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% A.
correct
Dropping
0% B. Slide on ramp (no friction)
0% C. Swinging down
1
70%D. All the same
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
Force
Section
Conservative forces
Gravity
5-6
Spring force
6-2
Nonconservative forces
Friction
6-1
Tension in a rope, cable, etc.
6-2
Forces exerted by a motor
7-4
Forces exerted by muscles
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.
1 2
U  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  U f  Ui  Wc (i  f )
Gravitation: U g =U f Ui  Wgrav (i  f )  mgy f  mgyi
Spring: Us  U f Ui  Wsp (i  f )  ( 12 kx f 2  12 kxi 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.
U1  K1  U 2  K2
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% 1.
Dropping
1
0% 2. Slide on ramp (no friction)
20% 3. Swinging down
correct
60% 4. All the same
Conservation of Energy
Kinitial + Uinitial = Kfinal+Ufinal
0 + mgh = ½ m v2final + 0
vfinal = 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?
Ei  E f
0
0
Ki  U i  K f  U f
1 2
mv  mgh
2

2.12 m

2
h
v2
s  0.23m
h

2 g 2  9.8 m 2
s
2
v
h
2g
1.06 m 

s

2  9.8 m
2
s2
 0.057m
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) vC>vA>vB>vD
c) vA=vB=vC=vD
b) vA>vB=vC>vD
e) vC<vA<vB<vD
d) vA<vB=vC<vD
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?
K1 + Ug1 = K0 + Ug0
½mv12+mgy1 = ½mv02+mgy0
v1 = [v02 + 2gy1]½
= [(2.0 m/s)2+2(9.80 m/s2)(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.
time to fall 1.50 m:
1 2
y  v y 0t  gt
2
2y
2 1.5m
t

 0.55s
m
g
9.8 2
s
Use the time to find the vx as she leaves slide
x 2.50m
vx  
 4.52 m
s
t 0.553s
Now use energy conservation to find h
Etop  Ebottom
1
mgh  mv 2
2
2
v
h
2g
4.52 m 

s

2  9.8 m
2
s2
 1.04m
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?
x
Use energy conservation to find the speed
at the bottom of the slide.
From previous problem – time to fall 1.5 m
Etop  Ebottom
0
t
U top  Ktop  U bottom  Kbottom
1 2
1 2
mgh  mvtop  mvbottom
2
2
2
vbottom  2 gh  vtop

vbottom  2  9.8 m
vbottom  7.94m / s
 3.2m   0.54 m 

s
s
2
2y
2 1.5m

 0.55s
g
9.8 m 2
s

x  vxt  7.94 m
2
x  4.37m
s
  0.55s 