Conservation of Energy
Questions for this unit should be completed
as progress is made through the material.
Work and Energy (pg 290-294)
By definition,
Energy is the ability to do work.
Basic concept upon which energy is based is work.
Definition of work is based on:
1. work is measurable,
2. work done on an object is to be a measure of the
object’s change in energy, W = ∆E
3. the definition of work should lead to conservation
of energy
4. work as a quantity should be scalar.
First attempt at definition of work,
Work is done when a force moves an object, W = F • d ,
all directions are removed to represent a scalar quantity
Consider a 1.0 kg mass raised 1.0 m in the air from
some reference position.
d = 1.0 m
m = 1.0 kg
a = 9.81 m/s2
W= F•d = m•a•d
= 1.0 kg • 9.81 m/s2 • 1.0 m
= 9.8 J, joule
This definition works when F and d are in the same
direction, but fails when:
1. the object doesn’t move in the direction of the
force,
2. the object is lowered, the object loses potential
energy (Ep); therefore need negative work if W = ∆E
Consider a 70 kg mass on a frictionless surface moved
forward a distance d by a 400 N force pulling at 60O
above the horizontal.
F = 400N
m = 70 kg
This block will not lift
into the air, but move
forward,
θ = 60O
d
FR = 0
the component of F
moving the block
forward is Fx
Fx is determined by Fx = F cos θ
therefore, W becomes
W = Fx • d , Fx is replaced by F cos θ
= F cos θ • d
= F d cos θ
How does the definition of work give
conservation of energy, ∑ ∆E = 0 ?
Consider,
If an object is lifted up, then there is positive
work done on the object, but when the object is
lowered, gravity does negative work of equal
magnitude on the object, for the system
∑ ∆E = +W + -W = 0
The law of conservation of energy
Units for Energy
∆E = W = F • d • cos θ
= N • m = J , joule
One joule is the work done by a 1 N force
that moves an object 1 m in the direction
of the force.
Complete:
Practice Problems pg 294
Graphic Analysis for Work
When F and d are in the same direction, work can be
represented graphically to show positive work
For a constant F over a distance d
F
force
Area = F • d = W
distance
d
When F is in the opposite direction from d, work can be
represented graphically to show negative work
For a constant F over a distance d
force
0
distance
Area = -F • d = -W
-F
d
For a uniform change in force over a
distance d
Ff
force
Area = [(Ff + Fi) / 2] • d
= Fave • d = W
Fi
distance
d
Potential Energy, EP
(pg 295-298)
Potential energy is the ability to do work because of the relative
positions of the interacting bodies in a system and can only be
defined for conservative forces.
Conservative forces are forces where the total work done by that
force around a closed loop is zero
Consider the work done by gravity on a mass as the mass moves.
d2
d3
d1
d4
Fg
Lifting the object up, work done by Fg is
Wg1 = Fg d1 cos 180 , since Fg is down and d1 is up the work is
negative
Carrying the object across at constant height, work done by Fg is
Wg2 = Fg d2 cos 90 = 0
Lowering the object, work done by Fg is
Wg3 = Fg d3 cos 0 , since Fg is down and d3 is down the work is positive
Carrying the object across at constant height to the starting point, work
done by Fg is
Wg4 = Fg d4 cos 90 = 0
Total Wg in the full cycle is 0, therefore gravity is a conservative force
(text pg 319).
Conservative forces include:
i)
ii)
iii)
iv)
v)
gravity, mg∆h
elastic, ½ kx2
electric (Physics 30)
magnetic (Physics 30)
mass energy, E=∆mc2 (Physics 30)
Non-conservative force is friction
Friction always opposes the direction of motion, therefore the work
done by friction, WF may be determined by
WF = FR d cos 180 = negative work
If the object returns to the start position
WF = FR d cos 180 = negative work ,
the total WF for the cycle ≠ 0 , so friction is a non-conservative force
(text pg 320).
