11. Law of Conservation of Energy

8/11/2019 11. Law of Conservation of Energy

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LAW OF CONSERVATIONOF ENERGY


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Energy is needed in order to carry out changes on anobject. Physically, this change is expressed in thechange of displacement due to an applied force on the

object. Thus, we formally define energy as the capacityto do work.

Moreover, the extremely important principle is thelaw of conservation of energy which states that:

Energy can neither be created nor destroyedbut can be converted from one form toanother.


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The only thing that changes is the form inwhich energy appears.

In a mechanical energy, its total mechanical

energy ME is the sum of its kinetic andpotential energies:

ME = KE + PE

ME = mv2+ mgh


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When there is no transfer of energy into or out of thesystem, the total energy of a system remains constantand is said to be conserved:

ME before= MEafter

PEbefore+ KEbefore= PEafter+ KEafterThis gives us the expression,

KE= - PE

which means that, of the total mechanical energyconserved, the changes in the kinetic and potentialenergy must be equal but opposite in sign. In other

words, a gain in one must be matched by acorresponding loss in the other.


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For example, a ballthrown upward wouldgain potential energy asit continues to rise but

as a consequence, it willlose its kinetic energy asit slows down.


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Consider the motion of a pendulum. When you pull the bobsideways to position A, work is done on the bob. This work

transforms part of your bodysenergy to GPE of the bob. Asthe bob swings from A to B, thebobsPE changes to KE. As it

continues to swing to C, the bobsKE at B transforms againto PE. The motion of a pendulum is an example of KE and PEexchanges or energy transformation.


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KE +PE = constant

Kinetic and potential energy are forms of mechanicalenergy. If friction is negligible, the sum of the kineticand potential energies of a body remains constant. Inequation,

KE + PE = constant

This applies to the motion of a pendulum as shown inthe figure.


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It also applies to any freelyfalling body. Consider theexample in the figure.

At the highest point, A

(PE)A= mgh; (KE)A= 0

At any point in its path, say, B

(PE)B+ (KE)B= constant

= mghAt the lowest point, C

(PE)C = 0

(KE)C= (PE)A= mgh


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A stone with mass 0.1 kg isdropped from a height of80 m. Calculate its PE andKE (a)at the time it was

dropped, (b) after it hastraveled 20 m, and (c) atthe instant it touches theground. Assume that air

friction is negligible andthe reference point is theground.

Solution:a. PE =mgh

= (0.1kg)(9.8m/s2)(80m)

PE = 78.4 J; KE = 0

b. PE = mgh

= (0.1kg)(9.8m/s2)(60m)

KE= 58.8 J; KE = 19.6 J

c. PE = 0;KE = 78.4 J


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The conservation law also applies to vibrating bodiessuch as the spring. The figure shows differentconditions of a spring. At A, the spring is atequilibrium position.

When you stretch the spring to position B, you dowork on the spring. The work done on the spring isstored as EPE of the spring. When you release thespring, it compresses to C and then stretches back toB, changing its PE to KE. Then the spring iscompressed again, changing its KE to PE.


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The spring attains maximum PE at C. From there, it stretches again to B,where maximum PE is also attained, and so on. Thus, in a vibratingspring, there is a continuous exchange of KE and PE. But the total energyremains constant if friction is negligible. It is commonly observed thatafter a few minutes of oscillating, the spring eventually stops. This is due

to friction.


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All the examples we have given are based on theassumption that there is negligible fiction. In reality,some amount of energy is often lost due to friction, asin pendulum. After several swings, the pendulum will

not be able to swing back to its maximum height. Itwill gradually slow down until it stops. This is due tofriction.


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Important points to remember

1. When work is done on an object, the objectgains energy; when work is done by theobject, the object loses energy.

2. When a body loses a certain amount of energy,another body gains this energy.

3. If a bodyskinetic energy is lost, this energy hasbeen converted to potential energy or some other

form of energy.


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Example 1

Suppose you throw anobject with mass 0.2 kgand at an initial speed of50 m/s, applying the lawof conservation of

mechanical energy,(a) what will be the

maximum GPE theobject can attain and

(b) how high will it go?Assume that the stones

initial GPE is zero andair friction is negligible.

Solution: (PE)i= 0; (KE)f= 0a. (PE)i+(KE)i= (PE)f + (KE)f

0 + (KE)i= (PE)f +0

(KE)i= mvi2

= (0.2kg)(50 m/s)2

(KE)i= 250 JNow, (PE)f= (KE)i

(PE)f= 250 Jb. PE =mgh

250 J = 0.2 kg(9.8 m/s2) h

127.55 m = h


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A 30-kg package slides from restdown a frictionless ramp from aheight of 5.0m. At the bottom ofthe ramp is a spring of forceconstant 400 N/m. The package

slams into the spring, compressingthe spring a distance x beforestopping momentarily. Find the

value of x.

