The Law of Conservation of Energy
Definition:
The law of conservation of energy is a law of science that states that energy cannot be created or destroyed, but only changed from one form into another or transferred from one object to another.
Mathematical Representation:
Abbreviated Version - Ei (initial) = Ef (final)
Extended Version- KEᵢ + PEᵢ + Wₙc + OEᵢ = KEf + PEf + OEf
Extended Version- KEᵢ + PEᵢ + Wₙc + OEᵢ = KEf + PEf + OEf
Different Units for Energy/Work:
SI Units:
-Joules
-Watt (SI unit for Electrical Energy in U.S)
American Units:
-Foot-pound (ft x lb)
-British Thermal Unit (BTU)
-Calories (cal)
In the video a 3D simulation of a roller coaster displays the concept of the conservation of energy. As the ride begins the rollercoaster has a constant level of potential energy, and a constant kinetic energy level due to how it is traveling on a flat part of the track. At the 0:30 mark, the rollercoaster begins to go up on the track, the potential energy is increasing (height increase) and the kinetic energy decreases (cart goes slower as it moves up). It travels along a flat path for a short time, then it begins to head down the track on a downward slant at the 0:35 mark. At this point, the potential energy is decreasing (height decrease), and the kinetic energy is increasing (cart is moving faster downward). When the cart is at its highest point, potential energy is at its highest. When the cart is at its lowest point on the track, the potential energy is at its lowest. The video goes on to show a few more upward and downward parts of the track. Overall, in the video when the cart is heading upward the potential energy is increasing and the kinetic energy is decreasing. The opposite happens when the cart is heading downward, the potential energy is decreasing and the kinetic energy is increasing. Throughout the ride kinetic energy and potential energy are being transferred back and forth.
BELOW ARE 2 SAMPLE PROBLEMS SO YOU CAN TRY IT YOURSELF
PEi = mgh
Below is the link to a video on the subject matter:
In the video a 3D simulation of a roller coaster displays the concept of the conservation of energy. As the ride begins the rollercoaster has a constant level of potential energy, and a constant kinetic energy level due to how it is traveling on a flat part of the track. At the 0:30 mark, the rollercoaster begins to go up on the track, the potential energy is increasing (height increase) and the kinetic energy decreases (cart goes slower as it moves up). It travels along a flat path for a short time, then it begins to head down the track on a downward slant at the 0:35 mark. At this point, the potential energy is decreasing (height decrease), and the kinetic energy is increasing (cart is moving faster downward). When the cart is at its highest point, potential energy is at its highest. When the cart is at its lowest point on the track, the potential energy is at its lowest. The video goes on to show a few more upward and downward parts of the track. Overall, in the video when the cart is heading upward the potential energy is increasing and the kinetic energy is decreasing. The opposite happens when the cart is heading downward, the potential energy is decreasing and the kinetic energy is increasing. Throughout the ride kinetic energy and potential energy are being transferred back and forth.
BELOW ARE 2 SAMPLE PROBLEMS SO YOU CAN TRY IT YOURSELF
Conservation of Mechanical Energy During a hurricane, a large tree limb, with a mass of 22.0 kg
and a height of 13.3 m above the ground, falls on a roof that is 6.0 m above the ground.
a. Ignoring air resistance, find the kinetic energy of the limb when it reaches the roof.
b. What is the speed of the limb when it reaches the roof?
ANALYZE and SKETCH the PROBLEM
Known: Unknown:
m = 22.0 kg g = 9.8 m/s^2 PEi = ? KEf = ?
hlimb = 13.3 m vi = 0.0 m/s PEf = ? Vf = ?
hroof = 6.0 m KEi = 0.0 J
SOLVE for the UNKNOWN
Sample Problem #1
a. Set the reference level as the height of the roof.
Solve for the initial height of the limb relative to the roof.
h = hlimb - hroof
h = 13.3 m - 6.0 m
h = 7.3 m
Solve for the initial potential energy of the limb
PEi = mgh
PEi = (22.0 kg)(9.8 m/s^2)(7.3 m)
KEf = 1.6 x 10³ J or 1,600 J
Identify the kinetic energy of the limb
KEf = 0.0 J
The kinetic energy of the limb when it reaches the roof is equal to
the initial potential energy because energy is conserved.
the initial potential energy because energy is conserved.
KEf = PEi
= 1.6 x 10³ J or 1,600 J
b. Solve for the speed of the limb.
KEf = 1/2 (m)vf ^2
Vf = √2KEf /m
Vf = √2(1,600 J) / 22.0 Kg
Vf = 12 m/s
Known: Unknown:
Vf = √2(1,600 J) / 22.0 Kg
Vf = 12 m/s
Sample Problem #2
Layla is snow skiing with her cousin at a local ski resort. She stops and stands at the top of a small mountain, at rest, to take a quick picture. All of a sudden her cousin, being the little rebel she is, jokingly shoves her. Layla is sent zooming down the 12 m mountain. How fast will she be moving at the bottom of the mountain if she and her ski gear have a combined weight of 48 kg?
Known: Unknown:
m = 48 kg Vf = ?
h = 12 m
g = 9.8 m/s^2
Vi = 0 m/s
Solve:
(mv^2 / 2)initial + mghintial = (mv^2 / 2)final + mghfinal
(48 x 0)^2 / 2 + (48)(9.8)(12) = (48 x v^2 / 2) + (48)(9.8)(12 m)
(48 x 0)^2 / 2 + (48)(9.8)(12) = (48 x v^2 / 2)
5644.8 = (48 x v^2 / 2) 5644.8 = (48 x v^2 / 2)
x2 x2
2 (5644.8) = 48 x v^2
2 (5644.8) = 48 x v^2
48 48
2 (5644.8) / 48 = v^2
√2 (5644.8) / 48 = √v^2
√2 (5644.8) / 48 = Vf
15.34 m/s = Vf
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Here in this first image, the skater´s potential energy is at it´s highest, seeing as the skater is at the highest point on the ramp. Kinetic energy is pretty much 0, seeing as the skater has not begun to pick up full speed, kinetic energy will start to increase as the skater heads down, and the potential energy will decrease since the skater´s height from the ground is decreasing. At the starting point however, the skater has 0 m/s as initial velocity.
Calculations:
Max Height:
0 m/s = Vf
At this point in time, the skater is at the 2nd highest point on the ramp, the potential energy is higher than the kinetic energy , seeing as kinetic energy decreased as the skater headed up the right side of the loop. When the skater makes it around the loop, kinetic energy will increase as he heads down the right side of the loop, towards the end of the track. Kinetic energy will begin to increase, and potential will surely decrease.
Calculations:
Top of the Loop:
mgh1 = mgh2 + mv^2 / 2
gh1 = gh2 + v^2 / 2
gh1 - gh2 = v^2 / 2
√2 (gh1 - gh2) = √v^2
√2 (gh1 - gh2) = v
√2 ((9.8)(8) -(9.8)(6)) = v
6.26 m/s = Vf
As the rider heads out of the loop, kinetic energy increases, and becomes greater than potential energy. The potential energy begins to decrease as the skater heads down the final ramp, and kinetic energy begins to increase, as the skater is picking up a substantial amount of speed. By the time the skater is at the very end of the final ramp, the kinetic energy is at it´s highest at the end of the ramp, while potential energy is at it´s lowest.
Calculations:
Fastest:
mvi / 2 +mgh1 = mvf^2 / 2 + mgh2
mgh1 = mvf^2 / 2
gh1 = vf^2 / 2
9.8 (8) = vf^2 / 2
2 (78.4) = vf^2
156.8 = vf
√156.8 = √vf^2
12.52 m/s = Vf