Calculating ∆Ep
Change in Gravitational Potential Energy, ∆Epg
The ∆Epg = W done by an external agent in moving the object from
an initial position to a final position (with no overall change in EK)
rest
hf
Wagent
= ∆Epg in no FR and no ∆EK
= FA d cos 0 , but FA = Fg = m g ,
d = ∆h and cos 0 = 1
FA
= m g ∆h
rest
hi
m
= m g hf – m g hi
If calculating a ∆Epg then a reference level is not required, but if
calculating Epg, then must specify a reference position. (pg 297-298)
Consider,
Calculate the Epg of the
building = 5.0 m
ground
m = 4.0 kg
4.0 kg object if the reference
level is:
a) Ground
well = 10.0 m
b) Bottom of well
c) Top of building
a) Epg
=mgh= 0
From the ground Epg is 0.
since h = 0, the object or mass is located at ground level.
b) Epg
=mgh
= 4.0 kg x 9.81 m/s2 x 10.0 m
= 3.9 x 102 J
From the bottom of the well Epg is
3.9 x 102 J relative to the bottom of the
well.
c) Epg
=mgh
= 4.0 kg x 9.81 m/s2 x -5.0 m
= -2.0 x 102 J From the top of the building the Epg
is -2.0 x 102 J
in this case, the work by the agent to move the object, or mass, to a
new position from the reference point is negative.
Complete:
Practice Problems pg 296, 298
Robert Hooke (pg 299)
Studied elastic materials, proposed the question,
“What is the relationship between the extent to
which an elastic material stretches and the applied
force or resulting restoring force toward the
equilibrium position?” Using elastic materials,
Hooke observed:
FA
Beyond the
elastic limit of
the material
Elastic
material
breaks
or
Fg
x
Complete:
Hooke’s Law and an Elastic Body Lab
Potential Energy in an Elastic
Body
Consider a horizontal elastic body, when an elastic body
is displaced from its equilibrium position, the elastic body
gains elastic potential energy, EP. If the equilibrium
position is the reference level where EP = 0, then can
obtain a formula for EP,
x
Fx = k x
equilibrium postion,
EP = 0
EP = ∆E = work, W, done by some agent to reposition the spring into
position x, where W = F x d , but the force used to displace the
spring changes over x, therefore use Fave for F,
Fx
Force, F
0
Position, x
x
Fave = (0 + k x) / 2 , where 0 is the value of F at the equilibrium
position and k x is the value of Fmax at the x position, therefore
EP = Fave x d = ((0 + k x) / 2) x , where d = x , therefore,
elastic EP = ½ k x2
Example. Determine the elastic EP stored
in a spring, which spring constant is 40
N/m if the spring is stretched 6.0 cm from
its equilibrium position.
EP = ½ k x2 = ½ x 40 N/m x (0.060 m)2
= 0.072 J
Elastic potential
stored is 0.072 J.
Elastic Body Problems
Example 1. A vertical spring stretches from 10.0 cm to
15.0 cm when a 250 g mass is attached to the end of the
spring. Determine the elastic potential energy, Epe
stored in the spring.
x = 15.0 cm – 10.0 cm = 5.0 cm
Fg = m g = k x , solve for k
k = m g / x = (0.250 kg • 9.81 m/s2) / 0.050 m = 49.05 N/m
Epe = ½ k x2 = ½ • 49.05 N/m • (0.050 m)2 = 0.0613 J
The elastic potential energy in the spring is 0.0613 J.
Example 2. A vertical spring is 15.0 cm from the
equilibrium position when a 21.0 N force is
applied. If the force is increased to increase the
position to 25.0 cm from the equilibrium position,
determine the change in elastic potential energy,
Epe stored in the spring.
FA = k x , solve for k
k = FA / x = 21.0 N/ 0.150 m = 140 N/m
ΔEpe = Epef - Epei = ½kxf2 - ½kxi2
= ½k(xf2 - xi2)
= ½ • 140 N/m • ((0.250 m)2 - (0.150 m)2 )
= 2.80 J
Complete:
Practice Problems pg 301
Translational Kinetic Energy, EK
(pg 302-304)
Definition: the energy a body has due to its linear
motion.
symbol: EK , scalar quantity
formula: EK = ½ mv2, where v is speed (no direction)
To derive EK = ½ mv2, must remember that EK is an
energy form, hence the unit is J, joules.