Solution:(GPE)i+(EPE)i= (GPE)f+(EPE)f

(GPE)i+0 = 0 +(EPE)f(GPE)i= +(EPE)f

mgh = kx2

x2= 2mgh

k

x2= 2(30kg)(9.8m/s2)(5m)

400 kg/s2

x2= 7.35 m2

x = 2.71 m

Example 2


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With what speed should thebaseball be thrown if itshould reach a height of 20m?

Given:vi =?

vf= 0; at max. height

(PE)i= 0; at lowest point

(KE)f= 0; at highest point

Solution:(PE)i+(KE)i= (PE)f + (KE)f

0+ mvi2= mghf+ 0

1/m [ mvi2= mghf]

( vi2= ghf) x 2

vi2 = 2ghvi= 2(9.8m/s2)(20m)

vi= 19.8 m/s

Example 4


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Consider the roller coasterillustrated below. What is thespeed of the cart at the lowestpoint if the initial speed is 25m/s?

Given:

vi= 25 m/shi= 15 m

hf= 3m

Solution:(PE)i+(KE)i= (PE)f + (KE)f

mghi+ mvi2= mghf+ mvf

2

( ghi+ vi2= ghf+ vf

2) x 2

2ghi+ vi2= 2ghf +vf

2

vf2= 2ghi+vi

22gh2

= 2(9.8m/s2)(15m)+(25 m/s)2

2(9.8m/s2) (3m)

vf2= 860.2 m2/s2

vf= 29.33 m/s

Example 5


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A large chunk of ice with mass15 kg falls from a roof 8.0 mabove the ground.

a. Ignoring air resistance, findthe KE of the ice when itreaches the ground?

b. What is the speed of the icewhen it reaches theground?

Given:m = 15 kgh = 8.0 mKEi= 0; at the highest pointPEf= 0; at the lowest point

Solution:a. (PE)i+(KE)i= (PE)f + (KE)f(PE)i+ 0 = 0 + (KE)f

(PE)i= (KE)f

(PE)i = mgh= 15 kg(9.8 m/s2)(8m)

(PE)i= 1,176 J = (KE)fb. KE = mv2

v2= 2 KE = 2(1,176 J)m 15 kg

v2= 156.8 m2/s2

v = 12.51 m/s

Exercise 1


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Exercise 2

The outline of a roller coastertrack is shown in the figure.The roller coaster car has amass of 1000 kg. It startsfrom rest at point A on the

track.

Determine the KEs,PEs, andspeeds at points indicated.

Position PE(J) KE(J) V(m/s) PE

+KE(J)

A

B

C

D

E

Position PE(J) KE(J) V(m/s) PE

+KE(J)

A 1.47 x105 0 0 1.47 x105

B 9.8 x1 04 4.9 x104 9.9 1.47 x105

C 4.9 x104 9.8 x104 14 1.47 x105

D 2.94 x1 04 1.18 x105 15.34 1.47 x105

E 0 1.47 x105 17.15 1.47 x105


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A golf ball (mass =0.22 kg) was struckat the tee, leaving at a speed of 44m/s at an angle of 450.

a. What was the initial kinetic energyof the ball?

b. What was the horizontalcomponent of this initial speed?

c. What was the velocity of the ball atthe highest point?d. What was the maximum height

reached by the ball?

Given:m = 0.22 kg

vi= 44 m/svf = 0; at the highest point= 450

KEf= 0; at the highest pointPEi= 0; at the lowest point

Solution:

a. KEi= mv2= (.22kg)(44m/s)2

KEi= 219.96 J

b. vix= vicos = 31.11 m/s

c. vif= 0d. (PE)i+(KE)i= (PE)f + (KE)f

0 +(KE)i= (PE)f + 0

(KE)i= mghf

mg mg

__219.96 J __ = hf(0.22kg)(9.8m/s2)

102 m = hf

Exercise 3


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A 60-kg high-diver startshis/her dive at a height of 20m.

a. What is his/her totalmechanical energy at thestart of his dive?

b. What is his/her velocityhalfway into the dive?

c. What is his/her velocity justbefore hitting the waterbelow?

Solution:a. MET= PEi+ KEi= mgh + 0= 60kg(9.8m/s2)(20m)= 11,760 J

b. (PE)i+(KE)i= (PE)f + (KE)f

mghi + 0 = mghf+ mv2

( mv2 = mghf+ mghi)1/m v2= ghf ghi

v2= g(hfhi)v = 14 m/s

c. vf2= vi2+ 2gdvf

2= 0 + 2gdvf

2= 2(9.8m/s2)(20m)v = 19.8 m/s

Exercise 4


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ASSIGNMENT

Answer Test Yourself Problem Nos. 9-11 on p. 223.Reference: Breaking Through Physics

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11. Law of Conservation of Energy