Transcript Conservation of Mechanical Energy

Sect. 8-3: Mechanical Energy & It’s Conservation
GENERALLY
In any process, total energy is neither
created nor destroyed.
• The total energy of a system can be transformed from
one form to another & from one object to another, but
the total amount remains constant
 The Law of Conservation
of Total Energy
In Ch. 7, we learned the Work-Energy Principle:
The net work W done on an object =
the change in the object’s kinetic energy K
Wnet = K
 The Work-Energy Principle
(As we’ve said, this is Newton’s 2nd Law in Work-Energy language!)
Note: Wnet = The work done by the net (total) force.
• If several forces act (both conservative & non-conservative),
The total work done is:
Wnet = WC + WNC
WC = the work done by all conservative forces
WNC = the work done by all non-conservative forces
• The Work-Energy Principle still holds:
Wnet = K
• For conservative forces (by the definition of potential energy U!):
WC = -U
or:

K = -U + WNC
WNC = K + U
 So, in general,
WNC = K + U
The Total Work done by all
Non-conservative forces
= The total change in K
+ the total change in U
Sect. 8-3: Mechanical Energy & Its Conservation
• In general, we just found that:
WNC = K + U
• In the
(Very!) SPECIAL CASE where there are
CONSERVATIVE FORCES ONLY!

WNC = 0
or
K + U = 0
 the Principle of Conservation of
Mechanical Energy
• Note: This ISN’T (quite) the same as the Law of Conservation of (total)
Energy! It’s a special
case of this law (where all forces are conservative)
For the SPECIAL CASE of
CONSERVATIVE FORCES ONLY!
K + U = 0
 the Principle of Conservation
of Mechanical Energy
Note Again! This is NOT (quite) the same as
the Law of Conservation of (total) Energy!
The Principle of Conservation of
Mechanical Energy is a special case of this law
(where all forces are conservative)
For CONSERVATIVE FORCES ONLY!
K + U = 0
 Principle of Conservation of Mechanical Energy
Note! This came from the Work-Energy Principle & the definition of
Potential Energy for conservative forces!
Since the Work-Energy Principle is N’s 2nd Law in Work-Energy language,
& since Conservation of Mechanical Energy came from this principle,
Conservation of Mechanical Energy
is Newton’s 2nd Law in Work-Energy language,
in the special case of Conservative Forces only!
Conservation of Mechanical Energy
So, for conservative forces ONLY! In any process
K + U = 0
Conservation of Mechanical Energy
• Definition of Mechanical Energy: E  K + U
Conservation of Mechanical Energy
 In any process with only conservative forces acting,
E = 0 = K + U
or
E = K + U = Constant
So, in any process with only conservative forces acting,
the sum K + U is unchanged
The mechanical energy changes from U to K or K to U, but
the sum remains constant.
If & ONLY if there are no nonconservative forces,
the sum of the kinetic energy change K & the potential energy
change U in any process is zero.
The kinetic energy change K & the potential energy change U
equal in size but opposite in sign.
The Total
are
Mechanical Energy is defined as:
Conservation of Mechanical Energy has the form:
Summary:
The Principle of Conservation of
Mechanical Energy
If only conservative forces are acting, the total
mechanical energy of a system neither increases
nor decreases in any process.
It stays constant; it is conserved.
• Conservation of Mechanical Energy

K + U = 0
E = K + U = Constant
or
For conservative forces ONLY (gravity, spring, etc.)
• Suppose, initially:
& finally:

E = K1 + U1
E = K2+ U2
E = Constant
K1 + U1 = K2+ U2
A very powerful method of calculation!!
Conservation of Mechanical Energy