Any change in EK is work energy done on an object and
therefore,
EK = W = F • d , but F = m • a , therefore, EK = m • a • d
Know that if vi = 0
i) a = (vf – vi) / t = vf / t
ii) d = ½ (vf - vi) t = ½ vf • t
vf
velocity
(m/s)
Area = d = (vf - vi)/2 • t
0
time (s)
t
Substitute vf / t for a and ½ vf • t for d
into
EK = m • a • d to obtain,
EK = m • vf / t • ½ vf • t , simplify to obtain
EK = ½ mv2
Complete:
Practice Problems pg 303, 304
and
Check and Reflect pg 305
Work-Energy Theorem (pg 306-309)
Changes in EK are related to the net work done by the
Fnet acting on the object. Fnet is how much unbalanced
force is on an object, therefore a net force does one
thing, Fnet accelerates the object to change the object’s
EK
vi
vf
FA
FR
m
Net Work done by net force
Wnet = Fnet • d • cos θ , but θ = 0 and
Fnet = m • a = FA + FR
Wnet = m a d
Know
vf2 = vi2 + 2ad , rearrange for ad
ad = (vf2 – vi2) / 2 , substitute into
Wnet = m a d to obtain
Wnet = m ((vf2 – vi2) / 2)
or
Wnet = ½ m vf2 – ½ m vi2 , useful when an object has
vx≠0
Wnet = ∆ EK , Work-Energy Theorem
Net Work is different from the work performed by individual
forces:
Wnet = Fnet • d • cos θ = Wapplied + Wfriction
Wfriction = Ff • d • cos θ
Wapplied = FA • d • cos θ
Applied forces on an object may cause changes in both
kinetic energy and potential energy; therefore, the workenergy theorem states that work done on a system is equal
to the sum of change in the kinetic and potential energies
of the system
W = ΔEk + Δ Ep ,
The Work-Energy Theorem
Complete:
Practice Problems pg 308
Mechanical (Total) Energy
and Projectiles (text pg 309)
Vertical only
Etotal = Epg = mgh
h
Etotal = Epgx + Ekx
= mghx + ½ mvx2
Projectile problems
require the recognition
of changes in potential
and kinetic energy
based on position.
When FR = 0, the energy
is conserved.
vz
Etotal = Ek = ½ mvz2
Horizontal and General
Projectiles
Etotal = Epg + Ek
= mgh + ½
Etotal = E′pg + Ekx
mv2
= mgh′ + ½ mvx2
vx
Etotal = E′pgz + Ekz
= mghz + ½
h
v
mvz2
h′
Etotal = E′pgz + Ekz
= mghz + ½ mvz2
Etotal = Epg + Ek
hz
= mgh + ½ mv2
h
Etotal = E′k
hz
v
= ½ mv′2
Etotal = E′k
= ½ mv′2
Example 1. A box is dropped from a height of 6.00m. Determine
a) The speed of the box just before impact.
b) The speed of the box 4.00 m above the ground.
Kinematics
a) vi = 0
a = 9.81 m/s2 [dn]
d = 6.00 m [dn]
vf = ?
vf2 = vi2 + 2ad , solve for vf
vf = √ (2ad)
Epg = E′k
mgh = ½ mv′2
v′ = √ (2gh)
= √ (2 (9.81 m/s2 x 6.00 m))
vf = √(2 (9.81 m/s2 [dn]) (6.00 m [dn])
vf = 10.8 m/s [dn]
a)
Energy
∑ E top = ∑ E bottom
Speed is
10.8 m/s.
= 10.8 m/s Speed is
10.8 m/s.
b) vi = 0
a = 9.81 m/s2 [dn]
d = 2.00 m [dn]
vf = ?
vf2
=
v i2
+ 2ad
vf = √ (2ad)
= √ (2 (9.81 m/s2 [dn] •
2.00 m [dn]))
= 6.26 m/s [dn]
Final speed is 6.26 m/s.
b) ∑ E top = ∑ E h = 4.00 m
Epg = E′pg + E′k
mgh = mgh′ + ½ mv′2
gh = gh′ + ½ v′2
v′ = √(2 g (h - h′))
= √ (2 (9.81 m/s2) (6.00 m
– 4.00 m))
= 6.26 m/s
Final speed is 6.26 m/s.
Example 2. A ball is kicked at 20.0 m/s, 60O above the
horizontal, determine how high the ball will rise.
Kinematics
viy = 20.0 m/s sin 60 [up]
ay = 9.81 m/s2 [dn]
vfy = 0 at the top
dy = ?
vf2 = viy2 + 2ad
d = – viy2 / 2a
= - (20.0 m/s sin 60 [up])2 /
2 (-9.81 m/s2 [up])
Energy
∑ E bottom = ∑ E top
Ek = E′k + E′pg
Ek in the x
plane.
½ mv2 = ½ mv′2 + mgh′
h′ = (½ (v2 – v′2)) / g
= (½ ((20.0 m/s)2 –
(20.0 m/s cos 60)2)) / 9.81 m/s2
= 15.3 m
= 15.3 m [up]
The ball will rise 15.3 m.
The ball will rise 15.3 m [up].
