Download CONSERVATION of

CONSERVATION OF
MECHANICAL ENERGY
INTRODUCTION
We already learned the different forms of energy
in the previous lessons and discussions. Now, our
report will discuss why energy transformation is the
unifying principle among these various forms of
energy.
“When energy changes from one form to another, the
amount of energy stays the same.”
-Law of Conservation of Energy
Law of Conservation of Energy
A bulb changes electrical energy into light and heat energy.
Telephones convert sound energy into electrical energy as
sound moves back and forth. A swing changes the energy of
the rider from potential energy into kinetic energy as the
swings moves back and forth.
From the given examples, you have seen that energy changes into
different forms, but the total amount of energy stays the same.
This is the application of this law which states that:
Energy can neither be created nor destroyed, it can only be
transformed from one form to another.
Computations

Potential Energy: The
Stored Energy
PEg = W = F x d = mgh

Kinetic Energy: The
Energy in Motion
Anything that is moving has
kinetic energy.
KE = ½ mv2
Law of Conservation of Mechanical Energy
The sum of the
kinetic energy
and potential
energy of a
system is
called
Mechanical
Energy.
In a conservative system, the total
mechanical energy is constant. In this
system, only conservative forces are
present and, therefore, a decrease in
potential energy is equal to an increase in
kinetic energy, and vice versa. This is now
what the law states:
The sum of the kinetic energy and potential
energy in a conservative system is constant
and equal to the total mechanical energy of
the system.
Sample Problem
A 50 kg box falls from a bridge and lands in the water 20 m
below. Find its (a) initial energy PE, (b) maximum KE, (c) KE and
PE 15 m above water and (d) velocity upon reaching the
water.
Given:
m = 50 kg
h = 20 m
g = 9.8 m/s2
Solution
a.
Initial PE is taken from the top.
Therefore,
PEi = mgh
=(50kg)(9.8m/s2)(20m)
= 9800 J
b. The maximum KE is equal to the
total PE at the top. Therefore,
KEmax = 9800 J
c. At the height 15m above the water,
PE = mgh
= (50kg)(9.8m/s2)(15m)
= 7350 J
KE = TE – PE
= 9800 J – 7350 J
= 2450 J
d. The velocity upon reaching the
water is
v = square root of 2 KE/ m
= square root of 2(9800 J)/50kg
= square root of 19600 kg x m2 /
s2 / 50kg
= 19.80 m/s
The KE used is 9800 J at the
bottom; the KE is maximum.
Thanks for paying attention to our
report! Any Questions? 
Reporters:
Ruaya, Rhobie A.
Paloma, Janey Anne M.
Agcol, Jacques Matthew E.
Salmayor, Jizelle M.
Ursabia, Joan D.
Musslewhite, Arthur M.