Chapter 7 Work and Kinetic Energy

Adapted from Hyde-Wright, ODU©James Walker, Physics, 2nd Ed. Prentice Hall

Energy• How are we here?

Nuclear energy in the sun, is converted into light, which is absorbed by chlorophyll in plants, making chemical energy from food, which undergoes many transformations, until our muscles convert the chemical energy into motion (kinetic energy).

• Energy can be defined (somewhat vaguely at this point) as the ability to create change.

• The book E=mc2 documents how it took over a century after Newton for the concept of energy to take hold.

Equations of constant acceleration

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• Net (constant) force, Fnet, times displacement, y-y0, equals change in kinetic energy.

• Kinetic Energy:

• Force times Displacement equals Work:

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Constant acceleration motion in two or more dimensions (e.g. projectile motion)

• Just add the equations for each coordinate

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Work done by net force = change in Kinetic energy

Kinetic Energy

An object in motion has kinetic energy:

m = massv = speed (magnitude of velocity)v2 = vx

2+ vy2 (Pythagoras)

The unit of kinetic energy is Joules (J).Kinetic energy is a scalar (magnitude only)Kinetic energy is non-negative (zero or positive)

221 mvK

Work done by a force

• For a constant force, F, the work done ON a mass m while the mass moves through a displacement r = r2r1is (switch from rr0to r2r1notation) W = Fx(x2- x1) + Fy (y2 –y1) = work done by force F Work = [x-component of Force] [x-component of displacement]

+ [y-component of Force] [y-component of displacement] …

• Total work is the work done by the Net force. WTotal = Fnet,x(x2- x1) + Fnet,y (y2 –y1)

• Total work = sum of the work done by each force Fnet = F1 + F2 + F2 +… WTotal = W1 + W2 + W3 +…

Work,in terms of magnitude & direction of force and displacement

• W = Fx(x2- x1) + Fy (y2 –y1) = work done by force F

• Pick a coordinate system with the x-axis along the direction of the force F (remember, because of initial velocity and/or other forces present, this force F does not have to be parallel to the displacement)

• The displacement r makes an angle with respect to the x-axis. Fx = F Fy = 0

Define d = | r | x2 x1 = d·cos

• W = (F) (d) cos

F

r2

r1

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F

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x

y

|r|=

d

• Work is a scalar Work has only magnitude, no direction. The value of W is independent of how we draw the

coordinate system (unlike the components of r or F)

• The SI unit of work is the Joule (J) 1 Joule 1 Newton·meter

1 J = 1 Nm = 1 kg m2/s2

• Note that if F is in the same direction as d, then = 0, and

W= Fd

• If 90º<<270º, then W = F d cos is negative.

Negative Work and Total Work

Work can be positive, negative or zero depending on the angle between the force and the displacement.

If there is more than one force, each force can do work. The total work is calculated from the total (or net) force:

Wtotal = (Ftotal cos)d = Ftotald cos

Walker Problem 9, pg. 194A 55-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 40.0° above the horizontal. If the tension in the rope is 125 N, how much work is done on the crate to move it 5.0 m?

a) How much work is done by the tension in the rope?

b) How much work is done by the floor (friction force)?

c) How much work is done by the net force?a) Wtension = (125 N) (5.0m) cos40.0º=625 Jb) Fnet = 0 (constant speed)

• 0 = Tx + Ffriction • Wfriction = (125N)(cos40.0)(5.0m) =625J

c) WTotal = 0

Work-Energy Theorem

The net (total) work done on an object by the total force acting on it is equal to the change in the kinetic energy of the object:

Wtotal = KE = KEfinal - KEinitial

Forms of Energy

• Kinetic Energy is one form of energy.

• Work is a transformation from Kinetic Energy to another form of energy (potential energy, dissipation into heat, see Chap 8…)

• 1 Calorie (nutrition) = 1kcal = 5.18·103 Joule

Typical Values of Work in our livesActivity Work or Energy (Joules)

Annual US energy use 8·1019

Mt St Helens (1981) 1018

Burning one gallon of gas 108

Human diet, one day 107

Melting an ice cube 104

100 watt light bulb for 1 hour 0.1 kiloWatt-Hr = 3.6·105

1 Human heart beat (pumping blood) 0.5

Turning page in book 103

Flea hop 103

One blue photon, absorbed on retina 5 ·1019

Breaking a bond in DNA 1020

Walker Problem 22, pg. 195A 65-kg bicyclist rides his 10.0-kg bicycle with a speed of 12 m/s.

(a) How much work must be done by the brakes to bring the bike and rider to a stop?

(b) How far does the bicycle travel if it takes 4.0 s to come to rest?

