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 𝑠 (𝑚)

velocity 𝑣 (𝑚𝑠−1)

acceleration 𝑎 (𝑚𝑠−2)

force 𝐹 (𝑁)

momentum 𝑝 (𝑘𝑔𝑚𝑠−1) torque 𝑇 (𝑁𝑚)

• Magnitude • Magnitude• Direction• Point of application

History


17th centuryVector mechanics

GallileiHuygensNewtonLeibniz 19th century

Scalar mechanics

DavyRitter

OerstedFaradayKelvin

It was observed that ‘natural forces’ could transform.19th century natural philosophers (later physicists)Searched for a conserved ‘natural force’.

This resulted in the new concept ‘ENERGY’

Concept of energy


IT

• Is transformed in all physical changes

• Is required to make anything happen

• Is puts limits to any process or activity

• Is the most fundamental concept in modern science

pixabay

Concept of work


StudentHotel

HollandISC

you (today)

𝑠 = 4.3𝑘𝑚 𝐹𝑓𝑜𝑟𝑤𝑎𝑟𝑑

Assume constant 𝑣𝑏𝑖𝑘𝑒 = 3 𝑚𝑠−1 so Σ 𝐹 = 0 ⇒

𝐹𝑓𝑜𝑟𝑤𝑎𝑟𝑑 = − 𝐹𝑟𝑜𝑙𝑙 + 𝐹𝑎𝑖𝑟 = 12𝜌𝐴𝐶𝐷 𝑣𝑏𝑖𝑘𝑒+𝑣𝑤𝑖𝑛𝑑

2 + 𝐶𝑟𝑜𝑙𝑙 ∙ 𝐹𝑁 =

𝐹𝑟𝑜𝑙𝑙

𝐹𝑎𝑖𝑟+

1

2∙ 1.29 ∙ 0.4 ∙ 1 ∙ 3 + 5 2 + 0.02 ∙ 70 ∙ 9.81 = 16.5 + 13.7 = 30𝑁

Work done = 𝑊 = 𝐹 ∙ 𝑠 = 30 ∙ 4.3 ∙ 103 = 1.3 ∙ 105𝐽𝑜𝑢𝑙𝑒

Work vs. Impulse


Why not use FORCE TIME as a measure of work?

𝐹𝑓𝑜𝑟𝑤𝑎𝑟𝑑


start

same force, same time,…same work?

Impulse is a vector: 𝐹 ∙ 𝑡 = 𝐼

Vector Scalar = Vector

Work is a scalar: 𝐹 ∙ 𝑠 = 𝑊

Vector Vector= Scalar

Vector ‘dot’ product

Force & displacement at an angle


𝐹

𝑠

𝜃

𝐹∕∕𝑠

𝐹⊥𝑠

𝑠

90°

𝐹∕∕𝑠 = 0

𝐹⊥𝑠 = 𝐹 𝐹

𝑠

𝜃

𝐹∕∕𝑠

𝐹⊥𝑠

𝑊 = 𝐹∕∕𝑠 ⋅ 𝑠 = 𝐹 ⋅ 𝐶𝑜𝑠𝜃 ∙ 𝑠 = 𝐹 ∙ 𝑠 ∙ 𝐶𝑜𝑠𝜃

Nett driving forceProvides energy

W > 0

Nett inhibiting forceWithdraws energy

W < 0

No driving forceNo energy transferred

W = 0

Non-constant forces


𝐹 pulling force

𝑠 𝑠

𝐹

𝑠𝑚𝑎𝑥

𝐹𝑚𝑎𝑥

𝐹𝑎𝑣𝑔

𝑊 = 𝐹𝑎𝑣𝑔 ∙ 𝑠

𝑊 = 0

𝑠

𝐹 ∙ 𝑑𝑠

𝑊 = 0

𝑠

𝐹 ∙ 𝐶𝑜𝑠𝜃 ∙ 𝑑𝑠

𝑊 = 0

𝑠

𝐹 ∙ 𝑑 𝑠

more general

more generalwith angle

complete formas vector dot product

W=Area

Work transforms energy


wikimedia

Work transforms energy


wikimedia

Work by the resistant force of the target transforms kinetic energy of the arrow to heat in the target.

1 2 3

𝐸𝑐ℎ 𝐸𝑝 𝐸𝑘 𝑄

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 Kim’s arm to elastic potential energy in the string.

1

Gravitational (potential) energy


+ℎ1

ℎ2

𝑚 ∙ 𝑔

𝐹

𝑊 = 𝐹 ∙ 𝑠 = 𝑚 ∙ 𝑔 ∙ ℎ2 − ℎ1 = ∆𝐸𝑝

To lift the object, the average lifting force needs to equal 𝑚 ∙ 𝑔The lifting force performs positive workGravitational potential energy is built up

Strictly 𝐸𝑝 = 0 at the Earth’s centre.

Since this is not practical, we set 𝐸𝑝 = 0 and ℎ = 0

at the lowest point in your context.

