Scalars in mechanics
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Transcript of Scalars in mechanics
Scalars in mechanics
Photo: Flickr
Quantities
Scalars in mechanics 2
Scalar quantities Vector quantities
mass m (kg)
energy E (J)
power P (W)
displacement π (π)
velocity π£ (ππ β1)
acceleration π (ππ β2)
force πΉ (π)
momentum π (ππππ β1) torque π (ππ)
β’ Magnitude β’ Magnitudeβ’ Directionβ’ Point of application
History
Scalars in mechanics 3
17th centuryVector mechanics
GallileiHuygensNewtonLeibniz 19th century
Scalar mechanics
DavyRitter
OerstedFaradayKelvin
It was observed that βnatural forcesβ could transform.19th century natural philosophers (later physicists)Searched for a conserved βnatural forceβ.
This resulted in the new concept βENERGYβ
Concept of energy
Scalars in mechanics 4
IT
β’ Is transformed in all physical changes
β’ Is required to make anything happen
β’ Is puts limits to any process or activity
β’ Is the most fundamental concept in modern science
pixabay
Concept of work
Scalars in mechanics 5
StudentHotel
HollandISC
you (today)
π = 4.3ππ πΉππππ€πππ
Assume constant π£ππππ = 3 ππ β1 so Ξ£ πΉ = 0 β
πΉππππ€πππ = β πΉππππ + πΉπππ = 12ππ΄πΆπ· π£ππππ+π£π€πππ
2 + πΆππππ β πΉπ =
πΉππππ
πΉπππ+
1
2β 1.29 β 0.4 β 1 β 3 + 5 2 + 0.02 β 70 β 9.81 = 16.5 + 13.7 = 30π
Work done = π = πΉ β π = 30 β 4.3 β 103 = 1.3 β 105π½ππ’ππ
Work vs. Impulse
Scalars in mechanics 6
Why not use FORCE TIME as a measure of work?
πΉππππ€πππ
πΉππππ€πππ
start
same force, same time,β¦same work?
Impulse is a vector: πΉ β π‘ = πΌ
Vector Scalar = Vector
Work is a scalar: πΉ β π = π
Vector Vector= Scalar
Vector βdotβ product
Force & displacement at an angle
Scalars in mechanics 7
πΉ
π
π
πΉββπ
πΉβ₯π
π
90Β°
πΉββπ = 0
πΉβ₯π = πΉ πΉ
π
π
πΉββπ
πΉβ₯π
π = πΉββπ β π = πΉ β πΆππ π β π = πΉ β π β πΆππ π
Nett driving forceProvides energy
W > 0
Nett inhibiting forceWithdraws energy
W < 0
No driving forceNo energy transferred
W = 0
Non-constant forces
Scalars in mechanics 8
πΉ pulling force
π π
πΉ
π πππ₯
πΉπππ₯
πΉππ£π
π = πΉππ£π β π
π = 0
π
πΉ β ππ
π = 0
π
πΉ β πΆππ π β ππ
π = 0
π
πΉ β π π
more general
more generalwith angle
complete formas vector dot product
W=Area
Work transforms energy
Scalars in mechanics 9
wikimedia
Work transforms energy
Scalars in mechanics 10
wikimedia
Work by the resistant force of the target transforms kinetic energy of the arrow to heat in the target.
1 2 3
πΈπβ πΈπ πΈπ π
3
Work by the elastic force transforms elastic potential energy in the bow to kinetic energy of the arrow.
2
Work by the pulling force transforms chemical potential energy in master Kimβs arm to elastic potential energy in the string.
1
Gravitational (potential) energy
Scalars in mechanics 11
+β1
β2
π β π
πΉ
π = πΉ β π = π β π β β2 β β1 = βπΈπ
To lift the object, the average lifting force needs to equal π β πThe lifting force performs positive workGravitational potential energy is built up
Strictly πΈπ = 0 at the Earthβs centre.
Since this is not practical, we set πΈπ = 0 and β = 0
at the lowest point in your context.
