Chapter 8: Conser ation of EnergChapter 8: Conservation of Energy

There are two types of forces:

ti ( it i f )• conservative (gravity, spring force)

• nonconservative (friction)( )

Conservative ForcesConservative Force – the work done by the force on an object moving from one point to another depends only on the initial and final positions and is independent of the particular path taken.

orConservative Force – the net work done by a force on an object

Consider the work done by gravity:

Co se vat ve o ce t e et wo do e by a o ce o a objectmoving around a closed path is zero.

y g y

∫∫ =⋅=22

cos dlmgldFW θ

( )∫

∫∫ ==

2

11

cos

y

GG dlmgldFW θ

( )∫ −−=−=1

12y

G yymgdymgW

Conservative ForcesA force is conservative if the work it does on an

object moving between two points is j g pindependent of the path taken.

⇒ work done depends only on ri and rf

⇒ If an object moves in a closed path (ri = rf)then total ork done bthen total work done by the conservative force is zero.

Nonconservative ForcesNonconservative ForcesConsider friction

⇒ work done by the force depends on theforce depends on the path

i⇒ non-conservative forces dissipateenergy

Potential Energy P i l E i i i d i h hPotential Energy in a system is associated with the position

or configuration of objects in the system.For example: When you lift a ball a distance y theFor example: When you lift a ball a distance y, the

gravitational force does negative work on the ball. (The new position of the ball relative to the earth is changed by y. The energy that was “stored” in this new configuration of the earth-ball system is called gravitational potential energy It turns out thatgravitational potential energy. It turns out that,

Wc = –ΔU= – (Uf –Ui)W = work done by a conservative forceWc work done by a conservative forceU = potential energy

The zero of potential energy is arbitrary. Only the p gy y ychange in potential energy is relevant.

Gravitational Potential Energy

The gravitational force is conservative!

( )0gW mg y mgy= − − = −

( )0 0g y yW U U U= − − = − +

yU mgy=

We call this potential energy because if this mass is released,and allowed to fall, it can do work. Potential energy is thus storedin the system and available to do work if releasedin the system and available to do work if released.

Example

When a 4-kg object is moved from the ground to a shelf 1 m high what is the change in its potential energy?1 m high, what is the change in its potential energy? What is the change in potential energy if the same object is moved from the 1 m shelf to a shelf 2 m high?j g

( )( )

1 4 9.81 1 0 39.24 J

4 9 81 2 1 39 24 Jf iU mgy mgy

U mgy mgy

Δ = − = × × − =

Δ = = × × =( )2 4 9.81 2 1 39.24 Jf iU mgy mgyΔ = − = × × − =

Spring Potential EnergSpring Potential Energy

The potential energy stored in a stretched or compressed spring is,p p g

U = ½ kx2 x = displacement from equilibrium positionequilibrium position

Notice the sign of x, i.e., + or – doesn’t matterg f

for the potential energy.

ExampleExample

An 82 kg mountain climber is in the final stage of the ascent of 4301 m Pikes Peak. What is the change in gravitational potential g g penergy as the climber gains the last 100 m of altitude?

L U 0 l lLet Ui = 0 at sea level,

mkgmkgmgymgyU if )4201)(m/s8.9)(82()4301)(m/s8.9)(82( 22 −=−=Δ

JUf

400,80=Δ

L t U 0 t 4301Let Ui = 0 at 4301m,

mkgmkgmgymgyU if )100)(m/s8.9)(82()0)(m/s8.9)(82( 22 −−=−=Δ

JU

gggygy if

400,80

))()(())()((

=Δ

Potential Energy Summarized• A potential energy is always associated with a

conservative force, and the difference in potential energy between two points is defined as the negative of the work done by that force.

• The choice of where U=0 is arbitrary and can be chosen wherever it is most convenientchosen wherever it is most convenient.

• Since a force is always exerted on one body bySince a force is always exerted on one body by another body potential energy is associated with the interaction of two or more bodies.

C ti f EConservation of Energy

E i i h d d d• Energy is neither created nor destroyed

• The energy of an isolated system of objects remains constant.

Mechanical EnergyMechanical energy E is the sum of the potential

Mechanical Energygy p

and kinetic energies of an object.E = U + KE U + K

The total mechanical energy in any isolated system of objects remains constant if the objectssystem of objects remains constant if the objects interact only through conservative forces:

EE = constantEf = Ei → Uf + Kf = Ui+ Ki

ExampleExampleA 5.00-kg rock is dropped and allowed to fall freely. Find the speed of the ball after it has dropped 2 m.

