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 K = -U + WNC

or: 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

• 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)

Sect. 8-3: Mechanical Energy & Its Conservation

For the SPECIAL CASE of

CONSERVATIVE FORCES ONLY!

K + U = 0 the Principle of Conservation

of Mechanical EnergyNote 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 EnergyNote! 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 Energyis Newton’s 2nd Law in Work-Energy language,

in the special case of Conservative Forces only!

Conservation of Mechanical EnergySo, 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 = ConstantSo, in any process with only conservative forces acting,

the sum K + U is unchangedThe 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 are equal in size but opposite in sign.

The Total 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

or E = K + U = Constant For conservative forces ONLY (gravity, spring, etc.)

• Suppose, initially: E = K1 + U1

& finally: E = K2+ U2

E = Constant

K1 + U1 = K2+ U2

A very powerful method of calculation!!

Conservation of Mechanical Energy

K + U = 0

or 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

U1 = mgh, K1 = 0

U2 = 0, K2 = (½)mv2

K + U = same as at points 1 & 2

the sum K + U remains constant

K1 + U1 = K2 + U2

0 + mgh = (½)mv2 + 0v2 = 2gh

all U

half K half U

all K

Sect. 8-4: Using Conservation of Mechanical Energy

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?

K1 + U1 = K2 + U2 = K3 + U3 0 + mgh = (½)mv2 + 0

= (½)m(v3)2 + mgy3

Example 8-3: Falling Rock

U3 = mgy3

K3 = (½)m(v3)2

U1 = mgy1, K1 = 0

U2 = 0, K2 = (½)m(v2)2

Example 8-3What is it’s speed at y = 1.0 m?

Conservation of

Mechanical Energy!

(½)m(v1)2 + mgy1

= (½)m(v2)2 + mgy2

(The mass cancels in the equation!)

y1 = 3.0 m, v1 = 0

y2 = 1.0 m, v2 = ?

Gives: v2 = 6.3 m/s

U only

Part U, part K

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 = hU1 = mghK1 = 0

y2 = y U2 = mgy K2 = (½)mv2

y

v

v

y3 = 0, U3 = 0 K3 = (½)m(v3)2

v

• Conservation of Mechanical Energy

K2 + U2 = K1+ U1

In general, for an initial velocity v1, solve for v:

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.

2 2f iv v g h y v =

(v1)2 + 2g(h – y)

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 shaped differently, but start 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 waterslides!

both starthere!

• 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.

Example – Grand Entrance

Free Body Diagrams

Actor Sandbag

at bottom

Step 1: To find actor’s speed at bottom, let yi = initial height above floor & use

Conservationof Mechanical Energy

Ki + Ui = Kf + Uf

or 0 + mactorgyi = (½)mactor(vf)2 + 0 (1)

mass cancels. From diagram,

yi = R(1 – cos θ)

So, (1) becomes: (vf)2 = 2gR(1 – cosθ) (2)

Step 2: Use N’s 2nd Law for actor at 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 = 60°