Law of Conservation of
Energy
chemical Light chemical Mechanical
chemical Mechanical
Electrical Light
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Energy transformations
Energy can be converted from one form to another form in
different systems, machines or devices.
Each change transfers energy or transforms energy from one
form to another.
Example
A bow and arrow transform potential energy in a stretched bow into
energy of motion (i.e., kinetic energy) of an arrow.
The Law of Conservation of Energy
Energy can neither be created nor be destroyed, it just
converted from one form into another. This is called the law of
conservation of energy.
The law of conservation of energy is one of the most important
laws in physics. It applies to all forms of energy.
Energy has to come from somewhere
The law of conservation of energy tells us that energy cannot
be created from nothing. If energy increases somewhere, it
must decrease somewhere else.
Once we know how energy flows and transforms, we have a
good understanding of how a system works.
Example
When we use energy to drive a car, that energy comes from
chemical energy stored in petrol. As we use the energy, the
amount left in the form of petrol decreases.
Conservation of Mechanical Energy
The mechanical energy i.e., the sum of potential and kinetic
energies is constant in the absence of any frictional forces.
A only PE
KE = 0
M.E. = K.E. + P.E. = Constant at
any point B Both PE & KE
C PE = 0
only KE
Let us take an example of free fall (here, the effect of air
resistance on the motion of the object is ignored).
Let us consider an object of A
m
mass ‘m’ at a certain height ‘h’
h
Let it is dropped from this height
from point A i.e., vA = 0 A PE = max
KE = 0
Mechanical energy at A,
EA = K.E. at A + P.E. at A
1 h
= mv + mghA
2
1
= 2 m (0)2 + mgh
EA = mgh ….. (1)
Let after a certain time ‘t’, it reaches
A PE = max
point B after covering a distance ‘x’.
KE = 0
Now, from third equation of motion, x
we have,
B PE ≠ 0
vB2 = vA2 + 2gx h KE ≠ 0
= (0)2 + 2gx
vB2 = 2gx ….. (2)
Mechanical energy at B,
A PE = max
EB = K.E. at B + P.E. at B KE = 0
1 x
= 2 mvB2 + mghB
1 B PE ≠ 0
= m (2gx) + mg(h – x)
2 h KE ≠ 0
= mgx + mgh – mgx
EB = mgh ….. (3)
Finally, the ball reaches the ground (at
A PE = max
point C) after covering a distance h.
KE = 0
Now, from third equation of motion,
x
we have,
vC2 = vA2 + 2gh B PE ≠ 0
= (0)2 + 2gh h KE ≠ 0
vC2 = 2gh ….. (4)
h-x
C
Mechanical energy at C,
A PE = max
EC = K.E. at C + P.E. at C KE = 0
1 x
= mv 2 + mghC
2 C
1 B PE ≠ 0
= m (2gh) + mg(0) h KE ≠ 0
2
EC = mgh ….. (5)
h-x
C
From equation. (1), (3) and (5), we get,
A PE = max
KE = 0
EA = EB = EC
x
This means total mechanical energy is
conserved during the free fall of an B PE ≠ 0
object. h KE ≠ 0
That is,
h-x
1
mv2 + mgh = constant
2
C
Question
An object of mass 20 Kg is dropped from a height of 4 m.
Compute the potential energy and kinetic energy at height 4m , 3
m , 2m , 1m and at the ground. u = 0 m/s
4m
Solution
At height of 4 m: 3m
2m
1m
0m
Solution 4m u = 0 m/s
At height of 3 m :
3m
2m
1m
At height of 2 m :
0m
Solution 4m u = 0 m/s
At height of 1 m:
3m
2m
At the ground: 1m
0m
Solution
Height at which Potential Kinetic Energy Total Energy
object is Energy (J) (J) (J)
located (m)
4 800 0 800
3 600 200 800
2 400 400 800
1 200 600 800
Just above the 0 800 800
ground
Homework
Question
If the water falls from a dam into a turbine wheel 19.6 m below,
then the velocity of water at the turbine is (g = 9.8 m/s2)
(A) 9.8 m/s (B) 19.6 m/s
(C) 39.2 m/s (D) 98.0 m/s
Solution
Ans. (B)
Question
A body is falling from a height h. After it has fallen a height h/2,
it will possess
(A) only potential energy
(B) only kinetic energy
(C) half potential and half kinetic energy
(D) more kinetic and less potential energy
Solution
Ans. (C)
