Energy and evolving force.


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1In an auto accident, which causes more prominent harm?. AA gigantic car.BA quick car.CA monstrous and quick moving auto.. . 2A auto slams into a settled divider. By what method can the harm be diminished?. AIt crashes at high speed.BIt has a gentler guard at the front with the goal that it stops in a more drawn out time.. . CIt has a hard guard so it stops in a brief timeframe..
Transcripts
Slide 1

Force and evolving energy

Slide 2

1 In a pile up, which causes more prominent harm? A A huge auto. B A quick auto. C A huge and quick moving auto.

Slide 3

2 An auto slams into a settled divider. In what manner can the harm be decreased? A It crashes at fast. B It has a milder guard at the front with the goal that it stops in a more drawn out time. C It has a hard guard with the goal that it stops in a brief span.

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1 What is energy? Encounter lets us know that: a substantial , quick auto causes more noteworthy harm than an overwhelming , moderate auto or a light , quick auto. Both mass and speed are vital physical amounts for concentrating on moving articles .

Slide 5

1 What is energy? Physicists have characterized an amount called energy ( 動量) of a moving article as... Force = mass  speed  = m v Unit: kg m s 1 A vector amount

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Example 1 Calculating energy Calculate the force of the accompanying items in kg m s –1 . ( an) A 20 g shot moving at 400 m s –1 . Energy of the projectile = 0.02  400 = 8 kg m s –1 ( towards the privilege )

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90 = 0.05  3.6 Example 1 ( b) A 0.05 kg tennis ball moving at 90 km h –1 . Energy of the tennis ball 0.05 kg 90 km h –1 = 1.25 kg m s –1 ( towards the left ) + ve

Slide 8

v  u = m ( ) t m v  m u = t 2 Momentum and constrain  is additionally identified with the compel on a protest . Newton\'s second law can be communicated as far as the adjustment in force . F = m an underlying  change in energy last 

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m v  m u = t 2 Momentum and drive change in energy Force = time  Newton\'s second Law can be restated as: The net constrain following up on a protest is equivalent to the rate of progress of energy of the question.

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Q1 Two rams battle... Two rams battle . On the off chance that smash A wins, Which of the accompanying must be valid? A m A > m B  A >  B C v A > v B  A  B m B m A v A v B

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Q2 Vincent & Esther... Vincent & Esther are skating. Who is right? A Vincent. B Esther. C Neither. We have an indistinguishable  from the results of our m & v are equivalent. 2.0 m s –1 2.5 m s –1 I don\'t think in this way, our headings are distinctive. 40 kg 50 kg Esther Vincent

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Q3 A 0.4-kg football... A 0.4-kg football is moving at 20 m s –1 . What is the force of the football? Force of the football = m v = ___ __ ____  _ __ ______ = _________ __ 0.4 20 8 kg m s –1

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= v 2 gs = v 2  10  1.8 Q4 A stone of mass 1.5 kg... A stone of mass 1.5 kg is dropped from a stature of 1.8 m. ( a) What is the adjustment in force of the stone when it achieves the ground? v 2  u 2 = 2 as = 6 m s –1 Momentum change of the stone = m v = __________= _________ 9 kg m s –1 1.5  6

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Q4 A stone of mass 1.5kg... A stone of mass 1.5 kg is dropped from a stature of 1.8 m. ( b) Why does the energy of the stone change? It is on account of a ____________ drive follows up on the stone for a time of _________. gravitational time

Slide 15

3 Impact When a tennis ball is hit by a racket, a huge compel follows up on the ball in a brief timeframe. How might we think about the constrain amid effect? Reproduction

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30 m s  1 20 m s  1 0.05 kg Example 2 Average compel following up on tennis ball What is the normal constrain following up on the ball? time of effect t = 0.005 s + ve

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m v  m u F = t 0.05  30  0.05  (  20) = 0.005 Example 2 Average constrain following up on tennis ball mass m = 0.05 kg u =  20 m s  1 v = 30 m s  1 t = 0.005 s + ve = 500 N to one side 500 N

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Experiment 8a Investigating the effect of compel Set up the accompanying mechanical assembly : Start information logging . Settle a little spring on the compel sensor. marginally push it from rest so it moves down the runway & crashes on the compel sensor.

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Experiment 8a Investigating the effect of drive From the v-t diagram, discover the speed of the trolley prior and then afterward affect. From the F-t diagram, discover the region under the chart. Contrast this territory and the adjustment in energy . Video

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a Force-time chart of effect v changes when trolley crashes on drive sensor speed Result of analysis 8a: time (s) max. F drive (N) F  F  v. brief time interim 1.65 s 1.60 s

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a Force-time diagram of effect When a tennis ball is being hit by a racket, the ball is distorted . Why does the compel shift amid effect? The compel of effect  to max. at the point when the ball is d eformed generally . As the ball recaptures its shape , the constrain  .