Gravitational (potential) energy
Scalars in mechanics 12
β1
β2
π β π
πΉ
π = 10πππ = 9.81ππ β2
β1 = 5.0πβ2 = 7.0π
To lift the object from the floor to β1, the lifting force performs work:π = π β π β β1 = 10 β 9.81 β 5.0 = 4.9 β 10
2π½
Arrived at β1the object has a potential energy πΈπ = 4.9 β 102π½
To lift the object from β1 to β2, the lifting force performs work:π = π β π β β2 β β1 = 10 β 9.81 β 2.0 = 2.0 β 102π½
Arrived at β2the object has a new potential energy: πΈπ = 4.9 β 10
2π½ + 2.0 β 102π½ = 6.9 β 102π½
Example
πΈπ = 4.9 β 102π½
πΈπ = 6.9 β 102π½
πΈπ = 0
4.9 β 102π½Work done
2.0 β 102π½Work done
Elastic (potential) energy
Scalars in mechanics 13
π
πΉ
πΉπππ₯
πΉ = πΆ β π
π
πΉπππ₯2
π = πΉππ£π β π =1
2πΆ β π β π =
1
2πΆπ 2 = βπΈπ
To pull the object, the average pulling force equals 1
2πΆ β π
The pulling force performs positive workElastic potential energy is built up
π
Elastic (potential) energy
Scalars in mechanics 14
πΉ = πΆ β π
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0 1 2 3 4 5 6 7 8 9 10
F (N
)
s (cm)
Example with spring πΆ = 0.20π/ππ
The spring starts without potential energy: πΈπ = 0
To extend the spring 4.0 ππ, the pulling force does work:
π = πΉππ£π β π =0 + 0.8
2β 0.040 = 0.016π½
The spring then has a potential energy: πΈπ = 0.016π½
To extend the spring 3.0 ππ more, the pulling force does work:
π = πΉππ£π β π =0.8 + 1.4
2β 0.030 = 0.033π½
The spring then has a new potential energy: πΈπ = 0.016 + 0.033 = 0.049π½
Area =1st W
Area =2nd W
Kinetic energy
Scalars in mechanics 15
π‘
π£1
π‘1 π‘2
π£2
βπ£
βπ‘
Total work done on an object β‘Work done by the overall force !
Ξ£π = Ξ£πΉ β π = π β π β π
= π βπ£2 β π£1βπ‘
βπ£2 + π£12
β βπ‘
= π ββπ£
βπ‘β π£ππ£π β βπ‘
=π
2β π£2 β π£1 β π£2 + π£1
=1
2ππ£2
2 β1
2ππ£1
2 = βπΈπΞ£π = ΞπΈπ
2nd law of Newton (rephrased)
Scalars in mechanics 16
Vector mechanics: the acceleration of an object equals the overall force divided by its mass
Scalar mechanics: the increase of objectβs kinetic energy equals the overall work done on the object
Ξ£π = ΞπΈπ
Ξ£ πΉ = π π
Energy transfer & efficiency
Scalars in mechanics 17
Photo: wikimedia
Aggregate
GeneratorEfficiency π2
πΈπβ
π
π1
πΈπ
π2
Chemical 100J Work 30J
Heat 70J
Heat 8J
Electric 22J
Flows out of the system
The engine converts chemical energy to work with an efficiency
π1 =π
πΈπβ=30
100= 0.30 (30%)
The generator converts work to electrical energy with an efficiency
π2 =πΈππ=22
30= 0.73 (73%)
Overall efficiency =π = π1 β π2 =
0.30 β 0.73 = 0.22 (22%)
EngineEfficiency π1
Energy conservation
Scalars in mechanics 18
EngineEfficiency π1
GeneratorEfficiency π2
πΈπβ
π
π1
πΈπ
π2
Chemical 100J Work 30J
Heat 70J
Heat 8J
Electric 22J
Surrounding air
Aggregate
Closed system
Loss for the aggregate
First law of thermodynamics (energy conservation law): In a closed system the total energy is conserved
Conservation of energy β Example I
Scalars in mechanics 19
A ball is thrown vertically upwards with a velocity of 8.0 m/s. Which height will the ball reach (above the point of release)?
