1. ENERGY
 LAW OF CONSERVATION OF
ENERGY
 CONSERVATION OF MECHANICAL
ENERGY
(BY GRAVITATIONAL FORCE)

2. LAW OF CONSERVATION OF ENERGY
The law of conservation of energy states that
energy can neither be created nor be destroyed but
can be transformed from one form to the other. The
total energy before and after the transformation
remains the same.
Our bodies convert the chemical energy in our
food into mechanical energy for us to do work.
The chemical energy in battery gets converted
into light energy in torches.
The electrical energy gets converted into
heat energy for bread toasts.

3. • Mechanical energy =
Potential energy + Kinetic energy
• Suppose a ball of mass ‘m’ is dropped from a
height(H).
• Let the velocity of ball at point A = v(A)=0 m/s.
• The velocity of the ball at point B=v(B).
• The velocity of the ball at ground level (C)=v(C).
m
v(A)=0 A
C
H
CONSERVATION OF MECHANICAL ENERGY
B

4. MECHANICAL ENERGY AT POINT ‘A’
• At present the ball is at point ‘A’ at a
height H from the ground level. The ball is
at rest in point ‘A’.
• Since the ball is at rest the v=0. Thus,
the KE is also equal to 0.
• Therefore, the ME possessed by the ball
at point ‘A’ is equal to the PE possessed by
the ball at ‘A’
PEA=mgH
KEA=0 (v=0)
MEA=PEA+KEA
MEA=mgH+0=mgH
m
v(A)=0 A
C
H

5. MECHANICAL ENERGY AT POINT ‘B’
• At present the ball is at point ‘B’ fallen from height H to have fallen
through (H-h).
• The ball at present has gained velocity[v(B)].
• Therefore, at point ‘B’ the ball possesses partial kinetic energy and
partial potential energy.
PEB=mgh
KEB=1/2m [v(B)]2
v2-u2=2gh
[v(B)]2-0(because it has fallen from rest)=2g(H-h)
[v(B)]2=2g(H-h)
KEB=1/2m[2g(H-h)]
KEB=mgH-mgh
MEB=PEB+KEB
=mgh+mgH-mgh
=mgH
 MEB=mgH
v(A)=0 A
C
H
B [v(B)]
H-h
h
m
PN:The red colour denotes cancellation.

6. MECHANICAL ENERGY AT POINT ‘C’
• The ball is now at point ‘C’ and has the highest velocity
i.e, v(C).
• The ball is now at ground level which implies that there is
no height due to which there is absence of PE.
• Therefore, the ME possessed by the ball at ‘C’ is equal
to the KE energy possessed by the ball at ‘C’.
PEC=0
KEC=1/2m[v(C)]2
v2-u2=2gh
[v(C)]2-0(fallen from rest)=2gH
[v(C)]2=2gH
KEC=1/2m(2gH)
KEC=mgH
MEC=PEC+KEC
=0+mgH
 MEC=mgH
v(A)=0 A
C
H
m
v(C)

7. CONCLUSION
• The sum of potential energy and kinetic
energy is same at all points, which implies:
PE+KE=constant
mgh+1/2mv2=constant
• The mechanical energy(ME) of the ball at all
points [A,B&C]=mgH which implies that the
mechanical energy is conserved.

8. SAMPLE PROBLEM
Use the law of conservation of energy (assume no friction) to fill in the
blanks at the various marked positions for a 1000-kg roller coaster
car.Find the value of A,B,C,D,E,F,G and H only.

9. SOLUTIONS
• A: h = 45.9 m
• B: v = 0 m/s
• C: KE = 250 000 J
• D: h = 20.4 m
• E: v = 22.4 m/s
• F: KE = 450 000 J
• G: PE = 0 J
• H: v = 30 m/s