Energy Conservation 1

Preflight 1

Imagine that you are comparing three different ways of having a ball move down through the same height. In which case does the ball get to the bottom first?

30% 0% 0% 70% A. Dropping C. Swinging down D. All the same correct B. Slide on ramp (no friction) 1 2 3 “The other balls have something acting on the acceleration of the balls. Ball A is the only one that is free falling.” “All of the balls will be going the same speed, because they all have an acceleration of -9.8 m/s/s. All of the balls will reach the bottom at the same time also because of the same reason, and they are dropped from the same height”

Conservative and Nonconservative Forces

Conservative force: the work it does is stored in the form of energy that can be released at a later time

Example of a conservative

force: gravity

Example of a nonconservative

force: friction Also: the work done by a conservative force in moving an object around a closed path is zero; this is not true for a nonconservative force.

Conservative Force: Gravity

Work done by gravity on a closed path is zero

Nonconservative Force: Friction

Work done by friction on a closed path is

not

zero

Table 8-1 Conservative and Nonconservative Forces

Section Force Conservative forces Gravity Spring force Nonconservative forces Friction Tension in a rope, cable, etc.

Forces exerted by a motor Forces exerted by muscles 5-6 6-2 6-1 6-2 7-4 5-3

Gravitational Potential Energy

• If we pick up an apple and put it on the table, we have done work on the apple. We can get that energy back if the apple falls back off the table; in the meantime, we say the energy is stored as potential energy.

The work done by a conservative force is equal to the negative of the change in potential energy.

When we lift the apple, gravity does negative work on the apple. When the apple falls, gravity does positive work.

U



mgh

Units: Joules (J) h is measured above an arbitrary 0 -- usually the lowest point in the problem

Elastic Potential Energy

• If we compress a ball bearing against a spring in a projectile gun, we have done work on the ball bearing and spring. We can get that energy back if the spring is released; in the meantime, we say the energy is stored as potential energy.

U

 1 2

kx

2 Units: Joules (J) x is the amount of stretch or compression beyond equilibrium length

Conservative Forces and Potential Energy

A potential energy can be associated with

any conservative force

.

U f



U i

 

W i c



f

) Gravitation: 

U g

=

U f



U i

 

W

grav (

i



f

) 

mgy f



mgy i

Spring: 

U s



U f



U i

 

W i

sp ( 

f

)   ( 1 2

kx f

2  1 2

kx i

2 ) Both are

location-dependent

and

reversible

potential energies.

Note that friction is

not

a conservative force and is

irreversible

.

Conservation of Energy • When there is no work done on a system by non-conservative forces (such as friction), the total mechanical energy of a system remains constant.

U

1 

K

1 

U

2 

K

2

Preflight 2

Imagine that you are comparing three different ways of having a ball move down through the same height. In which case does the ball reach the bottom with the highest speed?

20% 0% 20% 60% 1. Dropping 2. Slide on ramp (no friction) 3. Swinging down 4. All the same correct 1 Conservation of Energy K initial + U initial = K final +U final 0 + mgh = ½ m v 2 final + 0 v final = sqrt(2 g h) 2 3

56.

••

IP

At the local playground a child on a swing has a speed of 2.12 m/s when the swing is at its lowest point.

(a)

To what maximum vertical height does the child rise, assuming he sits still and “coasts”?

(b)

How do your results change if the initial speed of the child is halved?

E K i i

 

E U i f

0 

K f

0 

U f

1 2

mv

2

h



h



v

2 2

g v

2 2

g



mgh

   2.12

m s

2   1.06

m s

2

m s

2 

m s

2  0.23

m

 0.057

m

h

Slide ACT

A small child slides down four frictionless sliding boards. Which relation below describes the relative magnitudes of her speeds at the bottom?

a) v C >v A >v B >v D c) v b) v A >v B =v C >v D A =v B =v C =v d) v A D

Sled Energy

Christine runs forward with her sled at 2.0 m/s. She hops onto the sled at the top of a 5.0 m high, very slippery slope.

What is her speed at the bottom?

K 1 + U g1 = K 0 + U g0 ½ mv 1 2 +mgy 1 v 1 = [v 0 2 = ½ mv 0 2 +mgy 0 + 2gy 1 ] ½ = [(2.0 m/s) 2 +2(9.80 m/s 2 )(5.0 m)] ½ = 10.1 m/s

Energy Cons. ACT

What is the speed in the second situation?

v = 3 m/s

57.

•• The water slide shown in

Figure 8 –23

ends at a height of 1.50 m above the pool. If the person starts from rest at point A and lands in the water at point B, what is the height

h

of the water slide? (Assume the water slide is frictionless.) Use kinematics to determine speed at the bottom of the slide.

t

time to fall 1.50 m:  2

g y



y



v t y

0  1 2

gt

2 

m

9.8

m s

2  0.55

s

Use the time to find the v x as she leaves slide

v x t

2.50

m

0.553

s

 4.52

m s

Now use energy conservation to find h

E top



E bottom h



mgh v

2 2

g

  1

mv

2  4.52

2

m m s s

2 2   1.04

m

58.•• If the height of the water slide in Figure 8–23 is h=3.2 m and the person’s initial speed at point A is 0.54 m/s, at what location does the swimmer splash down in the pool?

Use energy conservation to find the speed x at the bottom of the slide.

E top



E bottom U top



K top mgh v bottom

 1  2

v bottom

  0

U bottom



K bottom

2

mv top

2  1 2  2

gh v top

2

mv bottom



m

 2 3.2

s



m

  0.54

m

From previous problem – time to fall 1.5 m

s

2 

x x t

  

v t x g

2

y

 

m

 9.8

m s

2  7.94

m s

 0.55

s

  0.55

s

 4.37

m v bottom

 7.94

/