situations Objectives
Identify in which conservation of mechanical energy is valid. Recognize the forms that conserved energy can take. Solve problems using conservation of mechanical energy.
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Concept Check Conservation of Energy
Three balls of equal mass start from rest and roll down different ramps. All ramps have the same height. Which ball has the greater speed at the bottom of its ramp? 4) same speed for all balls
1
2
3
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Concept Check Conservation of Energy
Three balls of equal mass start from rest and roll down different ramps. All ramps have the same height. Which ball has the greater speed at the bottom of its ramp? 4) same speed for all balls
1
2
3
All of the balls have the same initial gravitational PE, since they same are all at the same height (PE = mgh). Thus, when they get to the same bottom, they all have the same final KE, and hence the same same speed (KE = 1/2 mv2). speed
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Concept Check Conservation of Energy
Paul and Kathleen start from rest at the same time on frictionless water slides with different shapes. At the bottom, whose velocity is greater? 1. Paul 2. Kathleen 3. both the same
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Concept Check Conservation of Energy
Paul and Kathleen start from rest at the same time on frictionless water slides with different shapes. At the bottom, whose velocity is greater? 1. Paul 2. Kathleen 3. both the same Conservation of Energy:
i f
2
E = mgh = E = 1/2 mv mv 2
therefore: gh = 1/2 v gh
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Concept Check Conservation of Energy
Paul and Kathleen start from rest at the same time on frictionless water slides with different shapes. Who makes it to the bottom first? 1. Paul 2. Kathleen 3. both the same
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Concept Check Conservation of Energy
Paul and Kathleen start from rest at the same time on frictionless water slides with different shapes. Who makes it to the bottom first? 1. Paul 2. Kathleen 3. both the same Even though they both have the same final velocity, Kathleen is at a lower height than Paul for most of her ride. than Thus she always has a larger velocity during her ride and velocity therefore arrives earlier!
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Conserved Quantities
When we say that something is conserved, we mean that it remains constant.
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Mechanical Energy
Mechanical energy is the sum of kinetic energy and all forms of potential energy associated with an object or group of objects. ME = KE + PEg + PEe Mechanical energy is often conserved. MEi = MEf initial mechanical energy = final mechanical energy (in the absence of friction)
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Conservation of Mechanical Energy
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Conservation of Mechanical Energy
Energy of a falling 75g Egg
0.80 0.70 0.60 Energy (J) 0.50 0.40 0.30 0.20 0.10 0.00 0.00 0.10 0.20 Time (sec) 0.30 0.40 PEg (J) KE (J) ME (J)
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Practice E
1. h = 5.40 m m = 2.00 kg vi = 18.0 m/s vf = ?
KEi = 1 mvi2 2 PEg ,i = mgh
vx = 18.0 m s y = 5.40 m
KE f = mv
1 2
2 f
vx = 18.0 m s x
PEg , f = 0 J MEi = ME f KEi + PEg ,i = KE f + PEg , f
1 2
v y, f vf = ?
mvi2 + mgh = 1 mv 2 + 0 f 2
v f = vi2 + 2 gh =
( 18.0 m s )
2
+ 2 ( 9.81 m s 2 ) ( 5.40 m )
v f = 20.7 m s
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Practice E
2. h = 5. 0 m Fg = 755 N vi = 0.0 m/s vf = ?
MEi = ME f KEi + PEg ,i = f KE + PEg , f 0 J + mgh = 1 mv 2 + 0 J f 2 v f = 2 gh = 2 ( 9.81 m s 2 ) ( 5.0 m ) v f = 9.9 m s KEi = 0 J PEg ,i = mgh KE f = 1 mv 2 f 2 PEg , f = 0 J
Just before striking the water, h = 10.0 m
v f = 2 gh = 2 ( 9.81 m s 2 ) ( 10.0 m ) v f = 14 m s
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Practice E
3. h = 10.0 m Fg = 755 N vi = 2.00 m/s vf = ?
KEi = 1 mvi2 2 PEg ,i = mgh
MEi = ME f KEi + PEg ,i = KE f + PEg , f
1 2
KE f = 1 mv 2 f 2 PEg , f = 0 J
mvi2 + mgh = 1 mv 2 + 0 f 2
v f = vi2 + 2 gh =
( 2.00 m s )
2
+ 2 ( 9.81 m s 2 ) ( 10.0 m )
v f = 14.1m s
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Practice E
4. vi = 2.2 m/s vf = 0.0 m/s h=?
MEi = ME f KEi + PEg ,i = KE f + PEg , f
1 2
KE f = 0 J PEg , f = mgh KEi = 1 mvi2 2 PEg ,i = 0 J
2
mvi2 + 0 J = 0 J + mgh
vi2 ( 2.2 m s ) h= = 2 g 2 ( 9.81 m s 2 )
h = 0.25m
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Practice E
5. vi = 0.0 m/s vf = 1.9 m/s h=?
MEi = ME f KEi + PEg ,i = KE f + PEg , f 0 J + mgh = 1 mv 2 + 0 J f 2
KEi = 0 J PEg ,i = mgh
h
KE f = 1 mv 2 f 2 PEg , f = 0 J
h=
v2 f 2g
=
2 ( 9.81 m s 2 )
( 1.9 m s )
2
h = 0.18m
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Sample Problem
Conservation of Mechanical Energy Starting from rest, a child zooms down a frictionless slide from an initial height of 3.00 m. What is her speed at the bottom of the slide? Assume she has a mass of 25.0 kg.
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Sample Problem, continued
Conservation of Mechanical Energy 1. Define Given: h = hi = 3.00 m m = 25.0 kg vi = 0.0 m/s hf = 0 m Unknown: vf = ?
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Sample Problem, continued
Conservation of Mechanical Energy 2. Plan Choose an equation or situation: The slide is frictionless, so mechanical energy is conserved. Kinetic energy and gravitational potential energy are the only forms of energy present.
KE = PEg 12 mv 2 = mgh
The zero level chosen for gravitational potential energy is the bottom of the slide. Because the child ends at the zero level, the final gravitational potential energy is zero. PEg,f = 0
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Sample Problem, continued
Conservation of Mechanical Energy 2. Plan, continued The initial gravitational potential energy at the top of the slide is PEg,i = mghi = mgh Because the child starts at rest, the initial kinetic energy at the top is zero. KEi = 0 Therefore, the final kinetic energy is as follows:
1 KE f = mv 2 f 2
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Sample Problem, continued
Conservation of Mechanical Energy 3. Calculate Use the expressions to create an equation.
MEi = ME f
PEg ,i + KEi = PEg , f + KE f mgh = 1 mv 2 f 2
v f = 2 gh v f = 2 ( 9.81m s 2 ) ( 3.0m ) v f = 7.67 m s
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Sample Problem, continued
Conservation of Mechanical Energy 4. Evaluate The expression for the square of the final speed can be written as follows:
v2 = f 2mgh = 2 gh m
Notice that the masses cancel, so the final speed does not depend on the mass of the child. This result makes sense because the acceleration of an object due to gravity does not depend on the mass of the object.
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Mechanical Energy, continued
Mechanical Energy is not conserved in the presence of friction. As a sanding block slides on a piece of wood, energy (in the form of heat) is dissipated into the block and surface.
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Summarize
What are the key ideas from this lesson?
What connections can I make with other ideas?
What questions do I still have?
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