Download THINK ENERGY! KE = ½mv2 PE = mgh W = Fdcosθ

Tues Oct 6
Conservation of Energy
• Work, KE, PE, Mech Energy
• Power
• Assign 7 Fri
• Pre-class Thursday
To conserve total energy means
that the total energy is constant or
stays the same.
With Work, we now have way to
talk about getting started and
coming to a stop. W = F d cosθ
(KEF + PEF) = (KE0 + PE0) + WNC
m,v
m,h
THINK ENERGY!
PhET
F,d
KE = ½mv2
PE = mgh
W = Fdcosθ
Solving Problems
•
•
•
•
•
Energies? v? h? No time? (clues)
Draw - visualize
Decide on initial and final events
Choose reference level for PE
Any WNC?
•
WNC = ΔE
or
(KEi + PEi) + WNC = (KEf + PEf)
Suppose I drop a 0.2kg mass ball from a height 3.0m above the floor.
What is the speed of the ball just before it hits the floor?
Initial: Release ball
Final: Just before hit floor
Reference: floor h=0
EF = EI + WNC
No air resistance, etc… so Total Mech E
stays the same. (WNC=0)
KEF + PEF = KEI + PEI
½mv2 +
0 = 0 + mgh
Cancel m
gh = ½v2
2(9.8 m/s2)(3m) =v2
h=3m
h=0m
v=7.7 m/s
3 balls of the same mass are thrown from a cliff, all with a speed of
25m/s. A is thrown upward at an angle of 45°. B is thrown
horizontally. C is thrown downward at an angle of 45°. Which one is
traveling fastest just before it hits the ground?
1. A
2. B
3. C
4. All three are
traveling the same
speed
Same initial KE, same change in PE,
same increase in KE.
Does the mass matter?
Not in this case. It would cancel.
EF = E0 + WNC
KEF + PEF = KEI + PEI
½mv2 + 0 = ½m(25m/s)2+ mgh
Bowling Ball Pendulum
• Forces?
• WTENSION = ?
• Energy Lost? Gained?
A block starts from rest and slides down a frictionless ramp. How
does the speed at Point B compare to the speed at point G?
A
H
D
E
B
G
F
C
(1) vB < vG
(2) vB = vG
(3) vB > vG
PE same at points B, E and G.
If WNC = 0, then ETOT same and therefore KE same.
(KEF + PEF) = (KE0 + PE0) + WNC
Same
A particle, starting from point A in the drawing, is projected down the
curved runway. Upon leaving the runway at point B, the particle is
traveling straight upward and reaches a height of 4.00m above the
floor before falling back down. Ignore friction and air resistance. Find the speed of the particle at point A. Initial: Leave point A
Final: Max Height
Ref Level: Table Top
(KEF + PEF) = (KE0 + PE0) + WNC
No non-conservative forces
(0 + mg(4.0 m)) = (½mv02 + mg(3.0 m)) + 0
Can cancel mass, rearrange and solve
(9.8 m/s2)(4.0 m – 3.0 m) = ½v02
v0 = 4.4m/s
Four identical balls roll off four tracks. The tracks are of the same
height but different shapes, as shown. For which track is the ball
moving the fastest when it leaves the track?
(1) Yellow
(2) Red
(3) Green
(4) Blue
(5) All the same
Start at same height, end at same height, start with same KE
Same initial total energy
No loss of mechanical energy
Same loss in Potential energy
So same final kinetic energy
If all leave track at same angle, same distance before hit the
ground
Four identical balls roll off four tracks. The tracks are of the same
height but different shapes, as shown. For which track does the trip
from start to finish take the least time?
(1) Yellow
(2) Red
(3) Green
(4) Blue
(5) All the same
Greatest acceleration early on
Note: Gain ability to handle strange shapes and paths
LOSE time information and direction information.
Get the magnitude of velocity, but not direction.
An 80-kg stuntman starts at rest and slides down a roof, flies through
the air, and lands on a large pad, which compresses 1.2m in order to
bring the stuntman to a stop. Assume it is an icy day and the roof is
frictionless. Ignore air resistance. What is the average force on the stuntman due to the pad?
Top
Initial: Top
Final: Rest
Ref Level:
Top of uncompressed
Pad
KE
PE
ETOT
0
mg(4m)
4mg
Top of Pad ½mv2 0
½mv2
Rest
-1.2mg
0
mg(-1.2m)
(KEF + PEF) = (KE0 + PE0) + WNC
(0+(– 1.2)mg) = (0 + 4mg) + F*d*cos(180º)
– 1.2mg = 4mg + F*(1.2)*(-1)
1.2F = 5.2mg
F = 3400 N
An 80-kg stuntman slides down a roof, flies
through the air, and lands on a large pad,
which compresses 1.2m in order to bring
the stuntman to a stop. Assume it is an icy
day and the roof is frictionless. Ignore air
resistance. What is the speed of the stuntman just
(1) 4.4 m/s
before he hits the pad?