Example 3. An arrow is shot from the edge of a
25.0 m cliff. The arrow leaves at 30.0 m/s, 40.0O
above the horizontal. Determine:
a) the impact speed of the arrow
b) how far from the base of the cliff the arrow will land.
a) For impact speed, ∑ Etop = ∑ Ebottom , therefore
Epg + Ek = E′k
mgh + ½ mv2 = ½ mv′2 , m cancels, solve for v′
v′ = √ [2((9.81 m/s2) (25.0 m) + ½ (30.0 m/s)2)]
v′ = 37.3 m/s
The impact speed is 37.3 m/s.
b) use kinematics to determine dx
from y plane determine t
viy = 30.0 m/s sin 40.0 [up]
d = 25.0 m [dn]
a = 9.81 m/s2 [dn]
t=?
d = viyt + ½ at2,
25.0 m [dn] = -30.0 m/s sin 40.0 [dn] t + ½ 9.81 m/s2 [dn] t2
or
0 = 4.904 t2 + -30.0 sin 40.0 t + - 25.0,
use quadratic or combination of kinematics equations to
solve for t
t = [-(-30.0 sin 40.0) ± √ ((-30.0 sin 40.0)2 – (4 • 4.905 •
-25.0))] / (2 • 4.905)
t = 4.959358…s
determine dx by dx = vx t
dx = 30.0 m/s cos 40.0 [fd] • 4.959358…s
dx = 114 m [fd]
The arrow will land
114 m [fd] from the cliff base.
Complete:
Practice Problems pg 309
Check and Reflect pg 310
Conservation of Energy Problems:
Projectiles
and
Conservation of Energy Homework Check
Conservation of Energy
1. Energy can not be created or destroyed.
2. Initial energy must equal final energy in
an isolated system: Em = Em’.
3. The change in energy for an isolated
system must equal zero, ΔEm’ = 0
Law of Conservation of Energy (pg 311-313)
In an isolated system, Etotal is the mechanical energy, Em,
of the system and can be shown as
∑ Em at t = 0 = ∑ Em at t
Epe + Epg + Ek = E′pe + E′pg +E′k (no friction)
Problems
Elastic Bodies
Video
Example 1. A 1.00 kg ball is placed on a vertical spring.
The spring has a spring constant of 710 N/m and is
compressed 2.00 m below the equilibrium position.
Determine
a) the velocity of the ball at the equilibrium position.
b) the maximum height above the equilibrium position
reached
by the ball.
Solution
a) Em= Em‘, therefore Epe + Epg = Ek’
½ kx2 + mgh1 = ½ mv’2 , rearrange for v’
√(2(½ kx2 + mgh1)/m) = v’
√(2(½(710N/m●(2.00m)2)+(1.00kg●9.81m/s2●-2.00m))/1.00kg) = v’
52.92220706 m/s = v’
The speed is 52.9 m/s .
b) Em= Em‘, therefore Epe + Epg = Epg’
½ kx2 + mgh1 = mgh2 , rearrange for h2
(½ kx2 + mgh1)/(mg) = h2
h2 = ((½●710N/m●(2.00m)2) + (1.00kg●9.81m/s2●-2.00m))
/(1.00 kg●9.81m/s2)
142.7502548 m = h2 Maximum height is 143 m.
Does the question change if two springs of equal
length were used, with one spring placed inside
the other spring or placed next to each other in a
parallel arrangement, but with spring constants of
310 N/m and 400 N/m?
Example 2. A 500 g metal cylinder is at the end
of a horizontal spring, which spring constant is
35 N/m. The system is set vibrating with an
amplitude of 30.0 cm. Determine the cylinder’s
speed when the displacement of the metal
cylinder from the equilibrium position is 10.0 cm.
Solution:
Em= Em‘, therefore Epe = Epe’ + Ek’
½ kA2 = ½ kx2 + ½ mv’2 , ½ cancels
kA2 = kx2 + mv’2 , rearrange for v’
kA2 - kx2 = mv’2
√((k (A2 – x2)) / m) = v’
√(35 N/m ((0.30 m)2 – (0.10 m)2) / 0.500 kg) = v’
2.36643… m/s = v’
Speed of the cylinder at x = 10.0 cm is 2.4 m/s.
How would the equation be developed to
solve for position, x, if the maximum speed,
vmax, and speed at position x, vx, were
given?
Complete:
Practice Problems pg 313
and
Conservation of Energy Problems:
Springs
Simple Penduli (pg 314-316)
For pendulum problems, mass and pendulum length are
not required, the height difference between the point of
pendulum bob release and the bottom of the pendulum
swing is needed.
