Lecture 16 (Walker: 7.3-4) - Physics 111

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Physics 111Lecture 16 (Walker: 7.3-4)

Work & PowerMarch 6, 2009

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Energy• Forms of energy:

– Mechanical• Kinetic, Potential; focus for now

– Thermal– Chemical– Electromagnetic– Nuclear

• Energy can be transformed from one form to another– Essential to the study of physics, chemistry,

biology, geology, astronomy• Can be used in place of Newton’s laws to solve certain

problems more simply• Energy units: SI Unit - Joule (J); Calorie (food

calorie) = 4.2 kJ; Kilowatt-hour = 3.6 MJ

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Kinetic Energy &The Work-Energy Theorem

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

(7-7)

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Work and Kinetic Energy• An object’s kinetic

energy can also be thought of as the amount of work the moving object could do in coming to rest– The moving hammer

has kinetic energy and can do work on the nail

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Example: Work & KE-Rocket Launch150,000 kg rocket launched straight up. Rocket engine generates thrust of 4.0 x 106 N. Rocket’s speed at height 500 m? (Ignore air resistance and mass loss due to burned fuel.)

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6 9thrust thrust ( ) (4.0 10 N)(500 m) 2.0 10 JW F y= ∆ = × = ×

4 2grav ( ) ( ) (1.5 10 kg)(9.80 m/s )(500 m)W w y mg y= − ∆ = − ∆ = − × =

1 2 9thrust grav2 0 1.26 10 JK mv W W∆ = − = + = ×

2 1.30 m/sKvm∆

= =

= - 0.74 x 109 J

Lecture 16 7/23

Graphical Interpretation of WorkIf the force is constant, we can interpret

the work done graphically:

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Work Done by a Variable ForceIf the force takes on several successive

constant values, we can add the areas of each of the “blocks”:

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Work Done by a Variable ForceWe can then approximate a continuously

varying force by a succession of constant values.

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Example:Work Done by a Varying Force

A force varies with x as shown. Find work done byforce on a particle that moves from x = 0.0 m to x = 6.0 m.

ˆxF F x=r

totalW A=

1total 1 2 2(5.0 N)(4.0 m) (5.0 N)(2.0 m)

20.0 J 5.0 J 25.0 JW A A A= = + = +

= + =

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Work and SpringsThe force needed to stretch a spring an

amount x is F = kx.Therefore, the work done in stretching the spring is

(7-8)

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Spring Work by Average Force

1 22 fW k x=

Force needed to stretch spring distance x: F = kx

Average force during stretch from x = 0 to x = xf: ½kxf

Work done during stretch:

W = FAvg ∆x cosθ = (½kxf)xf = ½kxf2

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Question

A spring-loaded gun shoots a plastic ball with a speed of 4.0 m/s. If the spring is compressed twice as far, what is the ball’s speed?

a) 2.0 m/s b) 4.0 m/s c) 8.0 m/s d) 16.0 m/s e) 32.0 m/s

4.0 m/s

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Example: Work Done on Block by Spring4.0 kg block on frictionless surface is attached to horizontal spring with k = 400 N/m. Spring is initially compressed to 5.0 cm. (a) Find work done on block by the spring as block moves from x = x1 = -5.0 cm to its equilibrium position of x = x2 = 0 cm. (b) Find the speed of the block at x2 = 0 cm.

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1 12 2 2 22 2

2f i f i

WW mv mv v vm

= − ⇒ = +

2 2(0.50 J) 0.50 m/s(4.0 kg)f

Wvm

= = =

W = FAvg ∆x cosθ = (-½k(-.05m))(.05m) = ½(400 N/m)(.05m)(.05m) = 0.50 J

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Example: Dragging a BlockSpring is attached to 2

kg block. The other end is pulled by a motorized toy train that moves forward at 5.0 cm/s. Spring constant is k=50 N/m and the coefficient of static friction between the block and the surface is µs=0.6. Spring is in equilibrium at t=0 s when the train starts to move.

At what time does the block start to slip?

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net( ) ( ) ( ) 0x sp x s x sp sF F f F f= + = − =∑s s spf mg F k xµ= = = ∆

2(0.60)(2.0 kg)(9/80 m/s )(50 N/m)

0.235 m 23.5 cm

smgxk

µ∆ = =

= =

(23.5 cm) 4.7 s(5.0 cm/s)

xtv∆

= = =

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Power (P)Power is a measure of the rate at which work

is done. If work W done during time t:

(7-10)

SI power unit: 1 J/s = 1 watt = 1 W

Also: 1 horsepower = 1 hp = 746 W

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Power*If an object is moving at a constant speed in

the presence of friction, gravity, air resistance, and so forth, the power exerted by the driving force can be written:

(7-13)

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Human Basal Metabolism 80W

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Example: Power of a MotorA small motor operates a lift that raises a load of bricks weighing 500 N to height of 10 m in 20 s at constant speed. Lift weighs 300 N.

What is the power output of the motor?

W = F d cosθ = (800N)(10m) = 8000J

P = W/t = 8000J / 20s = 400 W

(400 W = 0.54 hp)

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Before Monday, read Walker 8.1-2

Homework Assignments #7b should be submitted using WebAssign by 11:00 PM on Monday, March 9.

End of Lecture 16