Lecture Outlines Chapter 7 Physics, 3

(1)

This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning.

Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

Lecture Outlines Chapter 7

Physics, 3

^rd

Edition

James S. Walker

(2)

Chapter 7 Work and Kinetic Energy

(3)

Units of Chapter 7

• Work Done by a Constant Force

• Kinetic Energy and the Work-Energy Theorem

• Work Done by a Variable Force

• Power

(4)

7-1 Work Done by a Constant Force

The definition of work, when the force is parallel to the displacement:

(7-1)

SI unit: newton-meter (N·m) = joule, J

(5)

7-1 Work Done by a Constant Force

Om F = 82,0 N, d= 3,00 m (och eftersom kraften är parallell med förflyttningen) så blir det

uträttade arbetet

W = Fd = 82,0 N • 3,00 m = 246 Nm = 246 J (7-1)

Exercise 7-1 Darwin’s fink och hårt fröskal

F=205 N, d=0,40 cm, ger W = 0,82 J

(6)

7-1 Work Done by a Constant Force

(7)

Example 7-1 (p.181) Heading for the ER (m_p=72,0 kg, m_gurney = 15 kg, if d=2,5 m, W =?

Anta att vagnen rör sig friktionslöst)

(8)

7-1 Work Done by a Constant Force

If the force is at an angle to the displacement:

(7-3)

(9)

Figure 7-3

Force at an angle to direction of motion: another look

(10)

7-1 Work Done by a Constant Force

The work can also be written as the dot

product of the force and the displacement:

(11)

Example 7-2 (p.183) Gravity Escape System (d = 5,0 m, m = 4970 kg W_g = ?)

(Lägg märke till att θ är inte definierad på det vanliga sättet )

(12)

Conceptual Checkpoint 7-1(p.184)

Path Dependence of Work (W₁>W₂?, W₁>W₂?, W₁=W₂? (Lådan glider friktionlöst)

(13)

7-1 Work Done by a Constant Force

The work done may be positive, zero, or

negative, depending on the angle between the

force and the displacement:

(14)

7-1 Work Done by a Constant Force

If there is more than one force acting on an

object, we can find the work done by each force, and also the work done by the net force:

(7-5)

(15)

Example 7-3 A Coasting Car I

(16)

Example 7-3 A Coasting Car I

W_N = Nd(cos90º) =0 W_mg = mg(sinΦ)d

W_air = F_air (cos180º)d = - F_air d

W_total = W_N + W_mg + W_air = 0 + mgsin(Φ) d - F_air

(17)

Example 7-4 A Coasting Car II

(18)

Example 7-4 A Coasting Car II

Nu skall vi beräkna arbetet genom att beräkna skalärprodukten mellan F_totaloch förflyttningen d. Då blir

W_total = (mgsin(Φ) - F_air)d

Eftersom F_totalärgiven på föregående bild.

Vilket förstås är det uttryck som vi erhöll i Example 7-3

(19)

7-2 Kinetic Energy and the Work-Energy Theorem

When positive work is done on an object, its

speed increases; when negative work is done,

its speed decreases.

(20)

7-2 Kinetic Energy and the Work-Energy Theorem

Med hjälp av rörelseekvationen v

_f²

=v

_i²

+2ad och Newton’s lag a=F

_tot

/m fås v

_f²

=v

_i²

+2(F

_tot

/m)d

Rörelseenergin definieras nedan och är ALLTID positiv (eller = 0)

(7-6)

(21)

Table 7-2

Typical Kinetic Energies

(22)

Exercise 7-2 (p.188) A truck moving at 15 m/s has a kinetic energy of 420 000 J a) What is the mass of the truck?

b) If the speed is doubled? K → (factor)K?

a)m = (2K/v

²

) = 2 • 420 000 J/(15 m/s)

²

= 3700 kg b) v → 2v då går K → 4 K

(7-6)

(23)

7-2 Kinetic Energy and the Work-Energy Theorem

Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy.

(7-7)

(24)

Example 7-5(p.188) Hit the Books (m=4,10 kg, F_app= 52,7 N) a) Hur stort arbete uträttar F_app? b) Hur stort arbete uträttar gravitationen? c) v_f= ?