or
K + U = 0
E = K + U = Constant
• For gravitational U:
Ugrav = mgy
E = K1 + U1 = K2+ U2
 (½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2
y1 = Initial height, v1 = Initial velocity
y2 = Final height, v2 = Final velocity
all U
 U1 = mgh, K1 = 0
the sum K + U remains constant
K + U = same as at points 1 & 2
half K
half U
K1 + U 1 = K 2 + U 2
0 + mgh = (½)mv2 + 0
v2 = 2gh
 U2 = 0, K2 = (½)mv2
all K
Sect. 8-4: Using Conservation of Mechanical Energy
Example 8-3: Falling Rock
U1 = mgy1, K1 = 0
Original height of the rock:
y1 = h = 3.0 m
S
Speed when it falls to
l
y3 = 1.0 m
a
above the ground?
K 1 + U 1 = K2 + U 2 = K3 + U 3
0 + mgh = (½)mv2 + 0
= (½)m(v3)2 + mgy3
U3 = mgy3
K3 = (½)m(v3)2
U2 = 0, K2 = (½)m(v2)2
Example 8-3
What is it’s speed at y = 1.0 m?
Conservation of
Mechanical Energy!
 (½)m(v1)2 + mgy1
= (½)m(v2)2 + mgy2
 U only
 Part U, part K
(The mass cancels in the equation!)
y1 = 3.0 m, v1 = 0
y2 = 1.0 m, v2 = ?
Gives: v2 = 6.3 m/s
 K only
Example – Free Fall
• A ball is dropped from height
h above the ground.
• Calculate the ball’s speed at
distance y above the ground.
• Neglecting air resistance, the
only force is gravity, which is
conservative.
• So, we can use
Conservation of
Mechanical Energy
y1 = h
U1 = mgh
K1 = 0
y
v
y2 = y
U2 = mgy
K2 = (½)mv2
v
v
y3 = 0, U3 = 0
K3 = (½)m(v3)2
• Conservation of Mechanical Energy
K2 + U2 = K1+ U1
In general, for an initial velocity v1, solve for v:
= v 2
+g2g(h
v fv 
 h – yy)
2 2
(vi 1)
The result for v is, of course, consistent with the results
from the kinematics of Ch. 2. For this specific problem,
v1 = 0 & K1 = 0, since the ball is dropped.
Example 8-4: Roller Coaster
• Mechanical energy conservation! (Frictionless!)
 (½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2
(mass cancels!) Only height differences matter! Horizontal distance doesn’t matter!
• Speed at the bottom?
y1 = 40 m, v1 = 0
y2 = 0 m, v2 = ?
Gives: v2 = 28 m/s
• For what y is
v3 = 14 m/s?
Use: (½)m(v2)2 + 0
= (½)m(v3)2 + mgy3
Gives: y3 = 30 m
Height of hill = 40 m. Starts from rest at the
top. Calculate:
a. The speed of the car at the hill bottom.
b. The height at which it has half this speed.
Ii Take y = 0 at the bottom of the hill.
Conceptual Example 8-5: Speeds on 2 Water Slides
Two water slides at a pool are
both
start
shaped differently, but start
here!
lllll at the same height h.

Two riders, Paul & Kathleen,
start from rest at the same time on 
different slides. Ignore friction &
assume that both slides have the
same path length.
a. Which rider is traveling
FF faster at the bottom?
b. Which rider makes it to the
FF bottom first? WHY??
Demonstration!
frictionless
water
slides!
Example – Grand Entrance
• An actor, mass mactor = 65 kg, in a
play is to “fly” down to stage
during performance. Harness
attached by steel cable, over 2
frictionless pulleys, to sandbag,
mass mbag = 130 kg, as in figure.
Need length R = 3 m of cable
between nearest pulley & actor so
pulley can be hidden behind stage.
For this to work, sandbag can never
lift above floor as actor swings to
floor. Let initial angle cable makes
with vertical be θ. Calculate the
maximum value θ can have such
that sandbag doesn’t lift off floor.
Step 1: To find actor’s speed at bottom, let
yi = initial height above floor & use
Conservationof Mechanical Energy
Ki + U i = K f + U f
or 0 + mactorgyi = (½)mactor(vf)2 + 0 (1)
Free Body Diagrams
mass cancels. From diagram,
yi = R(1 – cos θ)

So, (1) becomes: (vf)2 = 2gR(1 – cosθ) (2)
Actor
Step 2: Use N’s 2nd Law for actor at
at bottom
bottom of path (T = cable tension).
Actor:
∑Fy = T – mactorg = mactor[(vf)2/R]
or
T = mactorg + mactor[(vf)2/R]
(3)
Step 3: Want sandbag to not move.  N’s 2nd Law for sandbag:
∑Fy = T – mbagg = 0 or
T = mbagg
(4)
Combine (2), (3), (4): mbagg = mactorg + mactor[2g(1 – cosθ)].
Solve for θ: cosθ = [(3mactor - mbag)/(2 mactor)] = 0.5 or, θ

Sandbag
= 60°