πΈπ΄ = πΈπ΅πΈππ΄ + πΈππ΄ = πΈππ΅ + πΈππ΅
0 +1
2ππ£π΄
2 = ππβπ΅ + 0
1
2π£π΄2 = πβπ΅
1
2β 8.02 = 9.81 β βπ΅
βπ΅ = 3.3π
Strategy:
Indicate 2 points:A = point of releaseB = highest point
Set β = 0 in A (lowest point)So, in A there is NO potential energy
In B there is no kinetic energy
Solve the energy conservation law
B
A
π£π΄
βπ΄ = 0
π£π΅ = 0
Solution:
Conservation of energy β Example II
Scalars in mechanics 20
πΈπ΄ = πΈπ΅
πΈππ΄ + πΈππ΄ = πΈππ΅ + πΈππ΅
0 +1
2ππ£π΄
2 = ππβπ΅ +1
2ππ£π΅
2
1
2π£π΄2 = πβπ΅ +
1
2π£π΅2
1
2β 8.02 = 9.81 β 2.0 +
1
2β π£π΅2
π£π΅ = 5.0ππ β1
Strategy:
Indicate again 2 points:A = point of releaseB = highest point
Set β = 0 in A (lowest point)So, in A there is NO potential energy
In B there is both potential and kinetic energy
Solve the energy conservation law
B
A
π£π΄
βπ΄ = 0
Solution:
A ball is thrown upwards with a velocity of 8.0 m/s at a certain angle.The ball climbs 2.0 m. Calculate the magnitude of the velocity at the highest point
π£π΅
βπ΅
Making work easier (but not less) - I
Scalars in mechanics 21
β1
β2
100ππ β 9.81ππ β2
πΉ1
ββ = 2.0ππΉ2 π
There is NO other force than weight.Only the vertical displacement requires work.
βπΈπ = πΉ1 β ββ = πΉ2 β π
981π β 2.0π = πΉ2 β 6.0π
πΉ2 = 3.3 β 102π
Feels like 33kg
Making work easier (but not less) - II
Scalars in mechanics 22
= 40π 120π
The pulley block carries the weight on 3 ropes.The load must be lifted 10 m up.
The weight is split in 3. One only needs to pull with 40π !
But!!
βπΈπ = 120π β 10π = 40π β 30π
You need to haul in 30 m of rope
Collisions
Scalars in mechanics 23
Elastic
Partlyinelastic
Before After Ξ£ π Ξ£πΈπΞ£πΈ
Conservation of
Totallyinelastic
+ heat
++ heat
Power
Scalars in mechanics 24
π =ΞπΈ
Ξπ‘
Power is the rate of energy conversion
1πππ‘π‘ = 1π½/π
π =Ξπ
Ξπ‘=πΉ β βπ
βπ‘= πΉ β
βπ
βπ‘= πΉ β π£
Power
Scalars in mechanics 25
πΉππππ€πππ
πΉπππ ππ π‘ππππ π£
A car travels at a constant velocity π£. Newton: πΉππππ€πππ = πΉπππ ππ π‘ππππ
The required power to keep the car driving at this velocity equals:
π =Ξπ
Ξπ‘=πΉππππ€πππ β βπ
βπ‘= πΉππππ€πππ β
βπ
βπ‘= πΉππππ€πππ β π£ = πΉπππ ππ π‘ππππ β π£
Power - Example
Scalars in mechanics 26
πΉππππ€πππ
πΉπππ ππ π‘ππππ π£
A BMW 7-series car has:Drag coefficient πΆπ· = 0.30Frontal area π΄ = 2.2π2
Calculate the mechanical power required to drive 160 km/h.
π£ = 160ππββ1 =160
3.6ππ β1 = 44.4ππ β1
Density of air = 1.29 πππβ3
πΉπππ =1
2ππ΄πΆπ·π£
2 =1
2β 1.29 β 2.2 β 0.30 β 44.42
πΉπππ = 841π
π£ = ππππ π‘πππ‘ β πΉππππ€πππ = πΉπππ
π = πΉπππ β π£ = 841 β 44.4 = 37 β 103π = 37ππ
Solution
Power & efficiency
Scalars in mechanics 27
π =πΈπ’π πππ’π
πΈπ‘ππ‘ππ=
πΈπ’π πππ’π π‘
πΈπ‘ππ‘ππ π‘=ππ’π πππ’π
ππ‘ππ‘ππ
Efficiency can be related to a ratio of energies or a ratio of powers. Look at the context and choose!
END
Scalars in mechanics 28
DisclaimerThis document is meant to be apprehended through professional teacher mediation (βlive in classβ) together with a physics text book, preferably on IB level.