(25)

Example 7-5 Hit the books

a) W_app = F_app Δy = 52,7 N 1,60 m = 84,3 J

b) W_g = mg(cos180)Δy = 4,10 kg 9,81 m/s² (-1) 1,60 m

= - 64,4 J

c) ma_y = F_app - mg

= 52,7 N - 4,10 kg 9,81 m/s² = 12,5 N

v_f² = v_i² + 2Δy a_y = 0 + 2 1,60 m 12,5N/(4,10 kg) v_f= 3,12 m/s

(26)

Example 7-6 (p.189) Pulling a Sled

(m = 6,40 kg, v_i = 0,50 m/s, ingen friktion, W = ?, v_f= ?)

(27)

Example 7-6 Pulling a sled

a) W_boy = F(cos 29,0º)d = 11,0 N (cos 29,0º) 2,00 m = 19,2 J

b) ΔK =m(v_f² - v_i²)/2 = 19,2 J

v_f² = v_i² + 2 19,2J/(6,40 kg) = (0,5 m/s)² + 6,00 m²/s² v_f= 2,50 m/s

(28)

Conceptual Checkpoint 7-2 (p.190) Compare the Work a) W₂ = W₁ b) W₂ = 2W₁ c) W₂ = 3W₁ d) W₂ = 4W₁ ?

(29)

7-3 Work Done by a Variable Force

If the force is constant, we can interpret the

work done graphically:

(30)

7-3 Work Done by a Variable Force

If the force takes on several successive constant

values:

(31)

7-3 Work Done by a Variable Force

We can then approximate a continuously varying

force by a succession of constant values.

(32)

Figure 7-9

Stretching a spring

(33)

7-3 Work Done by a Variable Force

The force needed to stretch a spring an amount x is F = kx.

Therefore, the work done in stretching the spring is

(7-8)

(34)

Example 7-7a+b (p.192-3) Flexing an AFM (atomic-force microscopy) Cantilever (kiselkonsol, 250 μm lång) Till att förlänga fjädern 0,10 nm krävs arbetet 1,2 10^-20 J. a) Hur stor är fjäderkonstanten k? b) Hur stort arbete krävs för att öka x till 0,20 nm från 0,10 nm?

a) k = 2W/x² = 2 1,2 10^-20 J/(0,10 10^-9 m)² = 2,4 N/m

b) W(0 → 0,2 nm) = kx²/2 = 4 W(0 → 0,1 nm) + W(0,1 nm → 0,2 nm) = = 4 W(0 → 0,1 nm) = 4,8 10^-20 J dvs W(0,1 nm → 0,2 nm)= 3,6 10^-20 J

(35)

Figure 7-11

Work done in stretching a spring: average force

(36)

Example 7-7b Flexing an AFM Cantilever Fixeringsbild! Allt OK??

(37)

Figure 7-12

The work done by a spring can be positive or negative

(38)

Active example 7-1 (p.194) A block

compresses a string. (m=1,5 kg, v

₀

=2,2 m/s, k = 475 N/m Vad blir x (<0) då v=0?

K

_i

= mv

₀²

/2 = 3,6 J och K

_f

= 0

Dvs fjädern utför ett negativt arbete på blocket eftersom ΔK < 0

Då blir - kx

²

/2 = - 3,6 J

x = ((2•3,6 J)/(475 N/m))

^1/2

= 0,12 m

(39)

7-4 Power

Power is a measure of the rate at which work is done:

(7-10)

SI unit: J/s = watt, W

1 (engelsk) horsepower = 1 hp = 746 W

1 (svensk) hästkraft = 75 kp m/s = 736 W

(40)

7-4 Power

(41)

Example 7-8 Passing Fancy

(42)

Example 7-8 (p.196) Passing Fancy

m

_car

= 1300 kg , v

_i

= 13,4 m/s till v

_f

= 17,9 m/s på t = 3,00 s. P=?

W = ΔK = mv

_f²

/2 – mv

_i²

/2 = 91,6 kJ

P = W/t = 91,6 kJ/3,00 s = 30,5 kW

(43)

Figure 7-13 Driving up a hill

(44)

Active Example 7-2 (p.197) Find the maximum (constant) speed.

F = 1280 N (m

_car

= 1500 kg, θ = 5,00°) P = 50,0 hp = 50,0 • 746 W = 37,3 kW

P = W/t = F • d /t = F v (om hastigheten är konstant)

v = P/F = 37,3 kW/1280 N = 29,1 m/s

(45)