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Ch07: Kinetic Energy and WorkEnergyKinetic EnergyWork-kinetic energy theoremWork done by constant forceWork done by gravitational forceWork Done by a Variable ForceWork done by spring force

Energy is one of fundamentals goals of physics to investigate. Many examples exist for energy like

Any type of motionFlying in the skyLifting materials

Nations spends a huge amounts of money to acquire and use energy.What the term Energy means?

7.1: What is Physics

Most common definition: Capacity of a system to do work (produce change). Other definitions: Energy is a scalar quantity associated with the state (or condition) of one or more objects.Some characteristics:

Energy can be transformed from one type to another and transferred from one object to another, The total amount of energy is always the same (energy is conserved).

7.2: What is Energy? 7.3: Kinetic EnergyKinetic energy K is energy associated with the state of

motion of an object. The faster the object moves, the greater is its kinetic energy.

the kinetic energy for an object of mass m and has speed v is

The SI unit of kinetic energy (and every other type of energy) is the joule (J),

1 joule = 1 J = 1 kgm2/s2.Example: duck of mass 3 kg flying at 2 m/s has kinetic

energy J6)2)(3(

21

21 22 === mvK

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Two train locomotives, are 6.4 km apart on a track, are accelerates from rest at 0.26 m/s². if each locomotive has a weight of 1.2×106 N, Find the total kinetic energy before collision.

Solution. To calculate K, we need the speed of each locomotive before collision. Also since both have same acceleration, they will collide at the middle of the track (3.2 km). Speed of one locomotive can be calculated by

7.3: Kinetic Energy-Example

For the two locomotives, the Kinetic energy before collision is

7.4: Work

Work W is defined as energy transferred to or from an object by means of a force acting on the object.

Energy transferred Change in Energy work is done.To change K for an object change v we need awe must apply force F work is done by force F.

Energy transferred to the object positive (+ve) workEnergy transferred from the object negative (-ve) work.

7.5: Work and Kinetic EnergyFinding work expression:

Consider a force acting on an object of mass m at angle and accelerates it in the x-direction as shown

Applying Newton’s law in the x-direction(1)

(2)From (1) solve for ax and sub. in (2), then rearrange

the kinetic energy changes (∆K) due to the force F. this change is equal to Fxd. note that Fy has no contribution to change the speed Fy does no work

Through a displacement the object change velocity from due to we can write

vv rr to0

; Fx=Fcos

7.5: Work and Kinetic Energy: work done by constant force

According to definition, the energy transfer due to the force is the work W done on the object by the force. It can be expressed by:

Work done by the force can be defined as: The product of the displacement magnitude (d) and the force component in the direction of the displacement.

Work has same definition of scalar or dot product we can write the work

SI unit of work is joule; 1 joule = 1 J = 1 kgm2/s2= 1N.m

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No energy transfer

Energy transfer from environment (force) to the object (object)

Energy transferred out of the system (from object to environment)

If = 0°

⁄⁄ (parallel)

W= Fd (+ve work)

If = 90°

┴

W= 0 J

If = 180°

and are anti-parallel

W= -Fd (-ve work)

7.5: Work and Kinetic Energy: work done by constant force

Energy transfer from the object or to the object can be decided from the sign (+ or -) of work done by F. The sign can be found from the angle (cos )

From

Ex: a 50 kg object pulled 25m by force F= 80 N on floor with friction, if μk=0.1, finda) Work done by each forceb) Total work done on the object

JFdWF 173030cos)25(80cos === φ

JdFmgnd

dfdfW

kk

kkfk

1120)25)(8.9)(50(1.0)30sin(

180cos

−=−=°−−=−=

−==

µµ

090cos)F and n( 0 a) g =°⊥== dJWWgFn

rrr

JWWWWWgk FnfFT 6100011201730 =++−=+++=b)

or JdfFddfFdFWW kkxT 61030cos)30cos()( =−=−===∑ ∑

7.5: Work and Kinetic Energy: work done by constant force - Example

Find a)Displacement magnitude d = ?and Force magnitude F = ?b) Work done on the object W = ?c) Work done in the y-direction =?

Ex:A particle moving in the xy plane undergoes a displacement d as a constant force F acts on the particle.

29²2²5

13²3²2)

=+=

=+=

F

da

JdFWb 16)3(2)2(5) =+=⋅=rr

( )( )Nj.i.F

mj.i.dˆ02ˆ05

ˆ03ˆ02

+=

+=r

r

JWc y 6)3(2) ==

7.5: Work and Kinetic Energy: work done by constant force - Example 7.5: Work-Kinetic Energy theorem

The change in kinetic energy (K) have been related to the work by the equation

Which can be written as

In general, If a net Force (Fnet ) cause a change in kinetic energy we can write

Work-Kinetic Energy Theorem

KKKWW ifnet ∆=−==∑ ∑+= WKKor if

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Two spies sliding a ground safe a displacement d = 8.5 m as shown. Spy 1 pushes with force F1=12N at 30° below horizontal and spy 2 pulls with force F2=10 N at 40° above horizontal. Find

(a) Net work done on the safe(b) Work done by Fg and FN and normal forces(c) the safe speed at the end of the displacement d if it is start from rest

7.5: Work-Kinetic Energy theorem: Example

Net work

(a)

FBD

Two spies sliding a 225 kg ground safe a displacement d = 8.5 m as shown. Spy 1 pushes with force F1=12N at 30° below horizontal and spy 2 pulls with force F2=10 N at 40° above horizontal. Find(b) Work done by Fg and FN and normal forces(c) the safe speed at the end of the displacement d if it is start from rest

7.5: Work-Kinetic Energy theorem: Example

(b)

(c) (vi = 0 Ki = 0)

a crate of crepe is sliding across a slick,oily parking lot through a displacementwhile a steady wind pushes against the crate with a force . Find

