conservation of energy november 21 2014. the conservation of energy. in a closed system, energy is...
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Conservation of Energy
November 21 2014
The conservation of energy In a closed system, energy is neither created nor destroyed.
Energy simply changes from one form to another.
Einitial = Efinal
The conservation of energy In a closed system, energy is neither created nor destroyed.
Energy simply changes from one form to another.
Einitial = Efinal
Mostly, we will use this in situations without friction to say:
PEinitial + KEinitial = PEfinal + KEfinal
The conservation of energy In a closed system, energy is neither created nor destroyed.
Energy simply changes from one form to another.
Einitial = Efinal
Mostly, we will use this in situations without friction to say:
PEinitial + KEinitial = PEfinal + KEfinal
We can also use this in situations where energy is converted to heat due to friction or air resistance to say:
Einitial – Eheat = Efinal
The conservation of energy In a closed system, energy is neither created nor destroyed.
Energy simply changes from one form to another.
Einitial = Efinal
Mostly, we will use this in situations without friction to say:
PEinitial + KEinitial = PEfinal + KEfinal
We can also use this in situations with friction / air resistance to say:
Einitial – Efriction = Efinal
Conservation of energy is a very useful tool to solve physics problems!
Many problems that can be solved using Newton’s laws can be solved much easier by applying conservation of energy.
h=0
N
300
h=0.5mv=?
u=0
Method 1: Newton’s law
hd =
sinθ
d
Determine the final velocity of a block that slides down a frictionless ramp.
What steps would you have to take?
h=0
N
300B
A
h=0.5mv=?
u=0
Method 1: Newton’s law
hd =
sinθ
d
Determine the final velocity of a block that slides down a frictionless ramp.
What steps would you have to take?Step 1: Draw FBDStep 2: Write Fnet equation and solve for aStep 3: Use motion equations to solve for v
h=0
N
300B
A
h=0.5mv=?
u=0
Method 1: Newton’s law
F mgsinθa = = = gsinθ
m mvf
2 = vi2 + 2ad = 2gsinθ
hd =
sinθ
d
h
sinθ
v = 2gh v = 3.2 m/s
Determine the final velocity of a block that slides down a frictionless ramp.
What steps would you have to take?Step 1: Draw FBDStep 2: Write Fnet equation and solve for aStep 3: Use motion equations to solve for v
h=0
N
300B
A
h=0.5mv=?
u=0
Method 1: Newton’s law
F mgsinθa = = = gsinθ
m mvf
2 = vi2 + 2ad = 2gsinθ
hd =
sinθ
d
h
sinθ
v = 2gh v = 3.2 m/s
Determine the final velocity of a block that slides down a frictionless ramp.
What steps would you have to take?Step 1: Draw FBDStep 2: Write Fnet equation and solve for aStep 3: Use motion equations to solve for v
Method 2: Energy conservation
As the object slides down the plane its PE becomes transformed into KE.
h=0
N
300
h=0.5mv=?
u=0
Method 1: Newton’s law
F mgsinθa = = = gsinθ
m mvf
2 = vi2 + 2ax = 2gsinθ
hd =
sinθ
d
h
sinθ
v = 2gh v = 3.2 m/s
Determine the final velocity of a block that slides down a frictionless ramp.
What steps would you have to take?Step 1: Draw FBDStep 2: Write Fnet equation and solve for aStep 3: Use motion equations to solve for v
Method 2: Energy conservation
As the object slides down the plane its PE becomes transformed into KE.
PEinitial = KEfinal
mgh = ½ mv2
h=0
N
300
h=0.5mv=?
u=0
Method 1: Newton’s law
F mgsinθa = = = gsinθ
m mvf
2 = vi2 + 2ad = 2gsinθ
hd =
sinθ
d
h
sinθ
v = 2gh v = 3.2 m/s
Determine the final velocity of a block that slides down a frictionless ramp.
What steps would you have to take?Step 1: Draw FBDStep 2: Write Fnet equation and solve for aStep 3: Use motion equations to solve for v
Method 2: Energy conservation
As the object slides down the plane its PE becomes transformed into KE.
PEinitial = KEfinal
mgh = ½ mv2
v = 2gh v = 3.2 m/s
We get the same answer, but the energy conservation answer is much easier.
Examples of Energy Conservation
Free fall in the absence of air resistance.
A 10 kg ball is dropped from a height of 102 m. What is its velocity when it hits the ground?
Initial E = mgh = 10kg * 9.8 m/s2 *102 m = 10000JFinal E = ½ mv2
= 10000J
v = √(2000) = 45 m/s
Free fall with air resistance.
A 10 kg ball is dropped from a height of 102 m. When it hits the ground, it has a final velocity of 40m/s. How much energy was ‘lost’ due to air resistance?Initial E = mgh = 10kg * 9.8 m/s2 *102 m = 10000JFinal E = ½ mv2
= 0.5*10kg*(40m/s)2 = 8000 J
Eheat = Initial E – Final E = 10000J – 8000 J = 2000 J
A child of mass m is released from rest at the top of a water slide, at height h = 8.5 m above the bottom of the slide. Assuming that the slide is frictionless because of the water on it, find the child’s speed at the bottom of the slide.
You Do
mgh = ½ mv2
v = 2gh = 13 m/s
m cancels on both sides
a baby, an elephant and you would reach the bottom at the same speed !!!!!
