work-energy -power - universe of ali ovgun · 2018-08-28 · power q work does not depend on time...
TRANSCRIPT
Physics 111Lecture 6
Work-Energy -PowerDr.Ali ÖVGÜN
EMU Physics Department
www.aovgun.com
October 7-13, 2013
Why Energy?q Why do we need a concept of energy?q The energy approach to describing motion is
particularly useful when Newton’s Laws are difficult or impossible to use.
q Energy is a scalar quantity. It does not have a direction associated with it.
October 7-13, 2013
Kinetic Energyq Kinetic Energy is energy associated with the
state of motion of an objectq For an object moving with a speed of v
q SI unit: joule (J)1 joule = 1 J = 1 kg m2/s2
2
21 mvK =
October 7-13, 2013
Kinetic Energy for Various Objects2
21mvKE =
October 7-13, 2013
Why ?2
21 mvK =
October 7-13, 2013
Work Wq Start with Work “W”
q Work provides a link between force and energyq Work done on an object is transferred to/from itq If W > 0, energy added: “transferred to the
object”q If W < 0, energy taken away: “transferred from
the object”
xFmvmv xΔ=− 20
2
21
21
October 7-13, 2013
Definition of Work Wq The work, W, done by a constant force on an
object is defined as the product of the component of the force along the direction of displacement and the magnitude of the displacement
n F is the magnitude of the forcen Δ x is the magnitude of the
object’s displacementn θ is the angle between
xFxFW !!Δ⋅=Δ≡ )cos( θ
and ΔF x! !
October 7-13, 2013
Work Unitq This gives no information about
n the time it took for the displacement to occurn the velocity or acceleration of the object
q Work is a scalar quantityq SI Unit
n Newton • meter = Joulen N • m = Jn J = kg • m2 / s2 = ( kg • m / s2 ) • m
xFxFW !!Δ⋅=Δ≡ )cos( θ
xFmvmv Δ=− )cos(21
21 2
02 θ
October 7-13, 2013
Work: + or -?q Work can be positive, negative, or zero. The
sign of the work depends on the direction of the force relative to the displacement
q Work positive: if 90°> φ> 0°q Work negative: if 180°> φ> 90°q Work zero: W = 0 if φ= 90°q Work maximum if φ= 0°q Work minimum if φ= 180°
sFsFW !! ⋅=≡ )cos( φ
November 3, 2008
Work done by a Gravitational Force
q Gravitational Forcen Magnitude: mgn Direction: downwards to the
Earth’s centerq Work done by Gravitational
Force
20
2
21
21 mvmvWnet −=
cosW F r θ= Δ = ⋅ΔF rr r
θcosrmgWg Δ=
October 7-13, 2013
Example: When Work is Zeroq A man carries a bucket of water
horizontally at constant velocity.q The force does no work on the
bucketq Displacement is horizontalq Force is verticalq cos 90° = 0
xFW Δ≡ )cos( θ
October 7-13, 2013
Example: Work Can Be Positive or Negative
q Work is positive when lifting the box
q Work would be negative if lowering the boxn The force would still be upward,
but the displacement would be downward
October 7-13, 2013
Work Done by a Constant Force
q The work W done a system byan agent exerting a constantforce on the system is theproduct of the magnitude F ofthe force, the magnitude Δr ofthe displacement of the pointof application of the force, andcosθ, where θ is the anglebetween the force anddisplacement vectors:
θcosrFrFW Δ=Δ⋅≡ !!
F!
II
F!
IIIr!Δ
F!
Ir!Δ
F!
IVr!Δ
r!Δ
0=IW
θcosrFWIV Δ=rFWIII Δ=
rFWII Δ−=
October 7-13, 2013
Work and Forceq An Eskimo returning pulls a sled as shown. The
total mass of the sled is 50.0 kg, and he exerts a force of 1.20 × 102 N on the sled by pulling on the rope. How much work does he do on the sled if θ = 30° and he pulls the sled 5.0 m ?
JmN
xFW
2
2
102.5)0.5)(30)(cos1020.1(
)cos(
×=×=
Δ=!
