Energy Transformationsand Conservation of Mechanical Energy

8.01W05D2

Today’s Reading Assignment: W05D2

Young and Freedman: 7.1-7.5,12.3

Experiment 3: Energy Transformations

Review: Potential Energy Difference

c

B

c

A

U d WΔ ≡− ⋅ =−∫F rr r

Definition: Potential Energy Difference between the points A and B associated with a conservative force is the negative of the work done by the conservative force in moving the body along any path connecting the points A and B.

cFr

Review: Examples of Potential Energy with Choice of Zero

Point(1) Constant Gravity:

(2) Inverse Square Gravity

(3) Spring Force

U ( y) =mgy

U ( y =0) =0

U (r) = −

Gm1m2

r

U (r0=∞) =0

U (x) =(1 / 2)kx2

U (x =0) =0

Review: Work-Energy Theorem: Conservative Forces

The work done by the total force in moving an object from A to B is equal to the change in kinetic energy

When the only forces acting on the object are conservative forces

then the change in potential energy is

Therefore

0

total total 2 20

1 1

2 2

fz

fzW d mv mv K≡ ⋅ = − ≡Δ∫ F r

r r

ΔU =−W =−ΔK

ΔU + ΔK =0

totalc=F F

r r

Forms of Energy

• kinetic energy• gravitational potential energy• elastic potential energy• thermal energy• electrical energy• chemical energy• electromagnetic energy• nuclear energy• mass energy

Energy Transformations

Falling water releases stored ‘gravitational potential energy’ turning into a ‘kinetic energy’ of motion.

Human beings transform the stored chemical energy of food into catabolic energy

Burning gasoline in car engines converts ‘chemical energy’ stored in the atomic bonds of the constituent atoms of gasoline into heat

Stretching or compressing a spring stores ‘elastic potential energy’ that can be released as kinetic energy

Energy Conservation Energy is always conserved

It is converted from one form into another, as the system transforms from an “initial state” to a “final state”, each form of energy can undergo a change

Energy can also be transferred from a system to its surroundings

ΔE ≡Efinal −Einitial

1 21

... 0N

ii

E E E=

Δ =Δ +Δ + =∑

system surroundings 0E EΔ +Δ =

Concept Question: Energy Transformations

1. The potential energy of the system increases. 2. The kinetic energy of the system decreases. 3. The earth does negative work on the system. 4. You do negative work on the system. 5. Two of the above. 6. None of the above.

You lift a ball at constant velocity from a height hi to a greater height hf. Considering the ball and the earth together as the system, which of the following statements is true?

Mechanical Energy

When a sum of conservative forces are acting on an object, the potential energy function is the sum of the individual potential energy functions with an appropriate choice of zero point potential energy for each function

Definition: Mechanical Energy

The mechanical energy function is the sum of the kinetic and potential energy function

Emech ≡K +U

U =U1 +U2 +⋅⋅⋅

Conservation of Mechanical Energy

When the only forces acting on an object are conservative

Equivalently, the mechanical energy remains constant in time

ΔK + ΔU =0

E

fmech =K f +U f =K i +U i =Ei

mech

(K

f−K i ) −(U f −U i ) =0

ΔEmech =E f

mech −Eimech =0

Non-Conservative Forces

Definition: Non-conservative force Whenever the work done by a force in moving an object from an initial point to a final point depends on the path, then the force is called a non-conservative force and the work done is called non-conservative work

rF

nc

Wnc ≡

rFnc

A

B

∫ ⋅drr

Non-Conservative Forces Work done on the object by the force depends on the path taken by the object

Example: friction on an object moving on a level surface

friction

friction friction 0k

k

F N

W F x N x

μμ

==− Δ =− Δ <

Change in Energy for Conservative and Non-conservative Forces

Force decomposition:

Work done is change in kinetic energy:

Mechanical energy change:

W =

rF ⋅d

rr

A

B

∫ = (rFc +

rFnc)

A

B

∫ ⋅drr =−ΔU +Wnc =ΔK

ΔK + ΔU = ΔEmech = Wnc

rF =

rFc +

rFnc

Concept Question: Energy and Choice of System

1. block 2. block + spring 3. block + spring + incline 4. block + spring + incline +

Earth

A block of mass m is attached to a relaxed spring on an inclined plane. The block is allowed to slide down the incline, and comes to rest. The coefficient of kinetic friction of the block on the incline is µk. For which definition of the system is the change in energy of the system (after the block is released) zero?

Worked Example: Block Sliding off Hemisphere

A small point like object of mass m rests on top of a sphere of radius R. The object is released from the top of the sphere with a negligible speed and it slowly starts to slide. Find an expression for the angle θf with respect to the vertical at which the object just loses contact with the sphere.

