Energy
Unit 5: Energy and Work
Energy
Defined as the ability to do work / cause a change
Energy is stored work (transfer of energy as an object is displaced)
Unit of energy is the Joule (J)
Energy is a scalar quantity - it may have positive/negative values, but it has no direction
Mechanical Energy
Which point has the MOST kinetic energy?
Which point has the LEAST kinetic energy?
KE = ½mv2 m = mass v = velocity (or speed)
Sample Problem 1: A child, m = 34 kg, runs with a velocity of 1.75 m/s. what is the child’s kinetic energy?
Sample Problem 2: A 500 kg horse has a kinetic energy equal to 7840 J. How fast is the horse moving?
KE = ½mv2
KE = ½(34)(1.752)
KE = 52 J
KE = ½mv2
7840 = ½(500)(v2)
7840 = 250v2
31.36 = v2
v = 5.6 m/s
Which point has the MOST GPE?
Which point has the LEAST GPE?
PE = mgh m = mass g = gravity (9.8 m/s2) h = height
Sample Problem: What is the gravitational potential energy of a 10 kg cat resting atop a 2.3 m refrigerator?
GPE = mgh
GPE = (10)(9.8)(2.3)
GPE = 225.4 J
Total Mechanical Energy
TME = KE + PE
Lesson Check 5.2
LAW OF CONSERVATION OF ENERGY
Unit 5: Energy & Work
Open vs Closed System
OPEN
CLOSED
Open vs Closed System
Applying this to the law of conservation of energy:
Open vs Closed System
If the system is closed, and there are no friction or other energy-loss processes, the amount of mechanical energy described in a problem must be constant
The Law of Conservation of Energy
The Law of Conservation of Energy
On the way up, the ball is losing speed
KE is decreasing
PE is increasing
Ball is stationary
KE = 0
PE is greatest
On the way down, the ball is gaining speed
KE is increasing
PE is decreasing
BASICALLY
The total mechanical energy at one moment is equal to the total mechanical energy at another moment
Energy Bar Charts
KEi + PEi = TME
KEf + PEf = TME
Energy Bar Charts
Total Mechanical Energy stays the same, but the motion/location of the object determines how much KE and PE there is at a given point.
Law of Conservation of Energy Formula
PEi + KEi = PEf + KEf
mghi + ½mvi2 = mghf+ ½mvf2
MEi = MEf
Sample Problem 1
A. Match the energy bar chart with the correct position of the skateboarder.
3
2
3
Sample Problem 1
B. Where on the path would the skateboarder represent chart 4?
Sample Problem 2
Astro, a cat whose mass is 5.45 kg, is napping on top of the refrigerator when he rolls over and falls. Astro has a KE of 85.5 J just before he lands on his feet on the floor. How tall is the refrigerator?
m = 5.45 kg
KE = 85.5 J
h = ?
MEi = MEf
PEi + KEi = PEf + KEf
at rest, no KE
on the floor, no PE
PEi = KEf
mgh = KEf
(5.45)(9.8)h = 85.5
53.41h = 85.5
h = 1.6 m
| PE | KE | TME |
initial | | | |
final | | | |
Sample Problem 3
A 20-kg rock sits on the edge of a cliff. The height of the cliff is 100 m. What will be the speed of the rock before landing?
m = 20 kg
KE = ?
h = 100 m
MEi = MEf
PEi + KEi = PEf + KEf
at rest, no KE
on the floor, no PE
PEi = KEf
mgh = ½mv2
(20)(9.8)(100) = ½(20)v2
19600 = 10v2
1960 = v2
v = 44.3 m/s
| PE | KE | TME |
initial | | | |
final | | | |
Sample Problem 4
Frank, a San Francisco hot dog vender, has fallen asleep on the job. When an earthquake strikes, his 300 kg hot dog cart rolls down Nob Hill and reaches point A (h=50 m) at a speed of 8 m/s. How fast is the hot dog cart going at point B (h=30.0 m) when Frank finally wakes up and starts to run after it?
A
B
hi = 50m vi = 8 m/s
hf = 30m vf = ??
MEi = MEf
PEi + KEi = PEf + Kef
mghi + ½mvi2 = mghf + ½mvf2
(300)(9.8)(50) + ½(300)(82) = (300)(9.8)(30) + ½(300)vf2
147,000 + 9,600 = 88,200 + 150vf2
156,600 = 88,200 + 150vf2
68,400 = 150vf2
456 = vf2
vf = 21.4 m/s
| PE | KE | TME |
initial | | | |
final | | | |
Lesson Check 5.3