Download Lecture 6 - January 31

Power
€
W
Pave =
Δt
•  Conservative Force:
dW
P=
dt
closed path is zero
→
Net work done by a conservative force on an object moving around every
€
•  Non-conservative Force:
→
P = F net • v
€
Forces
Net work done by a non-conservative
force on an object moving around every
closed path is non-zero
Potential energy: Energy U which describes the configuration (or spatial
arrangement) of a system of objects that exert conservative forces on each
other. It’s the stored energy in system.
ΔU = −W
Definition
Gravitational Potential energy: [~associated with the state of separation]
yf
ΔU grav = − ∫ (−mg)dy = mg( y f − yi ) = mgΔy
yi
€
If U grav (y = 0) ≡ 0
then U grav (y) = mgy
ΔUgrav ↑ if going up
ΔUgrav ↓ if going down
Elastic Potential energy: [~associated with the state of compression/tension
of elastic object]
ΔU spring = 12 kx 2f − 12 kx i2
€
€
If
€
U spring (x = 0) ≡ 0
then
U spring (x) = 12 kx 2
ΔUspring ↑ if x goes ↑ or ↓ (any displacement)
€
Don’t forget…
Work done by force (general)
 
  
W = ∫ F ( x ) • dx = F • d
Work by Gravitational force:
 
Wg = Fg • d
€
Wspring = 12 kx i2 − 12 kx 2f
Work by Spring force:
€
Wnet = ΔKE = KE f − KE i
Work-Kinetic Energy theorem:
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Potential energy (if conservative force):
ΔU grav = mgΔy
W = −ΔU
€ If U grav (y = 0) ≡ 0 then U grav (y) = mgy
ΔU spring = 12 kx 2f − 12 kx i2
€
€
€
If U spring (x = 0) ≡ 0
then U spring (x) = 12 kx 2
Mechanical Energy
Mechanical energy:
Emech = KE + U
•  Only conservative forces (gravity & spring) cause energy transfer (work)
Wnet = ΔKE
Sec. 7 - 3
Wconservative = −ΔU
• 
Sec. 8 -1
Isolated system: Assuming only internal forces (no external forces yet - Sec. 8.6)
–  No external force from outside causes energy change inside
€
ΔE mech = 0 = Δ (KE + U )
ΔKE = −ΔU
€
How to apply your knowledge to solve problems?
Step I: Choose a system
Step II: Check if forces are conservative
Step III: If so, apply
ΔE mech = 0
⇒ ΔKE = −ΔU
⇒
€
(KE1 + U1 ) = (KE 2 + U 2 )
How to
use it?
ΔE mech = 0
⇒ ΔKE = −ΔU
⇒
(KE1 + U1 ) = (KE 2 + U 2 )
Example: Pendulum
E mech = KE + U = constant
€
= [( 12 mv 2 ) + ( mgy)] at all times = constant
€
Assumes no friction
(non-conservative, external force)
Example of Conservation of Mechanical Energy
Consider a 100 kg bobsled starting with vo = 0 on a frictionless track. Compute the KE,
PE, and total E at the labeled points.
PE = mgh
Emech=KE+PE
KE =
KE
1 2
mv
2
PE
588,000 J
Emech = KE + PE
Height h
600 m
588,000 J
0 J
588,000 J
196,000 J
392,000 J 400 m
588,000 J
392,000 J
196,000 J 200 m
588,000 J
588,000 J
0 J
0 m
Example of Conservation of Mechanical Energy
Consider a 100 kg bobsled starting with vo = 0 on a frictionless track. Compute the KE,
PE, and total E at the labeled points.
PE = mgh
Emech=KE+PE
1
KE = mv 2
2
KE
PE
Emech = KE + PE
Height h
0 m
0 J
0 J
O
0 J
J
0 J
196,000 J
-196,000 J
00 J
J
392,000
392,000J
-392,000 J -400 m
0 J
588,000 J
-588,000 J
-200 m
-600 m
New Zero
Question
Question 8-2
Three identical projectiles are fired at different launch angles
and with different initial velocities. Which projectile has the
greatest potential energy when it’s at the peak in its trajectory?
1. A
2. B
3. C
4. All are equal
Can you tell me anything about the kinetic energies?
Problem 8-4: Roller Coaster
What is the speed of coaster at a) 
Point A
b) 
Point B
c) 
How high will it go on the last hill?
Problem 8-12: Roller Coaster
hmax
What is the speed of coaster at a) 
Point A
ΔU grav = mgΔy = 0 = −ΔKE
1
ΔKE = m(vA2 − v02 )
2
€
b) 
Point B
⇒
vA = v0
ΔU grav = mgΔy = mg(h /2 − h) = −ΔKE
ΔKE = 12 m (vB2 − v02 )
1
2
(v
2
B
− v02 ) = −g(−h /2)
vB = gh + v02
c) 
ΔU grav = mgΔy = mg(h max − h) = −ΔKE
How high will it go on the last hill?
ΔKE = 12 m(0 − v02 )
mg(h max − h) = − 12 m(0 − v02 )
€
hmax
€
v02
=
+h
2g
Question
Question 8-4
A block initially at rest slides down a frictionless ramp and attains a
speed of v at the bottom. To achieve a speed of 2v, how many times as
high must a new ramp be?
KE f = KEi + PEi − PE f = 0 + mgh
1
2
mv 2f = mgh
v f = 2gh
1. 
2. 
3. 
4. 
5. 
6. 
1
2
3
4
5
6
Question
Question 8-3
A young girl wishes to select one of the frictionless playground
slides below to give her the greatest possible speed when she
reaches the bottom of the slide. Which one should she choose?
B
A
1. 
2. 
3. 
4. 
5. 
A
B
C
D
Any of them
C
D
Demo
€
ΔE mech = 0
⇒ ΔKE = −ΔU
⇒
(KE1 + U1 ) = (KE 2 + U 2 )
Example
A block of mass 0.800 kg is given an initial velocity
Va= 1.20 m/sec to the right and collides with a light
spring of force (spring constant k=50.0 N/m) as seen
in the figure.
If the surface is frictionless, calculate the maximium
compression of the spring after the collision.
A small block of mass m can slide along the frictionless loop-the-loop.
The block is released from rest at point P, at a height h = 5R above
the bottom of the loop. a) 
b) 
c) 
d) 
How much work does the gravitational force do on the block as the block
travels from point P to point Q?
If U = 0 at the bottom, what is the potential energy when the block is at the
top of the loop? If, instead of being released, the block is given some initial speed downward
along the track, how do the above answers change?
What is the minimum height h so that the block makes it around the loop?
Problem
A small block of mass m can slide along the frictionless loopthe-loop. The block is released from rest at point P, at a
height h = 5R above the bottom of the loop. How much work does the gravitational force do on the block as
the block travels from point P to point Q?
 