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m v  m u F = t a Force-time diagram of effect Rearrange terms in  F t = m v  m u drive : result of compel & time amid which the constrain demonstrations Impulse = change in energy

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a Force-time chart of effect constrain/N This F-t diagram is a straight level line. steady compel Area u nder F-t diagram = F  t = drive time/s region = motivation

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a Force-time chart of effect constrain/N bend line drive shifts a progression of restricted rectangular bars time/s region of e ach bar gives the drive amid the time interim aggregate region gives the aggregate drive

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a Force-time chart of effect compel/N drive/N time/s time/s Area under F-t chart = motivation = change in energy

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F t b Force of effect For a similar change in m omentum , shorter time of effect  constrain  same zone same drive same  in  bigger F , littler t littler F , bigger t

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b Force of effect Impact time relies on upon the hardness of impacting articles. Harder question  shorter effect time e.g. • Golf ball\'s hardness >> tennis ball\'s  shorter effect time on being struck by golf club • Design of auto:  crash time in the event of accidence

Slide 28

b Force of effect Video

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Example 3 F - t chart of an effect A trolley hits and bounce back from a compel sensor and the F-t diagram is acquired. The zone under chart is 0.46 N s. 87654321 constrain (N) 1.75 1.80 1.85 1.90 1.95 2.00 time (s)

Slide 30

Example 3 F - t diagram of an effect ( a) What is the most extreme compel following up on the sensor amid effect? Greatest drive = 7.8 N range under the chart = 0.46 N s 87654321 7.8 N constrain (N) 1.75 1.80 1.85 1.90 1.95 2.00 time (s)

Slide 31

87654321 1.75 1.80 1.85 1.90 1.95 2.00 Example 3 F - t diagram of an effect ( b) Find the adjustment in force of the trolley. region under the chart = 0.46 N s compel (N) got by information logging program time (s) Area under the diagram = 0.46 N s Change in energy = 0.46 kg m s –1

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87654321 1.75 1.80 1.85 1.90 1.95 2.00  in force 0.46 0.3 time Example 3 F - t chart of an effect ( c) Hence locate the normal constrain acting amid effect.  in energy = 0.46 m s –1 drive (N) time of effect = 2.05 – 1.75 = 0.3 s time (s) Average compel = 1.53 N =

Slide 33

Example 4 Crumple zone of an auto An auto of mass 1500 kg moving at 20 m s  1 (72 km h  1 ) slams into a divider head-on and grinds to a halt. Figure the drive of effect on the auto in the accompanying cases. ( a) A auto has a fold zone in the front segment and it stops in 0.5 s. ( b) A auto has a solid guard in the front area and it stops in 0.02 s.

Slide 34

m v  m u F = t 0  1500  20 = 0.5 Example 4 Crumple zone of an auto ( a) A auto has a fold zone in the front area and it stops in 0.5 s. mass of auto = 1500 kg inverse course to auto\'s movement + ve =  6  10 4 N 20 m s  1

Slide 35

m v  m u F = t 0  1500  20 = 0.02 Example 4 Crumple zone of an auto ( b) A auto has a solid guard in the front segment and it stops in 0.02 s. mass of auto = 1500 kg 25 times more prominent than that in (a)! + ve =  1.5  10 6 N 20 m s  1

Slide 36

Example 5 Force of effect of a falling can ( a) Calculate the speed of effect of the can on the ground. ( g = 10 m s  2 ) 0.4 kg + ve 30 m time of effect = 5 ms

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Example 5 Force of effect of a falling can u = 0 a = g = 10 m s  2 v 2  u 2 = 2 as v 2 = 2  10  30 m v = 24.5 m s  1 ( or 88.2 km h  1 ) v = ? + ve

Slide 38

m v  m u F = t (0  0.40  24.5) = 0.005 s Force of effect 1960 = Weight of can 0.4  10 Example 5 Force of effect of a falling can ( b) Find the normal drive of effect on the ground if the bounce back speed is immaterial. time of effect = 5 ms speed of effect =24.5 m s  1 mass of can = 0.4 kg = 1960 N (i.e. upwards) + ve = 490 times!

Slide 39

Example 6 Thrust of a rocket A rocket pushes out 100 kg of hot gas every second. The speed of the launched out hot gas is 500 m s  1 . Figure the forward constrain (push) on the rocket. rate = 100 kg s  1 v = 500 m s  1

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m v  m u F = ( 100  500)  0 = t 1 Example 6 Thrust of a rocket Let F be the constrain on gas = 50 000 N By Newton\'s third law, F = push u = 0 = 50 000 N v = 500 m s  1 rate = 100 kg s  1

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Q1 Which of the accompanying circumstances... Which of the accompanying circumstances does NOT include affect drive? A Traveling in a lift climbing at a consistent speed. B Kicking a ball. C A ping pong ball bobbing up starting from the earliest stage.

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F/N 4 3 2 1 2 t/s 12 1 2 3 4 5 0 5 Q2 A kid pushes a ball... A kid pushes a ball and the F-t diagram is demonstrated as follows. Locate the normal compel following up on the ball. Pixie u lse = region under F-t chart = _____ ___ _ __ _ = ____ ___ Average drive following up on the ball = _______ = _______  (1+5)  4 12 N s 2.4 N

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Q3 A auto of mass 1000 kg... An auto of mass 1000 kg quickens from rest to 8 m s –1 in 4 s. ( a) What is the adjustment in force of the auto? Change in force = mv  mu 1000  (8  0) 8000 kg m s –1 = ________ __

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