physics chapter 5 work and energy work work - (if force is constant) – is the product of the force...
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Physics Chapter 5
Work and Energy
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WorkWork - (if force is constant) – is the product of the force exerted on an object and the distance the object moves in the direction of the force.W = F||d
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Work is a scalar quantity and can be positive or negative, but you don’t have to indicate direction/angle in your answer.
Unit is Joule (N•m)The object must MOVE in the DIRECTION
(positive or negative) that the force is exerted on THE OBJECT.Holding a flag in the air (no work on the flag)Walking forward while holding the flag in the
air (no work on flag)
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Work and Direction of Force
If force is applied at an angle, then the work must be computed using the component of force in the direction of the motion.Ex: you are pushing a lawnmower…
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Work and Direction of Force
Force (parallel to ground) doing work is FcosWork (W) = Fcosd= FdcosWork that the grass/lawn is doing on the
mower = Ffrictiond
F
dFf
Fcos
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ExampleHow much work is done on a
vacuum cleaner pulled 4.0 meters by a force of 45 N at an angle of 30.0 degrees above horizontal?
Here’s the equation: W = Fd = F cos d = Fdcos
W = Fd = F cos d = 45Ncos30X 4.0 m = 156 Joules
30
45N
Fcos
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D
mg
mgx
Wgrav = mg sin d
Or Wgrav = mgd if lifted vertically
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Gravity ExampleHow heavy is a barrel that is
lifted 45 meters vertically using 670 joules of work?
Formula for work done against gravity.
Wagainst grav = -Wgrav =-mgd670J = -mgdm=-670J/(gd) = -670/(-9.81m/s2X45m) = 1.5 kg
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Practice 5A page 170 and Section Review on
page 171
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Happy Thursday!
Please take out 5A and section review for me to check.
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Energy - an object has energy if it can produce a change in itself or its surroundings.
Kinetic Energy - energy due to the motion of an object.
KE = 1/2mv2
Kinetic Energy
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Doing Work to Change Kinetic Energy
Kinetic Energy is related to the amount of work done on the object.Work done equals the KE
gained by the object.
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Example pg. 173
A 7.0 kg bowling ball moves at 3.00m/s. How much kinetic energy does the bowling ball have? How fast must a 2.45 g ping pong ball move in order to have the same kinetic energy as the bowling ball?
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Known, unknowns and formulas
mbb=7.0kg
Vbb=3.00m/s
mppb=2.45g = 0.00245kg
KEbb = 1/2mbbvbb2 = KEppb
= 1/2mppbvppb2
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Solve for Kinetic Energy of Bowling Ball
KEbb = 1/2mbbvbb2 = (½(7.0kg)
(3.00m/s)2=
=31.5 J = 32J
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Solve for vppb
KEbb = 32J = KEppb = 1/2mppbvppb2
vppb = 32J/(1/2mppb)
vppb = 32J/[(1/2)(0.00245kg)]
vppb =160m/s
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Example 2
What is the speed of a 0.20 kg baseball that has a Kinetic Energy of 390Joules?
m= 0.20 kg, KE = 390J, v=?KE = ½ mv2 v= √2KE/mv= √2(390J/0.20kg) = 62m/s
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Example 3What is the mass of a
satellite traveling at 92 km/h with a kinetic energy
of 240,000 J
KE = 240,000, v=92km/h?????92km/hX1000/3600=26m/sKE=1/2mv2 , m=2KE/v2 m=2(240,000J)/(26m/s)2 = 710kg
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5B page 174
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Effect of Doing Work
As positive work is done on an object (lifting a box), energy is transferred to that object (it can now fall and crush something).
As negative work is done on an object (lowering a box), energy is transferred from the object to the one lowering the box.
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The Work-Energy TheoremThe net work done on an
object is equal to its change in kinetic energy.Wnet = KEf - KEi = KE
If the net work is positive, the kinetic energy increases.
If the net work is negative, the kinetic energy decreases.
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Example
On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.20 m/s. How far does the sled move if the coefficient of kinetic friction between the sled and the ice is 0.10?
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Known, unknowns and formulasms=10.0kg
Vs=2.20m/sk = 0.100KEf = 0Fnet = Fk…the only force working
once the sled is kicked, but it is NEGATIVE in direction!!
Fnet=-Fk=-k Fn = k mgWnet = -Fkd = KEf – KEi
-k mg d = -KEi = -1/2msvs2
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Solve for d
d =-1/2msvs2 /-(k mg)=
-1/2(10.0kg)(2.20m/s)2/-(0.10)(10.0kg)(9.81m/s2)
=2.46mIt still travels a positive distance, but it
slows down as it is traveling so its change in kinetic energy, friction and acceleration are negative.
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Practice 5C
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Potential Energy - energy stored in an object because of its state or position.
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Gravitational potential energy depends on an object’s position above a zero level.PEg = mghPEg = mghf - mghi = Work done
Potential energy can be stored in bent or stretched objects=Elastic Potential Energy
PEelastic=1/2kx2
x = distance compressed or stretchedk = spring constant
g is positive for potential energy!
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What is the Potential Energy of a 24.0 kg mass 7.5 m above the ground
PEg = mgh =
(24.0)(9.81)(7.5m) = 1.8X103 J
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What is the Potential Energy of a spring
stretched 5.0 cm that has a spring constant of 79
N/m?PEE = ½ kx2 =
½ (79N/m)(.05m)2 = = 0.099J
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Total Potential energy =Elastic Potential Energy + Gravitational Potential Energy
PETotal=1/2kx2 + mgh
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ExampleA 70.0 kg stuntman is attached to
a bungee cord with an unstretched length of 15.0 m. He jumps off a bridge spanning a river from a height of 50.0 m. When he finally stops, the cord has a stretched length of 44.0 m. Treat the stuntman as a point mass, and disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling?