The angle of the pendulum is not
important when energy is conserved ,
Energy total at A = Energy total at B
Position A
Em at A = EP
= mgh
h
Position B
Em at B = EK
= ½ mvinst2
Therefore, EP = EK and
mgh = ½ mv2 , mass cancels to obtain
gh = ½ v2 or v = √(2gh)
Example A pendulum of length 2.00 m was pulled aside
50.0O to vertical. Determine the speed of the pendulum
bob at the bottom of the pendulum swing.
y
50.0O
h
Need to determine y and subtract y from the length of
the pendulum to obtain h.
y = 2.00 m cos 50.0O
h = 2.00 m – ( 2.00 m cos 50.0O) = 0.7144…m
∑ E top = ∑ E bottom
Epg = Ek′
mgh = ½ mv′2 , mass cancels, rearrange for v′
v′ = √ (2gh) = √ (2 • 9.81 m/s2 • 0.7144…m)
= 3.74 m/s
The speed of the pendulum
is 3.74 m/s.
Example 2. A 1.50 m long pendulum is released from a
vertical height of 30.0 cm above the lowest swing
position. Determine the speed of the pendulum bob at
15.0 cm above the lowest swing position.
0.300 m
0.150 m
Complete:
Practice Problems pg 315 and 316,
Conservation of Energy Questions: Penduli,
Pendulum and Energy Lab,
and
Pendulum Projectile Lab
Systems of Bodies
In a system of bodies, assume massless ropes and frictionless
pulleys.
All bodies in the system must move uniformly, same speed for
both bodies at any point. Maximum speed for the system
occurs just before the system strikes the ground.
Assume conservation of energy for all bodies:
∑ E at t = 0 = ∑ E at t
Epg + Ek = E′pg +E′k
mAghA + 1/2mAvA2 + mBghB + 1/2mBvB2 = mAgh’A + 1/2mAv’A2 + mBgh’B + ½ mBv’B2
Example 1. Determine the maximum
speed for the illustrated system, which
begins from rest.
8.4 kg
3.9 m
2.8 kg
ground
∑ E at t = 0 = ∑ E at t
identify all energies in the system at rest and at maximum
speed
Epg8.4 = E’pg2.8 + E’k2.8 + E’k8.4
m8.4gh = m2.8gh + ½m2.8v’2 + ½m8.4v’2
group gravitational potential energies and factor out v’2
gh(m8.4 – m2.8) = v’2(½m2.8 + ½m8.4)
isolate v’2 by moving (½m2.8 + ½m8.4) and then square root
to obtain
( gh (m8.4  m2.8 ))
v'
1
( ( m2.8  m8.4 ))
2
substitute values to solve
(9.81m / s 2 x 3.9m  8.4kg  2.8 kg )
v'
1
( (2.8 kg  8.4 kg ))
2
6.185…. m/s = v’
6.2 m/s = v’
The maximum speed of the system is 6.2 m/s.
Example 2. Determine the speed for the
illustrated system at 1.8 m, which begins
from rest.
8.4 kg
1.8 m
2.8 kg
ground
3.9 m
Complete:
Conservation of Energy Questions:
Systems of Bodies
and
Energy Conservation in a Fletcher’s Trolley
Lab
In a non-isolated system, Etotal is the mechanical
energy, Em, of the system and can be shown as
∑ Em at t = 0 = ∑ Em at t
Epe + Epg + Ek = E′pe + E′pg +E′k + heat
(energy is lost due to friction)
where E′x is the altered quantity of energy for each
energy type at “t” and heat is due to friction
produced through time “t” and the energy lost due
to friction in determined by FR • d, therefore the
equation becomes
Epe + Epg + Ek = E′pe + E′pg +E′k + Ff • d
Note: heat does not appear on the left side of the
equation as “new” heat has not been produced at
the reference point of t = 0 (text pg 320).
Efficiency and Percent Efficiency
The energy retained by a system represents the
efficiency of a system:
Efficiency = energy output/energy input
or
Efficiency = energy retained/initial energy
Efficiency is a decimal less than 1.00; Percent
Efficiency is a percentage less than 100%.
Complete:
Energy Loss Lab
View the website,
https://www.thephysicsaviary.com/Physics/Programs/Labs/find.php.
Select Energy Loss Lab and Begin.