(c) What is the magnitude of the braking force?

a) Friction = only horizontal forceWnet = Kf-Ki <0Wnet = 0 – (1/2) m vi

2

Wnet = (1/2) (10kg+65kg) (12m/s)2

Wnet = 5400 kg m 2 /s 2 = 5400 J

b) Find acceleration firstv = v0+at0 = v0 + at, a= v0/ta = (12m/s)/(4s) = 3 m/s2

x= x0+v0 t+(1/2)at2

x= 0 + (12m/s)(4s) + (0.5)(-3m/s2)(4s)2

x= 48 m-24 m = 24 mc) Braking force

Wnet = Fnet d = FFriction d

FFriction = Wnet /d = ( Nm)/(24m) = 225 N

Force vs. Displacement Graph

The work done by a force can be found from the area between the force curve and the x-axis (remember, area below the x-axis is negative):

Work done by a variable forceWalker Problem 27, pg. 195

An object is acted on by the force shown in the Figure. What is the final position of the object if its initial position is x = 0.50 m and the work done on it is equal to (a) 0.12 J or (b) –0.29 J? (c) If the force acts on a 0.13-kg object and the initial velocity is 3 m/s, find the object’s speed at x = 0.85 m.

a) Area from 0.5m to 0.75 m = (0.4N)(0.25m) = 0.10 JW=0.12 J = 0.10J + 0.02 J = 0.10 J + (0.2N)(x-0.75m)0.02 J = (0.2 N)(x-0.75m) (x-0.75m) = (0.02 J) / (0.2 N) = 0.1m

x=0.85mb) Area from 0.5 m to 0.25 m = (0.8N)(0.25m)= 0.20J

W = 0.29 J = 0.20J + (0.6N)(x-0.25m)0.09J = (0.6N)(x-0.25m) (x-0.25m) = 0.15m

c) Area from 0.25m to 0.00 m = (0.6N)(0.25m) = 0.15J

Work done by a variable forceWalker Problem 27, pg. 195

(c) If the force acts on a 0.13-kg object and the initial velocity is 3 m/s, find the object’s speed at x = 0.85 m.

c) Work = (0.4 N)(0.75m-0.5m) + (0.2 N)(0.85m-0.75m)W = 0.10 J + 0.02 J = 0.12 JW = Kf – Ki Kf = Ki + W

Kf = (0.5) m[vi2] + W = (0.5) (0.13kg)(3m/s) 2 + 0.12J

Kf = 0.585 J + 0.120J = 0.705 J

Kf = (0.5) m[vf2]

vf2 = 2 Kf / m = 2 (0.705 kg m2/s2)/ (0.13 kg) =10.85 m2/s2

Vf =3.29 m/s

Work Done by a Hooke’s Law Force (Spring force)

If the force depends on the displacement, then it will not be constant. Example: spring F = kx

W = -½ kx2

Work done on mass m by a spring as the mass stretches (x>0) or compresses (x<0) the spring a distance x from equilibrium:

Area of triangle = (one half) (base times height)

W = (1/2)(x)(-kx)

F

xF= kx

Work & Spring-Force

• Text describes “Work done by a force to stretch a spring”

• If a spring is stretched a distance x (positive or negative) by some other force (e.g. your hand pushing) such that the mass on the end of the spring travels at constant velocity (a=0) then the net force = 0 and the pushing force = (spring force) = kx) = kx

• Work by external force = (1/2) x (kx)

• W = ½ kx2Fig 7-11§7-3

Fig. 7-13

• Fig 7-13 (motion of a car) does not explain that the backwards force labeled Ffriction is actually kinetic friction in the bearings of the axle etc. and rolling resistance of the tires as they flex.

• The figure also fails to explain that the forward force F is actually static friction between the tires and road (which is the action-reaction partner to the force of the tires pushing backwards on the road).

PowerAverage Power is defined as the time rate of

doing work :

The unit of power is the Watt (W).

1W = 1Joule/sec

Fvt

Fd

t

WP

PowerExample Power (Watt)

Total US Electrical Generation 1012 W = 1TW

Large nuclear power plant 109 W = 1GW = 1000 MW

SUV on highway (12 mpg) 1.5·105

175 hp engine (175hp)(746W/hp)= 1.3·105

100 W light bulb 100

Human metabolism (daily average) 80

Solar energy reaching 1 m2 of earth’s surface at Norfolk on a clear day

250

Walker Problem 39, pg. 196In order to keep a leaking ship from sinking, it is necessary to pump 10.0 lb of water each second from below deck up a height of 2.00 m and over the side. What is the minimum horsepower motor that can be used to save the ship?

2.2lb = (1kg)(9.81 m/s2) = 9.81 NForce applied to water =10 lb = (10 lb)(9.81 N)/(2.2 lb) =44.6 N Velocity of water = (2.00 m) / (1.0sec) = 2.00 m/sPower = F· v = (44.6 N) (2.0 m/s) = 89.2 WP = (89.2 W) / (746 W/hp) = 0.060 hp

Work = Change in Kinetic Energy Fd = ½ mvf

2 ½ mvi2

• A car of mass m travels with speed v. A braking force F brings the car to rest in distance d.

• The car’s speed is now 2v. The same braking force F is applied to bring the car to rest. What is the new stopping distance?

• d/2

• d

• 2d

• 4d