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Chapter 7 Work and Kinetic Energy
Which one costs energy?Which one costs energy?
Question: (try it)How to throw a baseball to give it large speed?How to throw a baseball to give it large speed?Answer: Apply large force across a large distance!Apply large force across a large distance!
Force exerted through a distance performsForce exerted through a distance performsmechanical work. 1
Units of Chapter 7• Work Done by a Constant Force
• Kinetic Energy
• The Work-Energy Theorem• The Work-Energy Theorem
• Work Done by a Variable Force (optional)
• Power
Read Chapter 8, Potential energy before the next lecture.We will finish Chapter 8 in the next lecture.
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7-1 Work Done by a Constant ForceWhen the force is parallel to the displacement:When the force is parallel to the displacement: Constant force in direction of motion does work W.
(7-1)
SI unit: newton-meter (N·m) = Joule JSI unit: newton-meter (N·m) = Joule, J1 J = 1 N.m
If F= 15 N, distance = 2 m , W=30 J3
7-1 Work Done by a Constant Force
1 J l1 Joule1 J
How much is that?
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If the force is at an angle to the motion, it does the following work:the following work:
(7-3)
θ is the angle between force and motion direction.g
Pulling at θ= 20 , F=15N, d=2 mW= Fd cos θ =15*2* cos 20 = 28 J
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7-1 Work Done by a Constant ForceTh k l b itt th d tThe work can also be written as the dot product of the force and the displacement:
θ is the angle between force and motion di tidirection.
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The work done may be positive, zero, or negative, depending on the angle between the force and the motion:
Here for F and d we only use their sizes (absolute value). The sign of the work is determined only by the angle between that force and motionby the angle between that force and motion.
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Special cases:
When force is perpendicular
When force is Opposite to motion directionp p
to motion direction, it does no work.
to motion direction, cos(180)=-1.Examples:
Examples:Normal force is always perpendicular to surface
Examples:Kinetic friction force does negative work. perpendicular to surface.
Tension of pendulum …g
Wfk = – fk dAlways!
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If there is more than one force acting on an object we can find the work done by each forceobject, we can find the work done by each force and add them together to find the total work.
(7-5)
Total work: Wtotal = W1 + W2 +W3+ …..the sum of the work done by each forces.
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Q: Is Work a scalar or a vector?Scalar!It l h h i dd d dIt only says how much energy is added or used,(positive or negative), but doesn’t indicate motiondirections or force directionsdirections or force directions.When we add work to get total work, we add aspositive and negative simple numbers. p g pWe don’t add work as vector arrows.
Ff x
If FPull= fk ; a=0 , v=constantFpullfk
x
WPull= FPull d ; Wfk= – fk d ; Wtotal = 0
If F > f a >0 If >0 ill increaseIf FPull > fk ; a >0 , If v >0, v will increase.Wtotal = FPull d – fkd = (FPull – fk) d = Fnet d 10
7-2 Kinetic Energy and the Work-Energy TheoremTheorem
When positive work is done on an object, its speed increases; when negative work is done, its speed decreases.
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7-2 Kinetic Energy and the Work-Energy TheoremTheorem
After algebraic manipulations of the equations ofAfter algebraic manipulations of the equations of motion, we find: The total work done to one object is always equal to the change of its ½mv2
Therefore, we define the kinetic energy:(7-6)(7-6)
Kinetic energy has SI unit: J gy(same dimension as Work) 1 kg m2/s2 = 1 (kg m/s2)m =1Nm =1 Joule 12
7-2 Kinetic Energy and the Work-Energy TheoremTheorem
Work-Energy Theorem: The total work done onWork Energy Theorem: The total work done on an object is equal to its change in kinetic energy.
(7-7)
It’s true for ALL MOTIONS,no only for constant a motion!!! 13
Problem solving strategy : (work problems)1 C t k f i di id l f fi t
m=1000kg, v0=0, µk=0.2φ d 0 5 Fi d 1.Compute work for individual forces first.
2. Add all work together as scalar numbers.3. Set equation
φ=30ο, d=0.5m,Find vf
mg= 9800 (N)If you know total work, you can solve v,if you know v, you can solve Work.
g ( )N=mgcosφfk=µkN=µkmgcosφ
f =0 2∗9800∗cos30=1697(Ν)fk=0.2∗9800∗cos30=1697(Ν)
Wmg=mg d cos 60 = 9800*0.5* cos60=2450 JWfk = – fk d = – 1697*0.5= – 849 J ; WN=0 ;fk k ; N ;Wtotal = 2450 – 849 = 1601 J Kf = Ki +Wtotal ½mvf
2 =Wtotal =1601J vf=1.79m/s14
Wmg=mg d cos 60 =2450 JW = f d 849 J
Work is a scalar. T t t t l k d b ll f dd k i
Wfk = – fk d = – 849 J ; Wtotal = 2450 + (– 849) = 1601 J
To get total work done by all forces, add work in Joules directly as simple numbers!You only need to worry about the angles betweenYou only need to worry about the angles between each force and actual motion when you calculate work done by each force.yAfter the work is calculated. Add them as simple numbers, no worry about direction any more. 15
Spring Force: Hooke’s Law Fspring= - k ∆xk :Hooke’s constant (how strong a spring is)∆x : distance of stretch/compressionForce direction:Always in the opposite direction of ∆ xAlways in the opposite direction of ∆ xSpring forcealways triesalways triesto recover its natural Length
Att tiAttention:This k is not K,Spring constantSpring constantis not Kinetic Energy.
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7-3 Work Done by a Variable ForceIf the force is constant, we can interpret the work done graphically:
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7-3 Work Done by a Variable Force
If the force takes on several successive constant values:
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7-3 Work Done by a Variable Force
We can then approximate a continuously varying force by a succession of constant values.
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7-3 Work Done by a Variable Force
The force needed to stretch a spring an amount x is F = kx.
Therefore, the work done in stretching the spring isthe spring is
(7-8)(7-8)
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7-4 Power
Power is a measure of the rate at which work is done:
(7-10)
SI it J/ tt WSI unit: J/s = watt, W
1 horsepower = 1 hp = 746 W
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7-4 Power
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7-4 Power
If an object is moving at a constant speed in the face of friction, gravity, air resistance, and soface of friction, gravity, air resistance, and so forth, the power exerted by the driving force can be written:
(7-13)
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Summary of Chapter 7• If the force is constant and parallel to the displacement, work is force times distance
• If the force is not parallel to the displacement,
Th t t l k i th k d b th t• The total work is the work done by the net force:
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Summary of Chapter 7
• SI unit of work: the joule, J
T t l k i l t th h i ki ti• Total work is equal to the change in kinetic energy:
wherewhere
Kinetic energy is either positive or 0, never negative.25
Summary of Chapter 7
• Work done by a spring force:
• Power is the rate at which work is done:
• SI unit of power: the watt, Wp
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