||
WORK – CONSTANT FORCE We are only
interested in the component of the force that is parallel to the direction of motion
or
cosFdW
dFW
JOULE
Work done by 1N of force to move an object 1 meter in the same direction
cosFdW
JJouleWmNWmNW
111
11
JAMES PRESCOTT JOULE December 24, 1818-
October 11, 1889 The mechanical
equivalent of heat 838 ft.lbf of work to
raise temperature of 1 lb of water by 1 degree farenheit
Led to the theory of conservation of energy
Helped Lord Kelvin develop the absolute scale of temperature
WORK – ZERO, NEGATIVE, POSITIVE When defining work
done, its always important to specify which force is acting on what object Work done by man Work done by
gravity Work done by
barbell
TOTAL WORK Compute work done
by forces individually
Then just add to get total work done on the object
Note: work is scalar
...21 WWWWtot
EXAMPLE Farmer hitches a tractor
with firewood and pulls it a distance 20m on level ground. Total weight of the sled and wood is 14700N and the tractor pulls with a constant force of 5000N at an angle 36.9o above the horizontal. There is a 3500N friction force opposing the motion. Find the work done by each of the forces and the total work done by all forces.
EXAMPLE
JW
W
dFWJW
WW
dFW
friction
friction
frictionfriction
tractor
tractor
tractor
Tractortractor
70000
)180cos()20)(3500(
cos8000079968
)9.36cos()20)(5000(cos
WORK DONE BY NON-CONSTANT FORCE Requires the use of integrals
ENERGY Energy is a hard to
define concept Simplified definition The ability of a
physical system to do work on another physical system
Many types of energy- these are much easier to define
KINETIC ENERGY Energy of motion When work is done
to an object the object moves
It also affects an objects speed W>0 – object speeds
up W<0 – object slows
down W=0 – no effect
KINETIC ENERGY
Newton’s 2nd LawFdW
madW
KINETIC ENERGY
20
2
20
2
20
2
20
2
20
2
20
2
21
21
2
2
2
2
2
mvmvW
vvmW
ddvvmW
dvva
advv
advv
madW
KINETIC ENERGY
Work done is the change in kinetic energy of an object
This is translational kinetic energy
20
2
21
21 mvmvW
2
21 mvK
WORK – ENERGY THEOREM Assuming mass is
constant
Unit of work is Joules Unit of energy is also
Joules
Note: Energy is also scalar
KW
mvW
2
21
EXAMPLE Farmer hitches a tractor
with firewood and pulls it a distance 20m on level ground. Total weight of the sled and wood is 14700N and the tractor pulls with a constant force of 5000N at an angle 36.9o above the horizontal. There is a 3500N friction force opposing the motion. Suppose it’s initial speed is 2.0 m/s, what is its final speed after travelling 20m.
EXAMPLE
sm
total
total
total
vv
v
v
vvmW
KEWJW
41633.4
2333.13
)2)(1500(2110000
)(21
10000
22
22
20
2
EXAMPLE A 15kg block is
placed on a 40o incline and allowed to slide for 5m. What is it’s final speed? kg15
POTENTIAL ENERGY Energy due to a
body’s configuration or surroundings.
Many different types Springs Electrical Gravitational
GRAVITATIONAL POTENTIAL An object held in the
air has the “potential” to do work once released.
Assume object at some height
After travelling some distance y
00 v
gyv
ygv
advv
2
))((2
2
2
2
20
2
GRAVITATIONAL POTENTIAL An object held in the
air has the “potential” to do work once released.
KE after travelling some distance y
mgyK
mgymvK
gyv
221
212
2
2
GRAVITATIONAL POTENTIAL An object held in the
air has the “potential” to do work once released.
Amount of potential work
mgyKWmgyK
K
00
GRAVITATIONAL POTENTIAL An object held in the
air has the “potential” to do work once released.
Note: choose your origin and be consistent
mgyUPEgrav
EXAMPLE- GIANCOLI 6-28 By how much does the gravitational potential
energy of a 64-kg pole vaulter change if his center of mass rises 4.0m?
EXAMPLE- GIANCOLI 6-28 By how much does the gravitational potential
energy of a 64-kg pole vaulter change if his center of mass rises 4.0m?
JUJU
mgUmgUmgyU
25002500
)4)(8.9)(64()4(0)0(0
WORK DONE EXAMPLE What is work done
to lift a block by 5 m?
If a 40o was used?kg15
CONSERVATIVE AND NON-CONSERVATIVE FORCE Conservative Force Work Done is
independent of the path taken Gravity Elastic Electric
You can “store” energy in these types of systems by doing work on the system
Non Conservative Force
Work done depends on the path taken Friction Air resistance Tension Push-Pull from a
person Cannot define
potential energy for these types of forces
CONSERVATION OF MECHANICAL ENERGY If only gravity is
acting on the object
Valid for all conservative forces
If only conservative forces are acting, the total mechanical energy of a system neither increase nor decrease in any process. It stays constant- it is conserved.00
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grav
CONSERVATION OF MECHANICAL ENERGY If a non-
conservative force is acting on the object
Most common non-conservative energy is friction
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grav
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NCgrav
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EXAMPLE – FROM OUR 2ND LECTURE A motorcycle stuntman rides over a cliff. Just
at the cliff edge his velocity is completely horizontal with magnitude 9.0 m/s. Find the motorcycles speed after 0.50s.