7.5: Work-Kinetic Energy theorem: Example

mid ˆ3−=r

NjiF )ˆ6ˆ2( −=r

(a) How much work does this force do on the crate during the displacement?(b) If the crate has a kinetic energy of 10 J at the beginning of displacement , what is its kinetic energy at the end of the displacement

(a)

(b)

7.6: Work done by gravitational forceAssume object is thrown upward with initial speed v0 .The work done by gravitational force is

For a rising (moving up) object ( )

Wg does –ve work v decreases K decreases

When the object falls back down ( )

Wg does +ve work v increases K increases

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7.6: Work done by gravitational force: Work Done in Lifting and Lowering an ObjectAssume we lift a particle at constant speed v by a

vertical applied force Fa=Fg in magnitudeaFr

Appling work-kinetic energy theorem total work

Since v is constant ∆K = 0

Work done by Fa in lifting an object

Same result is obtained when lowering an object at constant speed v by a vertical applied force aF

r

7.6: Work done by gravitational force: Example

1)

2) Since elevator descending at constant speed

3) The net work

4)

An elevator cab (m=500 kg) descending at speed v = 4 m/s. calculate1) Work done by gravitational force through a distance 12m2) Work done by cable tension through the 12m distance3) The net work done on the elevator4) The cabs kinetic energy at the end of the 12 m descending

7.6: Work done by gravitational force: Example

1)

2) We have downward acceleration

3) The net work

4)

An elevator cab (m=500 kg) descending at initial speed vi = 4 m/s. If theacceleration of the cab was g/5 downward, calculate

1) Work done by gravitational force through a distance 12m2) Work done by cable tension through the 12m distance3) The net work done on the elevator4) The cabs kinetic energy at the end of the 12 m descending

∑ ∆=⇒∆≈∆ xFWxFW xx

∑∑ ∆=∆=→∆→∆

f

i

f

i

x

xxx

x

xxxFWW

00limlim

( )∫ ∑∑ ==f

i

x

xxnet dxFWW

For more than one force acting upon a particle moving along x-axis

Consider force and displacement along the x-axis

7.8: Work done by variable force

In general for three dimensional motion

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W=area under graph

=rectangle area +triangle area

W=4(5)+1/2 (2)(5)

=20 + 5 = 25J

7.8: Work done by variable forceExample: from the graph below calculate the work doneto move a particle from x=0 to x=6

7.7: Work done by variable force: exampleForce , , with x in meters, acts on a particle, changing only the kinetic energy of the particle. How much work is done on the particle as it moves from coordinates (2 m, 3 m) to (3 m, 0 m)? Does the speed of the particle increase, decrease, or remain the same?

Since the force is in two dimensions x and y

Spring Constant

Spring always tends to restore his relaxed state restoring force always has direction opposite to displacement d

For the figure shown

7.7: Work done by a spring force: The spring force

22

21

21

fis

x

x

x

xss kxkxWkxdxdxFW

f

i

f

i

−=⇒−== ∫ ∫

For an arbitrary displacement of a block between xi and xf , by the force of a spring, the work done by the spring force on the block is

7.7: Work done by a spring forceTo find the work done by the spring on a block, we considered the

following two assumptions:

1) The spring is massless and 2) the spring is ideal (obeys Hook’s law exactly)

Assuming initial position xi = 0and final position xf = x

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( ) ( )∫∫∫ =−==f

i

f

i

f

i

x

x

x

xs

x

xaa dxkxdxFdxFW

22

21

21

ifa kxkxW −=

7.7: Work done by a spring force: the work done by an applied force

Assume an object is displaced compressing or expanding a spring by an external applied force from xi to xf at constant speed v Fa = Fs

Appling work-kinetic energy theorem total work

Since v is constant ∆K = 0

22

21

21

ifa kxkxW −=or

7.7: Work done by a spring force: example

222

21

21

21 kdkxkxW fis −=−=

a box of mass m = 0.4 kg slides across a horizontal frictionless table with speed v = 0.50 m/s. It then and compresses a spring of spring constant k =750 N/m. When the box is momentarily stopped by the spring, by what distance d is the spring compressed?

but

Ex: A 0.55 kg object is attached to a vertical hanged spring, If the spring is stretched 2 cm, find a) spring constant b) the work done by spring as it stretched.

mNd

mgk

mgdkmgkymgFs

/107.2102

)8.9(55.0)(

22 ×=

×==⇒

=−−=−⇒=⇒

−

m = 0.55 kg, d = 2 cm (y=-d=-2cm)

FBD

a) After streach, the object is at equilibrium

J

kd

kykyW fis

2

2222

22

104.5

)102)(107.2(21

210

21

21

−

−

×−=

×−×−=−=

−=b)

7.7: Work done by a spring force: example

The kilowatt-hour (kWh) is a unit of energy (or work) 1 kWh=(103 W) (3600 s)= 3.6×106 J

7.9: PowerThe time rate at which work is done by a force is said to be the power

due to the force.

and

The SI unit for the Power is Watt (W) Where 1 watt = 1 W = 1 J/s

In British system, they usually use the horsepower

where 1 horsepower = 1 hp = 550 ft.lb/s = 746W.

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7.9: Power: ExampleTwo forces F1 = 2N and F2 = 4Nacting on box as shown. At the instant the speed v is 3 m/s find(a)power of each force (b) total power (c) is net power changing at that instantAt v = 3m/s instantaneous power

(a) (b) The total power is

(c) net power does not change because net rate of energy transfer (Pnet) is zero. Also you can find net force in v direction (x-direction) Fx,net = 0

v is constant no change in Kinetic energy no energy transfer ( Pnet= 0)

removing energy

supplying energy

Review