Download - 11. Law of Conservation of Energy
-
8/11/2019 11. Law of Conservation of Energy
1/25
LAW OF CONSERVATIONOF ENERGY
-
8/11/2019 11. Law of Conservation of Energy
2/25
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.
-
8/11/2019 11. Law of Conservation of Energy
3/25
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
-
8/11/2019 11. Law of Conservation of Energy
4/25
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.
-
8/11/2019 11. Law of Conservation of Energy
5/25
For example, a ballthrown upward wouldgain potential energy asit continues to rise but
as a consequence, it willlose its kinetic energy asit slows down.
-
8/11/2019 11. Law of Conservation of Energy
6/25
-
8/11/2019 11. Law of Conservation of Energy
7/25
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.
-
8/11/2019 11. Law of Conservation of Energy
8/25
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.
-
8/11/2019 11. Law of Conservation of Energy
9/25
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
-
8/11/2019 11. Law of Conservation of Energy
10/25
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
-
8/11/2019 11. Law of Conservation of Energy
11/25
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.
-
8/11/2019 11. Law of Conservation of Energy
12/25
-
8/11/2019 11. Law of Conservation of Energy
13/25
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.
-
8/11/2019 11. Law of Conservation of Energy
14/25
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.
-
8/11/2019 11. Law of Conservation of Energy
15/25
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.
-
8/11/2019 11. Law of Conservation of Energy
16/25
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
-
8/11/2019 11. Law of Conservation of Energy
17/25
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
-
8/11/2019 11. Law of Conservation of Energy
18/25
-
8/11/2019 11. Law of Conservation of Energy
19/25
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
-
8/11/2019 11. Law of Conservation of Energy
20/25
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
-
8/11/2019 11. Law of Conservation of Energy
21/25
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
-
8/11/2019 11. Law of Conservation of Energy
22/25
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
-
8/11/2019 11. Law of Conservation of Energy
23/25
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
-
8/11/2019 11. Law of Conservation of Energy
24/25
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
-
8/11/2019 11. Law of Conservation of Energy
25/25
ASSIGNMENT
Answer Test Yourself Problem Nos. 9-11 on p. 223.Reference: Breaking Through Physics