Gravitational (potential) energy


ℎ1

ℎ2

𝑚 ∙ 𝑔

𝐹

𝑚 = 10𝑘𝑔𝑔 = 9.81𝑚𝑠−2

ℎ1 = 5.0𝑚ℎ2 = 7.0𝑚

To lift the object from the floor to ℎ1, the lifting force performs work:𝑊 = 𝑚 ∙ 𝑔 ∙ ℎ1 = 10 ∙ 9.81 ∙ 5.0 = 4.9 ∙ 10

2𝐽

Arrived at ℎ1the object has a potential energy 𝐸𝑝 = 4.9 ∙ 102𝐽

To lift the object from ℎ1 to ℎ2, the lifting force performs work:𝑊 = 𝑚 ∙ 𝑔 ∙ ℎ2 − ℎ1 = 10 ∙ 9.81 ∙ 2.0 = 2.0 ∙ 102𝐽

Arrived at ℎ2the object has a new potential energy: 𝐸𝑝 = 4.9 ∙ 10

2𝐽 + 2.0 ∙ 102𝐽 = 6.9 ∙ 102𝐽

Example

𝐸𝑝 = 4.9 ∙ 102𝐽

𝐸𝑝 = 6.9 ∙ 102𝐽

𝐸𝑝 = 0

4.9 ∙ 102𝐽Work done

2.0 ∙ 102𝐽Work done

Elastic (potential) energy


𝑠

𝐹

𝐹𝑚𝑎𝑥

𝐹 = 𝐶 ∙ 𝑠

𝑠

𝐹𝑚𝑎𝑥2

𝑊 = 𝐹𝑎𝑣𝑔 ∙ 𝑠 =1

2𝐶 ∙ 𝑠 ∙ 𝑠 =

1

2𝐶𝑠2 = ∆𝐸𝑝

To pull the object, the average pulling force equals 1

2𝐶 ∙ 𝑠

The pulling force performs positive workElastic potential energy is built up

𝑠

Elastic (potential) energy


𝐹 = 𝐶 ∙ 𝑠

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 𝐶 = 0.20𝑁/𝑐𝑚

The spring starts without potential energy: 𝐸𝑝 = 0

To extend the spring 4.0 𝑐𝑚, the pulling force does work:

𝑊 = 𝐹𝑎𝑣𝑔 ∙ 𝑠 =0 + 0.8

2∙ 0.040 = 0.016𝐽

The spring then has a potential energy: 𝐸𝑝 = 0.016𝐽

To extend the spring 3.0 𝑐𝑚 more, the pulling force does work:

𝑊 = 𝐹𝑎𝑣𝑔 ∙ 𝑠 =0.8 + 1.4

2∙ 0.030 = 0.033𝐽

The spring then has a new potential energy: 𝐸𝑝 = 0.016 + 0.033 = 0.049𝐽

Area =1st W

Area =2nd W

Kinetic energy


𝑡

𝑣1

𝑡1 𝑡2

𝑣2

∆𝑣

∆𝑡

Total work done on an object ≡Work done by the overall force !

Σ𝑊 = Σ𝐹 ∙ 𝑠 = 𝑚 ∙ 𝑎 ∙ 𝑠

= 𝑚 ∙𝑣2 − 𝑣1∆𝑡

∙𝑣2 + 𝑣12

∙ ∆𝑡

= 𝑚 ∙∆𝑣

∆𝑡∙ 𝑣𝑎𝑣𝑔 ∙ ∆𝑡

=𝑚

2∙ 𝑣2 − 𝑣1 ∙ 𝑣2 + 𝑣1

=1

2𝑚𝑣2

2 −1

2𝑚𝑣1

2 = ∆𝐸𝑘Σ𝑊 = Δ𝐸𝑘

2nd law of Newton (rephrased)


Vector mechanics: the acceleration of an object equals the overall force divided by its mass

Scalar mechanics: the increase of object’s kinetic energy equals the overall work done on the object

Σ𝑊 = Δ𝐸𝑘

Σ 𝐹 = 𝑚 𝑎

Energy transfer & efficiency


Photo: wikimedia

Aggregate

GeneratorEfficiency 𝜂2

𝐸𝑐ℎ

𝑊

𝑄1

𝐸𝑒

𝑄2

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

𝜂1 =𝑊

𝐸𝑐ℎ=30

100= 0.30 (30%)

The generator converts work to electrical energy with an efficiency

𝜂2 =𝐸𝑒𝑊=22

30= 0.73 (73%)

Overall efficiency =𝜂 = 𝜂1 ∙ 𝜂2 =

0.30 ∙ 0.73 = 0.22 (22%)

EngineEfficiency 𝜂1

Energy conservation


EngineEfficiency 𝜂1

GeneratorEfficiency 𝜂2

𝐸𝑐ℎ

𝑊

𝑄1

𝐸𝑒

𝑄2

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 – Example I


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)?

