physics unit 3 - andrews universityrwright/physics... · unit named after james watt who invented...

Physics Unit 3

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𝐽 =𝑘𝑔 ⋅ 𝑚2

𝑠2

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𝑊 = 𝐹𝑑 cos 𝜃𝑊 = 60 𝑁 100 𝑚 cos 30° = 5196 𝐽

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𝑊 = 𝐹𝑑 cos 𝜃𝑊 = 200 𝑁 100 𝑚 cos 90° = 0 𝐽

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F = 200 N (lift up)d = 2 m (down)

𝑊 = 𝐹𝑑 cos 𝜃 = 200 𝑁 2 𝑚 cos 180° = −400 𝐽

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𝑚 = 0.075 𝑘𝑔, 𝑠 = 0.90 𝑚, 𝑣0 = 0𝑚

𝑠, 𝑣𝑓 = 40

𝑚

𝑠𝑊 = 𝐾𝐸𝑓 − 𝐾𝐸0

𝐹 0.90 𝑚 =1

20.075 𝑘𝑔 40

𝑚

𝑠

2

−1

20.075 𝑘𝑔 0

𝑚

𝑠

2

𝐹 0.90 𝑚 = 60 𝐽𝐹 = 66.7 𝑁

4. Use 𝑣2 = 𝑣02 + 2𝑎 𝑥 − 𝑥0 which becomes 𝑣2 = 2𝑔ℎ

6. 010. It would jump higher

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PE changes to KE

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𝑊 = 𝐹𝑑𝑃𝐸𝑆 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑓𝑟𝑜𝑚 0 𝑡𝑜 𝑘𝑥 𝑏𝑦 𝐻𝑜𝑜𝑘𝑒′𝑠 𝐿𝑎𝑤 𝑥

𝑃𝐸𝑆 =1

2𝑘𝑥 ⋅ 𝑥

𝑃𝐸𝑆 =1

2𝑘𝑥2

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22

𝑎) 𝑊 = 𝐹𝑑

𝑊 =1

2𝑘𝑥2

𝑊 =1

22.22 × 105

𝑁

𝑚0.03 𝑚 2 = 99.9 𝐽

𝑏) 𝐾𝐸𝑓 + 𝑃𝐸𝑓 = 𝐾𝐸0 + 𝑃𝐸01

2𝑚𝑣𝑓

2 +1

2𝑘𝑥𝑓

2 =1

2𝑚𝑣0

2 +1

2𝑘𝑥0

2

1

20.050 𝑘𝑔 𝑣𝑓

2 + 0 = 0 +1

22.22 × 105

𝑁

𝑚0.03 𝑚 2

0.025 𝑘𝑔 𝑣𝑓2 = 99.9 𝐽

𝑣𝑓2 = 3996

𝑚2

𝑠2

𝑣𝑓 = 63.2𝑚

𝑠𝑐) 𝐾𝐸𝑓 + 𝑃𝐸𝐺𝑓 + 𝑃𝐸𝑆𝑓 = 𝐾𝐸0 + 𝑃𝐸𝐺0 + 𝑃𝐸𝑆0

At end of barrel1

2𝑚𝑣𝑓

2 +𝑚𝑔ℎ𝑓 + 0 = 0 + 0 +1

2𝑘𝑥0

2

1

20.050 𝑘𝑔 𝑣𝑓

2 + 0.050 𝑘𝑔 9.80𝑚

𝑠20.03 𝑚 =

1

22.22 × 105

𝑁

𝑚0.03 𝑚 2

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0.025 𝑘𝑔 𝑣𝑓2 + 0.0147 𝐽 = 99.9 𝐽