Strategy: Using Multiple IdeasEnergy principle: No non-conservative work

For circular motion, you will also need to Newton’s Second Law in the radial direction because no work is done in that direction hence the energy law does not completely reproduce the equations you would get from Newton’s Second Law

Constraint Condition:

Conservation

ΔK + ΔU = ΔEmech = Wnc = 0

r̂ : F

r= −m

vf2

R

0 at fN θ θ= =

Worked Example: Energy Changes

W

nc=0 =E f

mech −Eimech ⇒

Ki; 0

Ui =0

Eimech ; 0

K f =12

mvf2

U f =−mgR(1−cosθ f )

0 =0 −12

mvf2 −mgR(1−cosθ f )

⎛⎝⎜

⎞⎠⎟⇒

1

2mv f

2 =mgR(1−cosθ f )

W

nc=0

Worked Example: Free Body Force Diagram

Newton’s Second Law

Constraint condition:

Radial Equation becomes

2

ˆ : cosv

N mg mR

θ− = −r

θ̂ : mgsinθ =mR

d2θdt2

0 at fN θ θ= =

mg cosθ f = m

vf2

R

Worked Example: Combining Concepts

Newton’s Second Law Radial Equation

Energy Condition:

Combine Concepts:

mg cosθ f = m

vf2

R⇒

R

2mg cosθ f =

12

mvf2

1

2mv f

2 =mgR(1−cosθ f )

(1 cos ) cos2f f

RmgR mgθ θ− = ⇒ 12 2

cos cos3 3f fθ θ − ⎛ ⎞= ⇒ = ⎜ ⎟

⎝ ⎠

Modeling the Motion: Newton’s Second Law

Define system, choose coordinate system.

Draw free body force diagrams.

Newton’s Second Law for each direction.

Example: x-direction

Example: Circular motion

totalˆ : .2

x 2

d xF m

dt=i

2totalˆ : .r

vF m

R=−r

Modeling the Motion Energy Concepts

Change in Mechanical Energy:

Identify non-conservative forces.

Calculate non-conservative work

Choose initial and final states and draw energy diagrams.

Choose zero point P for potential energy for each interaction in which potential energy difference is well-defined.

Identify initial and final mechanical energy

Apply Energy Law.

final

nc ncinitial

W d .= ⋅∫ F rr r

W

nc=ΔK + ΔU =ΔEmech

Table Problem: Loop-the-Loop

An object of mass m is released from rest at a height h above the surface of a table. The object slides along the inside of the loop-the-loop track consisting of a ramp and a circular loop of radius R shown in the figure. Assume that the track is frictionless. When the object is at the top of the track (point a) it just loses contact with the track. What height was the object dropped from?

Demo slide: Loop-the-Loop B95

http://scripts.mit.edu/~tsg/www/index.php?page=demo.php?letnum=B 95&show=0

A ball rolls down an inclined track and around a vertical circle. This demonstration offers opportunity for the discussion of dynamic equilibrium and the minimum speed for safe passage of the top point of the circle.

Demo slide: potential to kinetic energy B97

http://scripts.mit.edu/~tsg/www/index.php?page=demo.php?letnum=B 97&show=0

This demonstration consists of dropping a ball and a pendulum released from the same height. Both balls are identical. The vertical velocity of the ball is shown to be equal to the horizontal velocity of the pendulum when they both pass through the same height.

Table Problem: Experiment 3 Cart-Spring on an Inclined Plane

A cart of mass m slides down a plane that is inclined at an angle θ from the horizontal. The cart starts out at rest. The center of mass of the cart is a distance d from an unstretched spring with spring constant k that lies at the bottom of the plane. Assume that the inclined plane has a coefficient of kinetic friction μ. Find an equation whose solution describes how far the spring will compress when the cart first comes to rest.

Experiment 3 Energy Transformation

Potential Energy and ForceIn one dimension, the potential difference is

Force is the derivative of the potential energy

Examples: (1) Spring Potential Energy:

(2) Gravitational Potential Energy:

U (x) −U (x0 ) =− Fx

A

B

∫ dx

F

x=−

dUdx

2spring

1( )

2U x kx=

2,

1

2x spring

dU dF kx kx

dx dx⎛ ⎞=− =− =−⎜ ⎟⎝ ⎠

1 2grav ( ) =

Gm mU r

r−

1 2 1 2, 2r gravity

Gm m Gm mdU dF

dr dr r r⎛ ⎞=− =− − =−⎜ ⎟⎝ ⎠

Energy DiagramChoose zero point for potential energy:

Potential energy function:

Mechanical energy is represented by a horizontal line since it is a constant

Kinetic energy is difference between mechanical energy and potential energy (independent of choice of zero point)

U (x) =

12

kx2 , U (x=0) =0

U (x =0) =0

Emechanical =K (x) +U(x) =12

mvx2 +

12

kx2

K =Emechanical −UGraph of Potential energy function

U(x) vs. x

Table Problem: Energy Diagram

The figure above shows a graph of potential energy U(x) verses position for a particle executing one dimensional motion along the x-axis. The total mechanical energy of the system is indicated by the dashed line. At t =0 the particle is somewhere between points A and G. For later times, answer the following questions. a)At which point will the magnitude of the force be a maximum?b)At which point will the kinetic energy be a maximum?c)At how many of the labeled points will the velocity be zero?d)At how many of the labeled points will the force be zero?

Next Reading Assignment: W05D3

Young and Freedman 7.1-7.5, 12.3, 12.6

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