Define the release point as U=0. Write down the conservation of energy theorem.gy

UKEUKE +=+

msmkgvkg

UKEUKE

f

ffii

)2)(/8.9)(5()5(00 2221 −+=+

+=+

smv f /3.6=

Pendulum

A 0.2-kg pendulum bob is swinging back and forth. If the speed of the bob at its lowest point is 0.65 m/s, how high does the bob go above its minimumdoes the bob go above its minimum height? h

Use conservation of energy,gy,

UKEUKE +=+

mghmv

UKEUKE

i

ffii

00221 +=+

+=+

gi2

( ) cmvh i 2.2)/89(2

m/s65.02 2

22

===g )m/s8.9(22 2

Energy Conservation with Dissipative Forces: Solving Problems

Remember the work-energy theorem:

KEWnet Δ=

Also, remember that the work done by conservative forces is:

UWc Δ−=

nccnet WWW +=But,

nccnet

ΔΔ WUKE

So,

⇒+Δ−=Δ ncWUKE UKEWnc Δ+Δ=

ExampleA 2 kg ball is dropped from a height of 25m. The work done on the ball by friction is Wf= -200 J. How fast is the ball fmoving when it strikes the ground?

)(2121 WmghmghWUmvmvKE +=+Δ==Δ

)200())25)(/8.9)(2(0(0)2(

)(22

21

21

21

Jmsmkgvkg

WmghmghWUmvmvKE

f

fifncif

−+−−=−

+−−=+Δ−=−=Δ

smv f /17=

Without air friction vf = 22m/s

PowerPower is defined as the amount of work done per unit time. The average power over a time interval t is,

tWP = SI Units: watt ≡ joule/secondt

Power can also be written in terms of velocity and displacement:Power can also be written in terms of velocity and displacement:

( )dW d dlP F l Fd d d

= = ⋅ = ⋅( )dt dt dt

FPWhat’s the condition on F here?

vFP ⋅=

Examples4. A 66.5 kg hiker starts at an elevation of 1500 m and climbs to the top of a 2660 m peak. (a) What is the hiker’s change in

i l ? (b) Wh i h i i k i d b hpotential energy? (b) What is the minimum work required by the hiker? (c) Can the actual work done be greater than this?

8. Air resistance can be represented by a force proportional to the velocity V of an object: F=-kv. Is this force conservative? Explain ’Explain.

13. In the high jump, the kinetic energy of an athlete is t f d i t it ti l t ti l ith t th id ftransformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 2.10 m and cross the bar with a speed of 0.70 m/s?

Examples Cont.Calculate the power required of a 1400 kg car under the following circumstances: (a) the car climbs a 10o hill at a steady 80 km/h; ( ) y ;and (b) the car accelerates along a level road from 90 to 110 km/h in 6 seconds to pass another car. Assume that the car has a retarding force of F = 700 N throughout (this is due to airretarding force of FR= 700 N throughout (this is due to air resistance and friction inside the wheel bearings).

ExampleA 1.9-kg block slides down a frictionless ramp, as shown in the Figure. The top of the ramp is 1.5 m above the ground; the bottom f th i h 0 25 b th d Th bl k l thof the ramp is h = 0.25 m above the ground. The block leaves the

ramp moving horizontally, and lands a horizontal distance d away. Find the distance d.

Use conservation of mechanical to find the velocityat the bottom of the ramp.

i i f fK U K U+ = +

210 02

2

mgh mv

v gh

+ = +

=

2

Now we have a projectile problem.12 0 25h t t

2v gh=

2

2 2

2 0.252

1 2 0.250 0.252 9 81to hit to hit

x gh t y gt

gt t− −

= = −

×= − =

( ) ( )

2 9.810.226 s

2 2 9 81 1 5 0 25 0 226 1 1 m

to hit to hit

to hit

g

t

d gh t− =

= = × × − =( ) ( )2 2 9.81 1.5 0.25 0.226 1.1 mto hitd gh t −= = × × − =

Additional Examples6. If U = 3x2 +2xy+4y2z, what is the force, F?

15. A 60 kg bungee jumper jumps from a bridge. She is ties to a bungee cord that is 12 m long when unstretched, and she falls a total of 31m. (a) Calculate the spring constant k of the bungee cord assuming Hooke’s law applies. (b) Calculate the maximum g g pp ( )acceleration felt by the jumper.

21. A pendulum 2.00m long is released (from rest) at an angle of 30°. Determine the speed of the 70 0 g bob (a) at the lowest point (θ=0°); (b) at θ=15 0° (c) θ= 15 0° (d)speed of the 70.0 g bob (a) at the lowest point (θ=0°); (b) at θ=15.0° (c) θ=-15.0° (d) Determine the tension in the cord at each of these three points. (e) If the speed of the bob is given an initial speed of vo=1.20 m/s when release at 30.0°, recalculate the speeds for parts (a), (b) and (c).

31. A skier traveling 11 m/s reaches the foot of a steady upward 17° incline and glides 12m up along the slope before coming to a rest. What is the average coefficent of kinetic friction?kinetic friction?

54. How long will it take a 1750 motor to lift a 285 kg piano to a sixth story window 16 m above?

58. What minimum horsepower must a motor have to be able to drag a 300 kg box along a level floor at a speed of 1.2 m/s if the coefficient of kinetic friction is 0.45?