0.025 𝑘𝑔 𝑣𝑓2 = 99.8853 𝐽

𝑣𝑓2 = 3995.412

𝑚2

𝑠2

𝑣𝑓 = 63.2𝑚

𝑠At top of path

𝐾𝐸𝑓 + 𝑃𝐸𝐺𝑓 = 𝐾𝐸0 + 𝑃𝐸𝐺01

2𝑚𝑣𝑓

2 +𝑚𝑔ℎ𝑓 =1

2𝑚𝑣0

2 +𝑚𝑔ℎ0

0 + 0.050 𝑘𝑔 9.80𝑚

𝑠2ℎ𝑓 =

1

20.050 𝑘𝑔 63.2

𝑚

𝑠

2

+ 0

0.49 𝑘𝑔 ⋅𝑚

𝑠2ℎ𝑓 = 99.9 𝐽

ℎ𝑓 = 204 𝑚

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𝑃𝐸0 + 𝐾𝐸0 = 𝑃𝐸𝑓 + 𝐾𝐸𝑓

𝑚𝑔ℎ0 +1

2𝑚𝑣0

2 = 𝑚𝑔ℎ𝑓 +1

2𝑚𝑣𝑓

2

1500 𝑘𝑔 9.8𝑚

𝑠250 𝑚 +

1

21500 𝑘𝑔 20

𝑚

𝑠

2

= 0 +1

21500 𝑘𝑔 𝑣𝑓

2

𝑣𝑓 = 37.1 𝑚/𝑠

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𝐸0 +𝑊𝑛𝑐 = 𝐸𝑓1

2𝑚𝑣0

2 +𝑚𝑔ℎ0 +𝑊𝑛𝑐 =1

2𝑚𝑣𝑓

2 +𝑚𝑔ℎ𝑓

1

2200 𝑘𝑔 0 2 + 200 𝑘𝑔 9.8

𝑚

𝑠20 +𝑊𝑛𝑐

=1

2200 𝑘𝑔 500

𝑚

𝑠

2

+ 200 𝑘𝑔 9.8𝑚

𝑠25000 𝑚

𝑊𝑛𝑐 = 2.50 × 107 𝐽 + 9.80 × 106 𝐽𝑊𝑛𝑐 = 3.48 × 107 𝐽

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𝐸0 +𝑊𝑛𝑐 = 𝐸𝑓1

2𝑚𝑣0

2 +𝑚𝑔ℎ0 +𝑊𝑛𝑐 =1

2𝑚𝑣𝑓

2 +𝑚𝑔ℎ𝑓

0 + 1500 𝑘𝑔 9.8𝑚

𝑠210 𝑚 +𝑊𝑛𝑐 =

1

21500 𝑘𝑔 12

𝑚

𝑠

2

+ 0

𝑊𝑛𝑐 = −39000 𝐽

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𝐸0 +𝑊𝑛𝑐 = 𝐸𝑓1

2𝑚𝑣0

2 +𝑚𝑔ℎ0 +𝑊𝑛𝑐 =1

2𝑚𝑣𝑓

2 +𝑚𝑔ℎ𝑓

0 + 0 + 800000 𝐽 =1

290 𝑘𝑔 𝑣𝑓

2 + 90 𝑘𝑔 9.8𝑚

𝑠250 𝑚

129.6𝑚

𝑠= 𝑣𝑓

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Unit named after James Watt who invented the steam engineIn the American system, horsepower is often usedOne horsepower is moving 550 pounds 1 foot in 1 second

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Power in the human body would be how quickly calories are being burnedLook at the table on page 166 to compare the power with the activity

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v0 = 0vf = 100 km/h = 27.78 m/st = 3.2 sm = 500 kg

𝑃 =𝑊

𝑡

𝑃 =

12𝑚𝑣𝑓

2 −12𝑚𝑣0

2

𝑡

𝑃 =

12 1000 𝑘𝑔 27.78

𝑚𝑠

2− 0

3.2 𝑠= 121000 𝑊

𝑃 = 1000𝑊 = 1 𝑘𝑊, 𝑡 = 2 𝑚𝑖𝑛 =1

30ℎ

𝑃 =𝑊

𝑡

1 𝑘𝑊 =𝑊

130 ℎ

𝑊 =1

30𝑘𝑊ℎ

𝑐𝑜𝑠𝑡 =1

30𝑘𝑊ℎ $0.10 = $0.0033

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From 2008 units are billions of kWh

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𝐹Δ𝑡 = 𝑚𝑣𝑓 −𝑚𝑣0

𝐹Δ𝑡 = 0.14 𝑘𝑔 60𝑚

𝑠− 0.14𝑘𝑔 −40

𝑚

𝑠= 14 𝑘𝑔 𝑚/𝑠

𝐹 =14 𝑘𝑔

𝑚𝑠

0.001 𝑠= 14000 𝑁

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𝐹𝑡 = 𝑚𝑣𝑓 −𝑚𝑣0

𝐹 0.01 𝑠 = 0.001 𝑘𝑔 0 − 0.001 𝑘𝑔 −15𝑚

𝑠

𝐹 0.01 𝑠 = 0.015 𝑘𝑔𝑚

𝑠𝐹 = 1.5 𝑁

𝐹 0.01 𝑠 = 0.001 𝑘𝑔 10𝑚

𝑠− 0.001 𝑘𝑔 −15

𝑚

𝑠

𝐹 0.01 𝑠 = 0.025 𝑘𝑔𝑚

𝑠𝐹 = 2.5 𝑁

Hailstones are usually more massive than raindrops so that the force is even greater.The rebounding adds force to the collision

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The person has more mass than the bullet.Since they are half the mass, they will move at twice the speed. Mass and speed ratios are reciprocals.

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Usually only two objects for linear momentum because very rarely do more than two object hit at the same time. It usually happens that two hit, and then one of those hits another.Internal forces the objects pushing each otherExternal forces gravity pulling the objects down

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F12 and F21 are action-reaction pair from Newton’s Third Law are equal and opposite

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Isolated system = no external forces

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𝑝0 = 𝑝𝑓𝑚1𝑣01 +𝑚2𝑣02 = 𝑚1𝑣𝑓1 +𝑚2𝑣𝑓2

0.17 𝑘𝑔 5𝑚

𝑠+ 0.5 𝑘𝑔 0

𝑚

𝑠= 0.17 𝑘𝑔 𝑣 + 0.5 𝑘𝑔 𝑣

0.85 𝑘𝑔𝑚

𝑠= 0.67 𝑘𝑔 𝑣

𝑣 = 1.27 𝑚/𝑠

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𝑝0 = 𝑝𝑓𝑚1𝑣01 +𝑚2𝑣02 = 𝑚1𝑣𝑓1 +𝑚2𝑣𝑓2