LIST THE GIVEN Origin is cliff edge a=-g=-9.80m/s2
At time t=0s
At time t=0.50s
0v
v00 x 00 y
?d?v
smv 0.90
SPLIT INTO COMPONENTS
0v
v
yDxD
DDD
y
x
yx
yx vvv
sm
xv 0.90
00 yv
CALCULATE COMPONENTS INDEPENDENTLY
0v
vxv
yv
sm
xx vv 0.90
sm
y
y
yy
v
gtv
gtvv
9.4
)5.0)(8.9(0
CALCULATE VELOCITY
0v
vxv
yv
sm
xv 0.9s
myv 9.4
sm
sm
yx
xvv
vvv
100.125.10)9.4()0.9( 22
22
NOT NEEDED
29o below the horizontal
0v
v
xv
yv
sm
xv 0.9s
myv 9.4
smxv 100.1
544.099.4tan
x
y
vv
2956.28
smxv 100.1
ALTERNATE SOLUTION
0v
vxv
yv
0
0.9
0
202
10
0
00
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vUKUK
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ALTERNATE SOLUTION
0v
vxv
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ALTERNATE SOLUTION
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vxv
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0
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22
20
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0
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10
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vv
v
gyvv
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PROBLEM – YOUNG AND FREEDMAN 7.14 A small rock with mass 0.12 kg is fastened to
a massless string with length 0.80 m to form a pendulum. The pendulum is swung so that it makes a maximum angle of 45o with the vertical. (a) What is the speed of the rock when it passes the vertical position? (b) What is the tension in the string when it makes an angle 45o with the vertical? (c) What is the tension in the string when it passes through the vertical?
PROBLEM – SERWAY 7.33 A crate of mass 10.0 kg is pulled up a rough
incline with an initial speed of 1.50 m/s. The pulling force is 100N parallel to the incline, which makes an angle of 20o with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline system due to friction. (c) How much work is done by the 100N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5m?
OTHER TYPES OF POTENTIAL ENERGY Elastic Potential For Ideal Springs If a spring is to be
stretched a certain distance x
Where k is the spring constant (the spring’s stiffness)
It’s me again
kxF
POTENTIAL ENERGY OF SPRINGS Restoring Force
Hooke’s Law – valid for small x
kxFs
POTENTIAL ENERGY OF SPRINGS Work done ON the spring
(from equilibrium)
NO Force is not constant We can still use average
force Luckily F varies linearly
with x
))(( xkxWFdWkxF
POTENTIAL ENERGY OF SPRINGS Work done ON the spring
(from equilibrium)
Where U is the elastic potential
221
221
21
021
0
))((
)(
0)0(
kxU
kxW
xkxWdFW
FFFkxFkF
ave
ave
CONSERVATION OF MECHANICAL ENERGY EXPANDED Conservative
With Non conservative
springgravspringgrav UUKUUK 000
NCspringgravspringgrav WUUKUUK 000
YOUNG AND FREEDMAN 7.20 A 1.20kg piece of
cheese was placed on a vertical spring of negligible mass and force constant k=1800 N/m that is compressed 15.0 cm. When the spring is released how high does the cheese rise from its original position?
POWER Rate at which work
is done
SI unit is called the Watt = 1J/s
Horsepower = 550ftlb/s = 746W
TimeWorkPAve
POWER Rate at which work
is done
Efficiency
TimeWorkPAve
aveAve FvtFdP
in
out
PPe
EXAMPLE GIANCOLI 6-58 How long will it take a 1750W motor to lift a
315 kg piano to a sixth story window 16.0m above?
EXAMPLE GIANCOLI 6-58 How long will it take a 1750W motor to lift a
315 kg piano to a sixth story window 16.0m above?
sPWt
JFdWTimeWorkPAve
2.28
4939216)8.9(315
1750
PROBLEM SERWAY 7.40 A 650 kg elevator starts from rest. It moves
upward for 3s with constant acceleration until it reaches its cruising speed of 1.75m/s. (a) What is the average power of the elevator motor during this period? (b) How does this compare when the elevator moves at cruising speed?
YOUNG AND FREEDMAN – 7.42 A 2.00 kg block is pushed against a spring
with negligible mass and force constant k= 400 N/m, compressing it 0.220m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0o. (a) what is the speed of the block on the horizontal surface after leaving the spring? (b) How far up the slope does the block travel before starting to slide back down?
GIANCOLI 6-56 A 280 g wood block is firmly attached to the
end of a horizontal spring. The block can slide along the table with a coefficient of friction of 0.30. A force of 22 N compresses the string 18 cm. if the spring is released, how far from the equilibrium position will it stretch at its first maximum extension.
GROUP WORK A 1500 kg rocket is to be launched with an
initial upward speed of 50.0 m/s. In order to assist the engines, the engineers will start it from rest on a ramp that rises 53o above the horizontal. The engines provide a constant forward thrust of 2000N and the coefficient of kinetic friction with the ramp is 0.05. At what height should the rocket start?