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Known, unknows and formulasms=70.0kg
hi=50.0m, hf=?...hf= 50m-44m=6.0mLi= 15.0mLf = 44.0mL = 29.0m =xk=71.8N/mPETotal=1/2kx2 + mghf
PETotal=1/2(71.8N/m)(29m)2+ (70.0kg)(9.81m/s2)(6.0m) =
3.43X104J
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Swing problem…
hyp adj
Height above zero point = hyp-adj
(hyp)cos = adj
swing
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Problems 5D, page 180 and section review
page 181
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Potential and Kinetic Energy
When a ball is thrown into the air, the kinetic energy you give the ball is transferred to potential energy and then back into the same amount of kinetic energy -- Total Energy is constant.ETOTAL = KET + PET
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Mechanical Energy
Mechanical Energy (ME) is the sum of KE and ALL forms of PE associated with an object or group of objects.
ME = KE + PE
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Law of Conservation of Energy
States that within a closed, isolated system, energy can change form, but the total amount of energy is constant.(Energy can be neither created nor destroyed).
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MEi=MEf
KEi + PEg,i+ PEE,i = KEf + PEg,f+ PEE,f
½ mvi2+mghi + ½ kxi
2= ½ mvf
2+mghf+½kxf2
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Example 1
Starting from rest, a child slides down a frictionless slide from an initial height of 3.00 m. What is his speed at the bottom of the slide? Assume he has a mass of 25 kg.
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hi = 3.00m, hf = 0.00m
Vi=0
m=25.0kgg=9.81m/s2
Vf=?
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Use the Conservation of Mechanical Energy Formula
(no PEE)KEi + PEg,i = KEf + PEg,f
mghi = ½ mvf2
vf=(ghi )/(½ )vf= (9.81)(3.00m)/(1/2) =7.67m/s
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Example 2. A brick with a mass of 2.1 kg falls from a height of 2.0 m and lands on a relaxed spring with a constant of 3.5 N/m. How far will the spring be compressed? Assume final position is 0.0 m and final KE is 0.
Given:
Find:
Formula:
Solution:
m = 2.1 kg hi = 2.0 m g = 9.81 m/s2 k = 3.5 N/m
x
k mgh , PE P PE 2fg,iE,, 2
1ifE, xsoEPE ig
m 4.9 N/m 3.5
m) )(2.0m/s kg)(9.81 2(2.1
k
2mgh
2i
x
No KE initial or final
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Practice 5E
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Conservation of Energy Lab
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Power
Work is not affected by the time it takes to perform it.
Power is the rate of doing work.P=Wnet/t = Fnet,||d/t OR P= Fnet,|| v
Remember… Wnet = KEF = maFriction/Resistance is a force that needs to be included in Fnet calculations
Units are in watts (W)One Watt = 1 Joule/secondKilowatt = 1000 watts
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Power
Example 1. How long does it take for a 1.70 kW steam engine to do 5.6 x 106 J of work? (assume 100% efficiency.)
Given:
Find:
Formula:
Solution:
P = 1.70 x 103 W W = 5.6 x 106 J
t
t
W P
s 10 x 3.3 W10 x 1.70
J 10 x 5.6
P
W t 3
3
6
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Power
Example 2. How much power must a crane’s motor deliver to lift a 350 kg crate to a height of 12.0 m in 4.5 s?
Given:
Find:
Formula:
Solution:
m = 350 kg h = 12.0 m t = 4.5 s g = 9.81 m/s2
P
t
mgh
t
dF P
net
W9200 s 4.5
m) )(12.0m/s kg)(9.81 (350
t
mgh P
2
P=Wnet/t = Fnetd/t
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Power
Example 3. A 1.2 x 103 kg elevator carries a maximum load of 900.0 kg. A constant frictional force of 5.0 x 103 N retards the elevator’s upward motion. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s?
Hint: The key to this problem is to find the required net force required in the vertical
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Fnet = Fw elevator + Fw load + Ffriction = g(me + ml) + Ff
Fnet = 9.81 m/s2(1.2 x 103 kg + 900.0 kg) + 5.0 x 103 N
Fnet = 25 601 N (without rounding)
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Given:
Find:
Formula:
Solution:
Fnet = 25 601 N = 3.00 m/s
P
netF P
W000 77 m/s) N)(3.00 601 (25 F P net
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Example 4
A 1450 kg car accelerates from rest to 5.60m/s in 5.00sec. If Fk is 560.0N, what is the power developed by the engine?
Given: m=1450kg, vi=0, vf=5.60m/s, t=5.00s, Fk=560.0N
Need: acceleration to determine net force (F=ma)! Then you need x (this is d) to find Work. Then you can calculate Power.
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a=v/ t=5.60m/s/(5.00s)=1.12m/s2
**engine has to overcome friction(Fk) and cause acceleration (F=ma)
Fnet=Fk+ma=560.0N+(1450kg)X(1.12m/s2) = 2184N
**use an old formula to find xx=1/2(vi+vf)Xt= ½(0+5.60m/s)X5s = 14m
**Now find powerP=W/ t = Fnetd/ t = 2184NX14m/5.00sP=6115Watts = 6.115kW
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Practice 5F
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Review problems in class
1,7,10,12,13,17,19,22,24,25,30, 33,35