θ
October 7-13, 2013
Work Done by Multiple Forcesq If more than one force acts on an object, then
the total work is equal to the algebraic sum of the work done by the individual forces
n Remember work is a scalar, sothis is the algebraic sum
=∑net by individual forcesW W
rFWWWW FNgnet Δ=++= )cos( θ
October 7-13, 2013
Problem Solving Strategyq Identify the initial and final positions of the body, and
draw a free body diagram showing and labeling all the forces acting on the body
q Choose a coordinate systemq List the unknown and known quantities, and decide
which unknowns are your target variablesq Calculate the work done by each force. Be sure to check
signs. Add the amounts of work done by each force to find the net (total) work Wnet
q Check whether your answer makes sense
October 7-13, 2013
Kinetic Energyq Kinetic energy associated with the motion of
an object
q Scalar quantity with the same unit as workq Work is related to kinetic energy
2
21 mvK =
xFmvmv netΔ=− 20
2
21
21
net f iW KE KE KE= − = Δ
October 7-13, 2013
Work-Energy Theoremq When work is done by a net force on an
object and the only change in the object is its speed, the work done is equal to the change in the object’s kinetic energy
n Speed will increase if work is positiven Speed will decrease if work is negative
20
2
21
21 mvmvWnet −=
KKKW ifnet Δ=−=
November 3, 2008
Potential Energyq Potential energy is associated with the
position of the objectq Gravitational Potential Energy is the
energy associated with the relative position of an object in space near the Earth’s surface
q The gravitational potential energy
n m is the mass of an objectn g is the acceleration of gravityn y is the vertical position of the mass
relative the surface of the Earthn SI unit: joule (J)
mgyPE ≡
November 3, 2008
Reference Levelsq A location where the gravitational potential
energy is zero must be chosen for each problemn The choice is arbitrary since the change in the
potential energy is the important quantityn Choose a convenient location for the zero
reference heightn often the Earth’s surfacen may be some other point suggested by the problem
n Once the position is chosen, it must remain fixed for the entire problem
November 3, 2008
Work and Gravitational Potential Energy
q PE = mgyq
q Units of Potential Energy are the same as those of Work and Kinetic Energy
figrav ity PEPEW −=
)(
0cos)(cos
if
fig
yymgyymgyFW
−−=
−=Δ= θ
November 3, 2008
Extended Work-Energy Theorem q The work-energy theorem can be extended to include
potential energy:
q If we only have gravitational force, then
q The sum of the kinetic energy and the gravitational potential energy remains constant at all time and hence is a conserved quantity
net f iW KE KE KE= − = Δ
figrav ity PEPEW −=
gravitynet WW =
fiif PEPEKEKE −=−
iiff KEPEPEKE +=+
November 3, 2008
Extended Work-Energy Theorem
q We denote the total mechanical energy by
q Since
q The total mechanical energy is conserved and remains the same at all times
q If there is friction force
PEKEE +=
iiff KEPEPEKE +=+
ffii mgymvmgymv +=+ 22
21
21
i i Lostbyfriction f f
Lostbyfriction
PE KE E KE PEE fd
+ = + +
=
November 3, 2008
Problem-Solving Strategyq Define the systemq Select the location of zero gravitational
potential energyn Do not change this location while solving the
problemq Identify two points the object of interest moves
betweenn One point should be where information is givenn The other point should be where you want to find
out something
November 3, 2008
Platform Diverq A diver of mass m drops
from a board 10.0 m above the water’s surface. Neglect air resistance.
q (a) Find is speed 5.0 m above the water surface
q (b) Find his speed as he hits the water
November 3, 2008
Platform Diverq (a) Find is speed 5.0 m above the water
surface
q (b) Find his speed as he hits the water
ffii mgymvmgymv +=+ 22
21
21
ffi mgyvgy +=+ 2
210
smgyv if /142 ==
0210 2 +=+ fi mvmgy
smmmsm
yygv fif
/9.9)510)(/8.9(2
)(22 =−=
−=
October 7-13, 2013
Powerq Work does not depend on time intervalq The rate at which energy is transferred is
important in the design and use of practical device
q The time rate of energy transfer is called powerq The average power is given by
n when the method of energy transfer is work
WPt
=Δ
October 7-13, 2013
Units of PowerqThe SI unit of power is called the watt
n 1 watt = 1 joule / second = 1 kg . m2 / s3
qA unit of power in the US Customary system is horsepowern 1 hp = 550 ft . lb/s = 746 W
qUnits of power can also be used to express units of work or energyn 1 kWh = (1000 W)(3600 s) = 3.6 x106 J
Problems
October 7-13, 2013
October 7-13, 2013
October 7-13, 2013
October 7-13, 2013
October 7-13, 2013