Download Notes - Fort Bend ISD

W = Fr cos θ
In the examples below, we compare the angles between F and r to see what
happens to the value of W. WE ARE ADDING THESE VECTORS HEAD TO TAIL.
WE ARE LOOKING AT THE ANGLE BETWEEN THE VECTORS.

F

F

F
θ
θ = 90o, cos90o = 0 NO WORK

r
θ

r
θ

r
θ = 60o, cos60o = 0.5 WORK
θ = 0o, cos0o = 1 MAXIMUM WORK
CAN WORK BE NEGATIVE ??
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WORK, ENERGY, POWER, CONSERVATION OF ENERGY
In this unit we will continue to study the MOTION of objects but this time we will
understand it in terms of work and energy. Knowing the work an object does, or the
energy it possesses allows you to predict KINEMATIC variables.
I. Definitions (next few pages are just definitions of work, PE, KE)
WORK- for an object to do work, the FORCE it applies must DISPLACE (move) the
other object. Work is the the product of two vectors, F and r (displacement).
WORK is a scalar.
A vertical force CANNOT cause a horizontal displacement.
A horizontal force CANNOT cause a vertical displacement.
If I lift a box, I apply and upward force and displacement. I did work.
When I walk with the box, I am still applying an upward force but the displacement is
horizontal, at a 90o angle to the force. NO WORK when carrying the box, no matter how
sweaty I get.
W = Fr cosθ
θ is the angle between the
F and r vectors.
SI UNIT FOR WORK is the Nm = Joule, J
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POTENTIAL ENERGY, PE, is the STORED ENERGY in an object.
There are two types we will define: GRAVITATIONAL and ELASTIC
GRAVITATIONAL refers to the PE an object possesses with respect to
to its VERTICAL displacement from the ground (how high is it?)
PE = mgh
h is the height from a surface (like the ground) defined as h = 0
the bigger h is, the more PE the object stores.
the more massive the object, the more PE it stores for a given
value of h
SI UNIT OF PE is
kg m2/s2 = JOULES, J
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Elastic objects, when stretched or compressed, store energy. A spring is a good example.
The force that a compressed spring can
exert depends on how much the spring
is compressed, x, and a value for the
spring called the spring constant, k

F = kx
x
THIS IS A STATEMENT OF HOOKE’S LAW
x = 0 there is no compression. PE = 0
1 2
PE = kx
2
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KINETIC ENERGY is the energy of MOTION
There are many types of motion so KE has several forms. In this class,
we are usually concerned with how an object TRANSLATES…how it
moves from point A to point B. The form of KE defined here is for
translational motion (you ought to remember this from Chem I)
1 2
KE = mv
2
SI UNIT IS,
kg m2/s2 = JOULE, J
Notice that KE is directly proportional to the square of the speed and to mass
Like work and PE, KE is scalar.
PE can be converted to KE and vice-versa.
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TOTAL MECHANICAL ENERGY is the sum of the PE and KE that an object possesses.
When work is done on an object, it acquires energy. The energy gained by the object
is MECHANICAL ENERGY.
An object with mechanical energy can DO WORK.
MEtot = PEgrav + PEspring + KE
Hammer pulled back: PE
Hammer moves toward nail: KE
Hammer strikes nail: FORCE
Nail moves in direction of force: WORK
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More on ME
The sum of the KE and PE at a given point for the ski jumper is his mechanical energy.
Notice how PE decreases as vertical displacement decreases
Notice how KE increases as PE decreases
Notice how ME is conserved…the same value from start to finish.
image from the physics classroom on-line
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POWER is the quantity of work done per unit time.
Did you notice that WORK has no time variable in its equation? The amount
of work done is independent of the time it takes to do it.
POWER, however, in inversely proportional to time. The more work done
in a given amount of time, the more power demonstrated.
Power is scalar.
W
P=
t
SI UNIT FOR POWER,
J/s = Watt, W
Power is also directly
proportional to the force
and velocity of an object
and
W = Fr
r
P=F
t
r
v=
t
P = Fv
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Conceptual understanding of power:
Two students are lifting weights. Student 1 lifts the 100-pound barbell over his head
10 times in one minute; Student 2 lifts the 100-pound barbell over his head 10 times
in 10 seconds. Which student does the most work? ______________ Which student
delivers the most power? ______________
Quantitative…
Suppose that a 40-horsepower engine could accelerate
the car from 0 mi/hr to 60 mi/hr in 16 seconds. If this
were the case, a car with four times the horsepower
could do the same amount of work in how much time?
The problem is asking how much time it would
take a 160 hp (40 x 4) engine to accelerate a car by
the same amount as a 40 hp engine. The work
is constant.
X times more power means 1/X times less to
do the work !
W
t
W1 = P1t1
W2 = P2t 2
P=
W1 = W2
P1t1 = P2t 2
P1t1
P2
40hp(16s)
t2 =
160hp
t 2 = 4s
t2 =
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II. Applying the definitions to analyze motion and work-energy
A. The relationship of WORK and ENERGY
SOME FORCES CAN CHANGE THE TOTAL MECHANICAL ENERGY OF AN OBJECT
SOME FORCES CAN ONLY CONVERT PE TO KE OR KE TO PE
WORK DONE ON AN OBJECT BY AN EXTERNAL FORCE CHANGES THE
OBJECT’S TOTAL MECHANICAL ENERGY
WORK DONE ON AN OBJECT BY AN INTERNAL FORCE CONSERVES THE
OBJECT’S MECHANICAL ENERGY BUT CONVERTS PE TO KE OR KE TO PE
INTERNAL FORCES (conservative)