(3) 6.3 m/s
(5) 8.8 m/s
Initial: top, Final: just before hit – No energy lost
(KEF + PEF) = (KE0 + PE0) + WNC
(½mv2 + 0) = (0 + 4mg) + 0
v = 8.85 m/s
(2) 4.8 m/s
(4) 7.7 m/s
(6) 78 m/s
At the same time that Stuntman A slides down the roof (starting from
rest), Stuntman B steps off a roof (starting at the same height) on the
other side of the street. Assume a 'no-friction' roof and negligible air
resistance. Which Stuntman is traveling faster when they make contact with the
pad? (1) A
(2) B
(3) both have the same speed
Start at same height and initial speed, end at same height,
same energy lost due to pad, no friction
At the same time that Stuntman A slides down the roof, Stuntman B
steps off a roof (starting at the same height) on the other side of the
street. Which Stuntman makes contact with the pad first?
(1) A
(2) B
(3) both make contact at
the same time
Conservation of energy doesn’t help us here.
A is not in free fall until leaves roof, so for a while
acceleration smaller than B’s acceleration.
Suppose the roof on which A is sliding is NOT frictionless. Which Stuntman is traveling faster when they make contact with the
pad?
(1) A
(2) B
(3) both have the same speed
Work due to friction reduces total energy of A.
A crate is given a kick by a spring such that it has 50J of kinetic energy. It travels
along a straight frictionless track until it hits a 10m stretch of sandpaper. For this
stretch, the frictional force is 4N. The track becomes frictionless again after the
sandpaper. At which point or points does the box has a velocity of zero?
E
A
B C
D
1.
2.
3.
4.
5.
6.
7.
8.
A (1/4)
B (1/2)
C (3/4)
D (Flat after friction)
E (Highest point on slope
after friction)
E and C
E and B
E and A
Can lose KE due to Work done by frictional force or by transforming to PE.
Total work done by friction = F*d*cos(180) = 4*10*(-1) = -40J
Has 10J left. Climbs up slope, stops momentarily, then comes back.
If W=F*d*cos(180); -10 = 4*d*(-1); d= 2.5m
Conservation of Energy in General
• So far, just mechanical energy
• Other Energy (OE) types:
– Thermal, Electrical, Chemical, Nuclear, Sound, Light,...
• "Energy can neither be created nor destroyed, but only converted
from one form to another.”
KEi + PEi + WNC + OEi = KEf + PEf + OEf
Power
• “Rate at which work is done”
• “Rate at which energy is expended”
– J/s – watt (W)
P=
Work
Time
P=
Change in Energy
Time
A motor has a power rating of 500W. How much energy can the
motor provide when run for 20 s?
(1) 20 J
(2) 25 J
500 W is 500 J/s.
Every second you get 500 J
500 J/s * 20s = 10,000 J
(3) 500 J
(4) 10,000 J
Work, Energy, and Power in Humans
Efficiency (eff) = Useful Energy or Work Output / total Energy Input
What is the minimum energy needed for you to climb a 20m-tall cliff?
Assume a mass of 70kg. Also assume that you are at rest before and
after the climb.
(1) 14.3 J
(3) 196 J
(5) 686 J
(2) 140 J
(4) 463 J
(6) 13700 J
KEi + PEi + WNC = KEf + PEf
Pick reference at base
0J + 0J + WNC = 0 J + mgh
WNC = (70kg) (9.8m/s2)(20m) = 13700J
What is the minimum power output if this climb is accomplished over
a time of 3 minutes?
(1) 76.2 W
(2) 2380 W
(3) 4570 W (4) 8290 W
(5) 12850 W (6) 41200 W
Power = Energy or Work / time
= 13700 J / 180 s
= 76.2 W
If your body is only 0.20 (20%) efficient at creating useful energy
from your energy input (food), what would be the energy input
needed for the climb?
(1) 2740 J
(3) 13700 J
(5) 68500 J
(2) 10960 J
(4) 17100 J
Eff = Useful Energy / Energy Input
Energy Input = Useful Energy / eff
= 13700 J / 0.20
= 68500 J
How much ‘non-useful’ energy would be produced?
(typically in the form of thermal energy)
Energy input = Energy useful + Energy not useful
ENOT USEFUL = 68500 J – 13700 J = 54800 J
Stuntman
Look for similarities!
Diver
Fool
Stop
Stop