Wg = Fg • d = mg(− ˆj ) • (Rˆj − 5Rˆj ) = mg(− ˆj ) • (−4 Rˆj ) = 4mgR
(
) (
) (
) (
)
If U = 0 at the bottom, what is the potential energy when the block is at the top of the loop? = 0
 
€
Wg = Fg €
• d = −ΔU
[(
)]
⇒ U top −U bottom = −Wg = − mg(− ˆj ) • (2Rˆj − 0Rˆj ) = 2mgR
) (
If, instead of being released, the block is given some initial speed downward along the track,
how do the above answers change?
Do the answers depend on velocity? NO -> Answers don’t change
€
What is the minimum height h so that the block ΔU = −ΔKE
makes it around the loop?
2
(mgh min ) − (mg(2R)) = 12 m (vtop2 − vinitial
)
From forces (see prior lecture)
1 2
 v2 
h
=
vtop − 0) + 2R
(
min
yˆ : −Fg = −( mg) = m−  or vmin = gr
2g
 r
1
5
h min = (gR) + 2R = R
2g
2
€
€
HW #12
Problem
A block of mass m is dropped onto a spring. The block
becomes attached to the spring and compresses it by
distance d before momentarily stopping.
While the spring is compressed, what work is done on the block by:
a) What is the speed of the block just before it hits the spring?
b) From what height h was the box dropped?
c) How high will it go back.
Potential Energy Curve
Plot of U(x), the potential energy as a function of the a system with 1-D
movement along x-axis:
E = KE(x) + U (x) = constant
mech
KE(x) = E mech −U (x)
dU (x)
F(x) = −
dx
Equilibrium Points
Turning point
€
€
Emech = 5 J
Emech = 4 J
Emech€= 3 J
Emech = 1 J
Equilibrium positions: where slope of U(x) curve is zero [i.e. F(x) = 0 ; NO FORCE ]
-> Neutral vs Unstable vs Stable Equilibrium
KE=0 ; F=0 & if move left or right  move back KE=0 ; F=0 but if move left or right  force to move away KE=0 ; Emech=U  stationary particle Finding a conservative force from potential energy 
U (x)
F (x)
x2
In 1-D: where F is a slowly varying, internal
force acting on a particle in system ΔU (x) = −Wcons = − ∫ Fdx
x1
€
€
ΔU (x) ≈ −F(x)Δx
then €
€
Now go backwards, say you know the change in potential energy at some
point and you want to know the force at that point…
(in the differential limit) Fgrav (y) = −
dU (x)
F(x) = −
dx
Fspring (x) = −
€
€
€
dU grav (y)
(mgy)
=−
= −mg
dy
dy
✔ ✔
(
=−
✔ ✔
dU spring (x)
dx
1
2
kx 2 )
dx
= −kx
Potential Energy Curve
Plot of U(x), the potential energy as a function of the a system with 1-D
movement along x-axis while a conservative force does work on it:
dU (x)
F(x) = −
dx
€
F(x) is negative slope of
tangent to U(x)
Problem #40: The figure shows a potential energy curve U vs position.
If UA=9.00J, UC=20.J, UD=24.0J. The particle is released the point on U at
the point between 1 and 3 m with a kinetic energy of 4J and UB=12. J.
What is the Kinetic energy and speed at x=3.5 m and x=6.5m? Where is
the turning point on the left and right.
16.0J
12.0J
Ek=mv2/2
9.0 J