Other examples of conservation of energy:Up and down the track
A B C D
Assuming that the track is frictionless, at what point(s) is the total energy of the system the same as in point A ? 1. B 2. C 3. D 4. All of the above 5. none of the above
Other examples of conservation of energy:Up and down the track
A B C D
Assuming the track is frictionless, at what height will the ball end up? 1. same height as A 2. lower height than A 3. higher height than A 4. impossible to determine
Other examples of conservation of energy:Up and down the track
A B C D
If the track does have friction, at what height will the ball end up? 1. same height as A 2. lower height than A 3. higher height than A 4. impossible to determine
Other examples of conservation of energy:Up and down the track
A B C D
At what point(s) does the ball have a combination of PE and KE? 1. A 2. B 3. C 4. D 5. B and C
Other examples of conservation of energy:Up and down the track
A B C D
At what point does the ball have the greatest speed? 1. A 2. B 3. C 4. D
Three balls are thrown from the top of the cliff along paths A, B, and C with the same initial speed (air resistance is negligible). Which ball strikes the ground below with the greatest speed?
1. A 2. B 3. C 4. All strike with the same speed
h
Think about it for a minute … then when I say to, show which answer is correct with a show of fingers
#4 is correct
All balls have the same initial energy, so all have the same KE when they hit the ground.
It depends ONLY on HEIGHT and initial SPEED, not mass, not path !!!!!
A pendulum swings back and forth. At which point or points along the pendulum’s path …
1) Is PE greatest?
2) Is KE greatest?
3) Is speed greatest?
An ideal pendulum would keep going forever. Why do real pendulums eventually stop?
Because some energy is ‘lost’ due to friction and air resistance.
G & A
D
D
A car (toy car – no engine) is at the top of a hill on a frictionless track. What must the car’s speed be at the top of the first hill if it can just make it to the top of the second hill?
40m
v1
v2 = 0
20m
We Do
What is our strategy?
A car (toy car – no engine) is at the top of a hill on a frictionless track. What must the car’s speed be at the top of the first hill if it can just make it to the top of the second hill?
40m
v1mgh1 + ½ mv1
2 = mgh2 + ½ mv22
v2 = 0PE1 + KE1 = PE2 + KE2
20m
We Do
A car (toy car – no engine) is at the top of a hill on a frictionless track. What must the car’s speed be at the top of the first hill if it can just make it to the top of the second hill?
m cancels out ; v2 = 0
40m
v1mgh1 + ½ mv1
2 = mgh2 + ½ mv22
gh1 + ½ v12 = gh2
v1 = 22 m/s
v2 = 0PE1 + KE1 = PE2 + KE2
20m v12 = 2g( h2 – h1)
We Do
A simple pendulum consists of a 2.0 kg mass attached to a string. It is released from rest at A as shown. Its speed at the lowest point B is:
2 2A A B B
1 1mv + mgh = mv + mgh
2 2
2B Av = 2gh = 2×9.80m/s ×1.85m = 6m/s
1.85 m
A
B
PEA + KEA = PEB + KEB
2A B
10 + gh = mv + 0
2
You Do
A simple pendulum consists of a 2.0 kg mass attached to a string. It is released from rest at A as shown. Its speed at the lowest point B is:
1.85 m
A
B
You Do
● Example: An object of mass 4.0 kg slides 1.0 m down an inclined plane starting from rest. Determine the speed of the object when it reaches the bottom of the plane if
a. friction is neglected b. constant friction force of 16 N acts on the object as it slides down.
What happens to energy in each case?
● Example: An object of mass 4.0 kg slides 1.0 m down an inclined plane starting from rest. Determine the speed of the object when it reaches the bottom of the plane if
a. friction is neglected b. constant friction force of 16 N acts on the object as it slides down.
all PE was transformed into KE Friction converts part of KE of the object into heat energy. This energy equals to the work done by the friction. We say that the frictional force has dissipated energy.
● Example: An object of mass 4.0 kg slides 1.0 m down an inclined plane starting from rest. Determine the speed of the object when it reaches the bottom of the plane if
a. friction is neglected b. constant friction force of 16 N acts on the object as it slides down.
all PE was transformed into KE Friction converts part of KE of the object into heat energy. This energy equals to the work done by the friction. We say that the frictional force has dissipated energy.
initial energy = final energy initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object
● Example: An object of mass 4.0 kg slides 1.0 m down an inclined plane starting from rest. Determine the speed of the object when it reaches the bottom of the plane if
a. friction is neglected b. constant friction force of 16 N acts on the object as it slides down.
all PE was transformed into KE Friction converts part of KE of the object into heat energy. This energy equals to the work done by the friction. We say that the frictional force has dissipated energy.
initial energy = final energy initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object
PEA = KEB PEA – Ffr d = KEB
● Example: An object of mass 4.0 kg slides 1.0 m down an inclined plane starting from rest. Determine the speed of the object when it reaches the bottom of the plane if
a. friction is neglected b. constant friction force of 16 N acts on the object as it slides down.
all PE was transformed into KE Friction converts part of KE of the object into heat energy. This energy equals to the work done by the friction. We say that the frictional force has dissipated energy.
initial energy = final energy initial energy – Ffr d = final energy
Wfr = – Ffr d decreases KE of the object
PEA = KEB PEA – Ffr d = KEB mgh = ½ mv2 mgh – Ffr d = ½ mv2
20 – 16 = 2.5 v2
v = 1.4 m/s
v = 2gh = 3.2 m/s