The location (“planet”), object mass and drop height may be
manipulated.
i) Set and record the object mass and drop height.
ii) Drop the object.
iii) Record the final vertical height.
iv) Calculate initial and final gravitational potential energies and
determine the loss of energy.
v) Repeat for a total of 5 replicates.
vi) Manipulate the drop height and repeat four times.
vii) Calculate the efficiency and percent efficiency for each drop
height.
Comment on the energy loss, is there any trend?
Ramps (pg 319-322)
If there is no friction, vf at the bottom of the ramp
is independent of the mass of the object and the
angle of ramp.
All four situations produce the same vf , since
Ff = 0 and vf is related to h.
drop
h
Example 1. A 5.0 kg object begins at rest
and slides down an 8.0 m long ramp, from
a position 2.0 m in height. Determine the
speed of the object at the bottom of the
ramp if
a) Ff = 0
b) Ff = 5.0 N
8.0 m
2.0 m
a) ∑ E top = ∑ E bottom
Epg = Ek′
mgh = ½ mv′2 , mass cancels, rearrange for v′
v′ = √ (2gh) = √ (2 • 9.81 m/s2 • 2.0 m) = 6.3 m/s
The speed of the object is 6.3 m/s.
b) ∑ E top = ∑ E bottom
Epg = Ek′ + heat
mgh = ½ mv′2 + Ffd , mass does not cancel, solve for v′
v′ = √(2(mgh – Ffd) / m)
= √(2(5.0 kg • 9.81 m/s2 • 2.0 m – 5.0 N • 8.0 m) / 5.0 kg)
= 4.8 m/s The speed of the object is 4.8 m/s.
Example 2. A roller coaster car train and contents have a
combined mass of 300 kg. If the roller coaster car train is
moving with constant velocity of 4.00 m/s at point A,
determine:
a) the speed of the roller coaster car train at point B if Ff = 0.
b) the average frictional force if the speed of the roller coaster
car train at point B is 5.0 m/s and the length of the track from
point A to point B is 250 m.
A
B
35.0 m
15.0 m
a) ∑ E at A = ∑ E at B
Epg + Ek = Epg′ + Ek′
mgh + ½ mv2 = mgh′ + ½ mv′2 , mass cancels, solve for v′
v′ = √ (v2 + 2g(h- h′))
= √ ((4.00 m/s)2 + 2 • 9.81 m/s2 (35.0 m – 15.0 m))
= 20.2 m/s The speed is 20.2 m/s.
b) ∑ E at A = ∑ E at B
Epg + Ek = Epg′ + Ek′ + heat
mgh + ½ mv2 = mgh′ + ½ mv′2 + Ffd , solve for Ff
Ff = (mgh - mgh′ + ½ mv2 - ½ mv′2) / d , simplify to
Ff = (mg(h - h′) + ½ m(v2 - v′2)) / d
= (300 kg • 9.81 m/s2 (35.0 m – 15.0 m) + ½ • 300 kg ((4.0 m/s)2
– (5.0 m/s)2)) / 250 m
= 230 N The average frictional force is 230 N.
Loop the Loop (no friction)
If Ff for a track is zero, determine the minimum height
from which an object must start to barely complete travel
through a loop of radius 4.00 m.
Point for minimum v = v′,
Fnet = Fc = Fg
Fc Fg
h
4.00 m
h′
∑ E at A = ∑ E at B
Epg = Epg′ + Ek′
mgh = mgh′ + ½ mv′2 , mass cancels, solve for h
h = (gh′ + ½ v′2) / g , need v′
From circular motion, Fnet = Fc = Fg , therefore
mv2 / r = mg , mass cancels, solve for v , and v = v′
v = √ rg , substitute for v′ to solve for h
h = (gh′ + ½ (√ rg )2) / g , g cancels to obtain
h = h′ + ½ r = 4/2 r + ½ r = 8.00 m + ½ • 4.00 m = 10.0 m
The minimum height is 5/2 r = 10.0 m.
Complete:
Ramps Lab
Conservation of Energy Problems: Ramps
and
Check and Reflect pg 323
Power: rate of work done (pg 324-328)
P=W/t
or
P = change in energy / time
P = (F • d) / t , but d / t = v , then
P=F•v
or by unit analysis
J / s = kg • m / s2 (m) (1/s) = N • (m/s) = F • v
Unit is J/s = Watt
Therefore, change in energy = P x t in W • s or
kW • h
Note: 3.6 x 10 6 W• s = 1 kW• h
Complete:
Practice Problems pg 328
Check and Reflect pg 330
Conservation of Energy Lab/Assignment
Quiz
and
Unit Exam
(Unit Exam Preparation Review pg 332,
333)