F grav , F spring
these forces conserve energy,
just change its form
EXTERNAL FORCES (non-conservative)
 
  
F app , F frict , F air , F tens , F norm
these forces cause an object to LOSE or
GAIN either PE or KE
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when an external force CHANGES the energy of an object by doing work on it,
Wext = ΔMEtot
the external work done on the object = the CHANGE of total mechanical energy of the object
refine our total mechanical energy definition:
ΔMEtot = (MEtotf − MEtoti ) = Wext
MEtotf = Wext + MEtoti
this means if I add work to the
initial ME, I get the final ME value
work may be + or -
KE f + PE f = Wext + KEi + PEi
f and i mean final and initial values
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B. Analyzing motion from external forces as it relates to work, energy, and power
If you apply 1000N of force to lift a box 0.25 m, what is the total mechanical energy of the
box at the new position? The initial total mechanical energy is1500 J.
MEtotf = Wext + MEtoti
MEtotf = ?
0.25 m

F app = 1000N
MEtoti = 1500 J
we need work. The displacement is in the
same direction as the applied force so the
angle between them is 0,
Wext = Fr cosθ
MEtotf = Wext + MEtoti
Wext = (1000N )0.25m cos 0 o
Wext = 250J
MEtotf = 250J + 1500J
MEtotf = 1750J
total mechanical energy at new position is 1750 J
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A car with 320,000 J of mechanical energy skids to a stop over 30 m. What
is the total mechanical energy of the stopped car?
A car applying breaks is analogous to have a force push AGAINST its direction
of motion. In other words, the displacement, 30 m, results from an applied force
180o (opposite) to the direction of the displacement. Expect – work.

F app = 8000N
30 m
MEtoti = 320,000 J
MEtotf = Wext + MEtoti
we need work. The displacement is in
the opposite direction as the applied
force so the angle between them is 180o
Wext = Fr cosθ
Wext = (8000N )30m cos180 o
Wext = −240, 000J
MEtofi = ?
MEtotf = Wext + MEtoti
MEtotf = −240, 000J + 320, 000J
MEtotf = 80, 000J
total mechanical energy at new position is
80,000 J
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In the previous examples, ME was not conserved. The following example
shows how ME is conserved and KE and PE are converted.
KE f + PE f = Wext + KEi + PEi
when Wext = 0
KE f + PE f = KEi + PEi
Revisit the ski jumper. There is no external force. The work is done by gravity.
At any point along the way, the sum of KE and PE will be CONSTANT.
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