5 𝑘𝑔 0 + 0.15 𝑘𝑔 0 = 5 𝑘𝑔 𝑣 + (0.15𝑘𝑔)(35 𝑚/𝑠)0 = 5 𝑘𝑔 𝑣 + 5.25 𝑘𝑔 𝑚/𝑠−(5 𝑘𝑔)𝑣 = −5.25 𝑘𝑔 𝑚/𝑠

𝑣 = 1.05 𝑚/𝑠

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4. One flies out5. Two fly out6. Three fly out7. Four fly out8. Can’t9. v10. mv is the same before and after

11. mv is the same before and after, but 𝐾𝐸 =1

2𝑚𝑣2 will not be the same because of

the v2

𝐾𝐸 = 𝐾𝐸1

2𝑚 2𝑣 2 =

1

2𝑚𝑣2 +

1

2𝑚𝑣2

2𝑚𝑣2 ≠ 𝑚𝑣2

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Demo energy lost to heat by smashing two steel balls together

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SUV crash test videoNASCAR videoCrash test humor

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Momentum𝑚𝑠𝑣0𝑠 +𝑚𝑐𝑣0𝑐 = 𝑚𝑠𝑣𝑓𝑠 +𝑚𝑐𝑣𝑓𝑐

0.1 𝑘𝑔 1𝑚

𝑠+ 0.05 𝑘𝑔 0 = 0.1 𝑘𝑔 𝑣𝑓𝑠 + 0.05 𝑘𝑔 𝑣𝑓𝑐

0.1 𝑘𝑔𝑚

𝑠= 0.1 𝑘𝑔 𝑣𝑓𝑠 + 0.05 𝑘𝑔 𝑣𝑓𝑐

𝑣𝑓𝑠 = 1𝑚/𝑠 − 0.5 𝑣𝑓𝑐Kinetic Energy

1

2𝑚𝑠𝑣0𝑠

2 +1

2𝑚𝑐𝑣0𝑐

2 =1

2𝑚𝑠𝑣𝑓𝑠

2 +1

2𝑚𝑐𝑣𝑓𝑐

2

1

20.1 𝑘𝑔 1

𝑚

𝑠

2

+ 0 =1

20.1 𝑘𝑔 𝑣𝑓𝑠

2 +1

20.05 𝑘𝑔 𝑣𝑓𝑐

2

0.05 𝐽 = 0.05 𝑘𝑔 𝑣𝑓𝑠2 + 0.025 𝑘𝑔 𝑣𝑓𝑐

2

𝑣𝑓𝑠2 + 0.5 𝑣𝑓𝑐

2 = 1𝑚

𝑠

2

1𝑚

𝑠− 0.5𝑣𝑓𝑐

2

+ 0.5 𝑣𝑓𝑐2 = 1

𝑚

𝑠

2

1𝑚

𝑠

2

− 1𝑚

𝑠𝑣𝑓𝑐 + 0.25 𝑣𝑓𝑐

2 + 0.5 𝑣𝑓𝑐2 = 1

𝑚

𝑠

2

− 1𝑚

𝑠𝑣𝑓𝑐 + 0.75 𝑣𝑓𝑐

2 = 0

𝑣𝑓𝑐 −1𝑚

𝑠+ 0.75 𝑣𝑓𝑐 = 0

𝑣𝑓𝑐 = 0 𝑜𝑟 1.33 𝑚/𝑠

𝑣𝑓𝑠 = 1𝑚

𝑠− 0.5 𝑣𝑓𝑐

𝑉𝑓𝑠 = 1 𝑚/𝑠 − 0.5(1.33 𝑚/𝑠) = 0.333 𝑚/𝑠

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Do an actual demonstrationCollision

𝑚𝑏𝑣0𝑏 +𝑚𝑤𝑣0𝑤 = 𝑚𝑏𝑣𝑓𝑏 +𝑚𝑤𝑣𝑓𝑤0.01 𝑘𝑔 𝑣0𝑏 + 0 = 0.01 𝑘𝑔 𝑣𝑓 + 3 𝑘𝑔 𝑣𝑓

0.01 𝑘𝑔 𝑣0𝑏 = 3.01 𝑘𝑔 𝑣𝑓After collision

ℎ = 0.5 𝑚 − 0.5 𝑚 cos 40° = 0.1170 𝑚𝑃𝐸0 + 𝐾𝐸0 = 𝑃𝐸𝑓 + 𝐾𝐸𝑓

0 +1

23.01 𝑘𝑔 𝑣𝑓

2 = 3.01 𝑘𝑔 9.8𝑚

𝑠2(0.1170 𝑚) + 0

1.505 𝑘𝑔 𝑣𝑓2 = 3.45 𝐽

𝑣𝑓 = 1.51 𝑚/𝑠

Combine

0.01 𝑘𝑔 𝑣0𝑏 = 3.01 𝑘𝑔 1.51𝑚

𝑠𝑣0𝑏 = 455 𝑚/𝑠

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