𝐸𝐴 = 𝐸𝐵𝐸𝑝𝐴 + 𝐸𝑘𝐴 = 𝐸𝑝𝐵 + 𝐸𝑘𝐵

0 +1

2𝑚𝑣𝐴

2 = 𝑚𝑔ℎ𝐵 + 0

1

2𝑣𝐴2 = 𝑔ℎ𝐵

1

2∙ 8.02 = 9.81 ∙ ℎ𝐵

ℎ𝐵 = 3.3𝑚

Strategy:

Indicate 2 points:A = point of releaseB = highest point

Set ℎ = 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

𝑣𝐴

ℎ𝐴 = 0

𝑣𝐵 = 0

Solution:

Conservation of energy – Example II


𝐸𝐴 = 𝐸𝐵

𝐸𝑝𝐴 + 𝐸𝑘𝐴 = 𝐸𝑝𝐵 + 𝐸𝑘𝐵

0 +1

2𝑚𝑣𝐴

2 = 𝑚𝑔ℎ𝐵 +1

2𝑚𝑣𝐵

2

1

2𝑣𝐴2 = 𝑔ℎ𝐵 +

1

2𝑣𝐵2

1

2∙ 8.02 = 9.81 ∙ 2.0 +

1

2∙ 𝑣𝐵2

𝑣𝐵 = 5.0𝑚𝑠−1

Strategy:

Indicate again 2 points:A = point of releaseB = highest point

Set ℎ = 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

𝑣𝐴

ℎ𝐴 = 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

𝑣𝐵

ℎ𝐵

Making work easier (but not less) - I


ℎ1

ℎ2

100𝑘𝑔 ∙ 9.81𝑚𝑠−2

𝐹1

∆ℎ = 2.0𝑚𝐹2 𝑠

There is NO other force than weight.Only the vertical displacement requires work.

∆𝐸𝑝 = 𝐹1 ∙ ∆ℎ = 𝐹2 ∙ 𝑠

981𝑁 ∙ 2.0𝑚 = 𝐹2 ∙ 6.0𝑚

𝐹2 = 3.3 ∙ 102𝑁

Feels like 33kg

Making work easier (but not less) - II


= 40𝑁 120𝑁

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 40𝑁 !

But!!

∆𝐸𝑝 = 120𝑁 ∙ 10𝑚 = 40𝑁 ∙ 30𝑚

You need to haul in 30 m of rope

Collisions


Elastic

Partlyinelastic

Before After Σ 𝑝 Σ𝐸𝑘Σ𝐸

Conservation of

Totallyinelastic

+ heat

++ heat

Power


𝑃 =Δ𝐸

Δ𝑡

Power is the rate of energy conversion

1𝑊𝑎𝑡𝑡 = 1𝐽/𝑠

𝑃 =Δ𝑊

Δ𝑡=𝐹 ∙ ∆𝑠

∆𝑡= 𝐹 ∙

∆𝑠

∆𝑡= 𝐹 ∙ 𝑣

Power



𝐹𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑣

A car travels at a constant velocity 𝑣. Newton: 𝐹𝑓𝑜𝑟𝑤𝑎𝑟𝑑 = 𝐹𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒

The required power to keep the car driving at this velocity equals:

𝑃 =Δ𝑊

Δ𝑡=𝐹𝑓𝑜𝑟𝑤𝑎𝑟𝑑 ∙ ∆𝑠

∆𝑡= 𝐹𝑓𝑜𝑟𝑤𝑎𝑟𝑑 ∙

∆𝑠

∆𝑡= 𝐹𝑓𝑜𝑟𝑤𝑎𝑟𝑑 ∙ 𝑣 = 𝐹𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 ∙ 𝑣

Power - Example



𝐹𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑣

A BMW 7-series car has:Drag coefficient 𝐶𝐷 = 0.30Frontal area 𝐴 = 2.2𝑚2

Calculate the mechanical power required to drive 160 km/h.

𝑣 = 160𝑘𝑚ℎ−1 =160

3.6𝑚𝑠−1 = 44.4𝑚𝑠−1

Density of air = 1.29 𝑘𝑔𝑚−3

𝐹𝑎𝑖𝑟 =1

2𝜌𝐴𝐶𝐷𝑣

2 =1

2∙ 1.29 ∙ 2.2 ∙ 0.30 ∙ 44.42

𝐹𝑎𝑖𝑟 = 841𝑁

𝑣 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⇒ 𝐹𝑓𝑜𝑟𝑤𝑎𝑟𝑑 = 𝐹𝑎𝑖𝑟

𝑃 = 𝐹𝑎𝑖𝑟 ∙ 𝑣 = 841 ∙ 44.4 = 37 ∙ 103𝑊 = 37𝑘𝑊

Solution

Power & efficiency


𝜂 =𝐸𝑢𝑠𝑒𝑓𝑢𝑙

𝐸𝑡𝑜𝑡𝑎𝑙=

𝐸𝑢𝑠𝑒𝑓𝑢𝑙 𝑡

𝐸𝑡𝑜𝑡𝑎𝑙 𝑡=𝑃𝑢𝑠𝑒𝑓𝑢𝑙

𝑃𝑡𝑜𝑡𝑎𝑙

Efficiency can be related to a ratio of energies or a ratio of powers. Look at the context and choose!

END


DisclaimerThis document is meant to be apprehended through professional teacher mediation (‘live in class’) together with a physics text book, preferably on IB level.

Scalars in mechanics

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Transcript of Scalars in mechanics