Laws of conservational energy: the principle that in a system that does not undergo any force from outside the system, the amount of energy is constant, irrespective of its changes in form
EQUATION FOR CONSERVATION ENERGY
Abbreviated: E=KE+PE
Extended: KE(before)+PE(before)=KE(after)+PE(after)
Units of energy
m= mass(kg)
g= gravity
h= distance/high (m)
v= velocity (m/s)
Different Units of Energy
- SI Unit: Joules (J)
- American Unit : Calories (Cal)
This video should give you a better understanding over the topic we are talking about.
In this video you get a better understanding of how to solve conservation of energy. During the video it gave a example were you had to draw how the function was moving. Also you had to determine if it was either gravitational potential energy or any other type of energy. Afterwards you also needed to find you known and unknown variables, to determine the type of formula you need. From there you solve the equation and get your answer.
EXAMPLE 1:
Point A is the initial point. Point A also has a velocity of 20 m/s and the mass of the cart is 60 Kg. What will be the Velocity of the cart when it gets to point B?
KNOW VARIABLES
m= 60Kg
g= 9.8 m/s^2
hf= 25m
hi= 10m
vi= 20 m/s
F= 0
UNKNOWN VARIABLES
vf=?
Before you begin solving the equation you need to first fix the formula to fit your known and unknown variables. So therefore you will end up with this formula
KEj= KEf+PE
(60)(10)^2i/2= (60)v^2f/2+(60)(9.8)(25)f
3000i= (60)v^2f/2+ 14700f
-11700= (60)v^2f
-23400= 60v^2f
-390=v^2f
The answer is vf=19.74m/s
EXAMPLE 2:
In this picture the skater is already on the ramp if you can tell the speed has helped to go on the loop without falling. Also there is no friction being used because his speed has helped to get the over the loop.
In this last picture the skater has grained even more speed but I also added friction but because the skater gained so much energy he was able to go on the second loop without falling.
KNOW VARIABLES
m= 60Kg
g= 9.8 m/s^2
hf= 25m
hi= 10m
vi= 20 m/s
F= 0
UNKNOWN VARIABLES
vf=?
Before you begin solving the equation you need to first fix the formula to fit your known and unknown variables. So therefore you will end up with this formula
KEj= KEf+PE
(60)(10)^2i/2= (60)v^2f/2+(60)(9.8)(25)f
3000i= (60)v^2f/2+ 14700f
-11700= (60)v^2f
-23400= 60v^2f
-390=v^2f
The answer is vf=19.74m/s
EXAMPLE 2:
A block having mass 2kg and velocity 2m/s slide on the inclined plane. If the horizontal surface has friction constant µ=0, 4 find the distance it travels in horizontal before it stops.
We use conservation of energy in solution of this problem.
Einitial=Efinal
Einitial=Ep+Ek=mgh+1/2mv² Efinal=0
Einitial=2kg.10m/s².8m+1/2.2kg. (2m/s) ² Work done by friction=Einitial Einitial=164joule
Wfriction=µ.N.X=0,4.2kg.10m/s².X=Ei
8. X=164joule X=20,5m
Block slides 20,5m in horizonta
 |
In this picture you can see how the skater is going with a very high speed down the ramp. There is no friction being added because the skater is going the very fast.  |
In this picture the skater is already on the ramp if you can tell the speed has helped to go on the loop without falling. Also there is no friction being used because his speed has helped to get the over the loop.
In this last picture the skater has grained even more speed but I also added friction but because the skater gained so much energy he was able to go on the second loop without falling.