Download 7TH CLASSES PHYSICS DAILY PLAN

9TH CLASSES
PHYSICS
DAILY PLAN
SUBJECT: Conservation of mechanical energy
AIM: To understand the law of conservation of
energy
DURATION: 80 min
REAL LIFE: railway in an amusing park
PRESENTATION:
When an object falls under only the influence of gravity: (no
friction):
h is decreasing, v is increasing
v
1
OR
h
PE is decreasing, KE is increasing
1
2
Suppose the object falls from A to B.
A
g
v0=0
v=?
Ex: Find the velocity of
the object in the position 520 cm
shown in figure. (No
friction)
200 cm
{*first way: take zero level at the bottom;
second way: Take zero level at the final position of the ball*}
Ex: An object is thrown upwards with
initial velocity v0. What maximum height
will it reach. Find a formula by using
conservation of mechanical energy.
v
0
v
h2
DATE:
4. If no external work the total enrgy in the first photo=total
enrgy in the second photo:
PEi + KEi = PEf + KEf
Since the object is falling down,
potential energy decreases by (mgy).
y
h
Since gravitational force does work
kinetic energy increases by (mgy).
Ex: What maximum
displacement will the
object experience over
the inclined plane, once
it is delivered an initial
push, giving it 8 m/s
velocity?
v = 8 m/s
L=?
37°
v0=0
Ex:
v2=?
1
B
v3=?
Therefore:
h1=5m
Increase in KE = Decrease in
PE
h
2
KE= - PE
KE + PE = 0
(KEf – KEi) + (PEf – PEi) = 0
h2=3.2m
Find v2 and v3 if the object starts from rest. (No friction
anywhere)
Solution:
v0=0
PEi + KEi = PEf + KEf.
If external work on an object is zero, total mechanical energy
of the object is conserved (=does not change):
h=1.8m
v2=?
v3=?
0
h1=5m
MEi=MEf .
h2=3.2m
Ex:
PE = 100J
KE = 0
ME = 100 J
PE = 75J
KE = 25J
ME = 100 J
PE = 50J
KE = 50J
ME = 100 J
h3=2m
h3=2m
PEi + KEi = PEf + KEf
mgh + 0 = 0 + ½ mv22
v22=2gh; v22=36; v2= 6 m/s
v0=0
v2=?
h=3m
PE = 0
KE = 100J
ME = 100 J
{*The following 6 examples are to be solved by using
PEi + KEi = PEf + KEf method*}
Ex: An object of mass m is released from 5 m above the
ground. Find its velocity just before it touches the ground.
{*Draw the picture*}
Problem solving method (1):
1. Choose a zero level for potential energy
2. Take two photographs (draw them) at any two instants you
want
3. Is there any energy loss (because of friction) between the
photographs?
v3=?
h1=5m
0
h2=3.2m
h3=2m
PEi + KEi = PEf + KEf
mgh + 0 = 0 + ½ mv32
v32=2gh; v32=60; v3= 7.74 m/s
HOMEWORK: wsh.21-22-23
MULTIMEDIA: serway1-serway11
DEMONSTRATION:
EXPERIMENT:
TEACHER:
DIRECTOR: