Transcript Physics Slide Show - Eastern Illinois University

Work & Energy
Energy is Conserved
• Energy is “Conserved” meaning it
can not be created nor destroyed
– Can change form
– Can be transferred
• Total Energy does not change with
time.
• This is a BIG deal!
Energy
• Forms
–
–
–
–
Kinetic Energy
Potential Energy
Heat
Mass (E=mc2)
Motion (Today)
Stored (Wednesday)
later
1161
• Units Joules = kg m2 / s2
Work
The definition of work, when the force is
parallel to the displacement:
(7-1)
SI work unit: newton-meter (N·m) = joule, J
Typical Work
Positive & Negative Work
Work done may be positive, zero, or negative,
depending on the angle between the force and the
displacement:
Force at an Angle Work
If the force is at an angle to the displacement:
(7-3)
Only the horizontal component of the force
does any work (horizontal displacement).
Work by Constant Force at Angle
• You pull a 30 N chest 5 meters across the floor
at a constant speed by applying a force of 50 N
at an angle of 30 degrees. How much work is
done by the 50 N force?
W = T cos q Dx
= (50 N) cos (30) (5 m)
= 217 Joules
50 N
30
f
N
Ty
mg
T
Tx
Where did the energy go?
• You pull a 30 N chest 5 meters across the floor at a
constant speed, by applying a force of 50 N at an angle
of 30o.
• How much work did gravity do?
W = Fcos q Dx
Dx
= 30 cos(90) x 5
90
=0
mg
• How much work did friction do?
T
N
X-Direction: SF = ma
T cos(30) – f = 0
f = T cos(30) = 43.3 N
W = f cos(180) Dx
= (43.3 N) cos(180) x 5
= -217 Joules
f
mg
Dr
f
180
Perpendicular Force and Work
A car is traveling on a curved highway.
The force due to friction fs points
toward the center of the circular
path.
How much work does the frictional
force do on the car?
Zero!
General Result: A force that is
everywhere perpendicular to the
motion does no work.
Work Done by Gravity
In lifting an object of weight mg by a height h, the
person doing the lifting does an amount of work W = mgh.
If the object is subsequently allowed to fall a distance h,
gravity does work W = mgh on the object.
W = mgh
Work By Gravity
When positive work is done on an object, its
speed increases; when negative work is done, its
speed decreases.
Kinetic Energy: Motion
• Apply constant net force along xdirection to an object m.
W = Fx Dx
= m a Dx
= ½ m (v2 – v02)
v 2 = v02  2a( x - x0 )
1 2 2
a  Dx = ( v - v0 )
2
• Work changes ½ m v2
• Voila – new concept:
• Define Kinetic Energy K = ½ m v2
W=DK
Work-Energy Theorem
Work-Energy Theorem: The total work
done on an object is equal to its change in
kinetic energy.
(7-7)
Falling Ball Example
A ball falls a distance 5 meters, what is
final speed?
Only force/work done by gravity
SW = DKE
Wg = ½ m(vf2 – vi2)
Fg h = ½m vf2
mgh = ½m vf2
Vf = sqrt( 2 g h ) = 10 m/s
mg
Work: Energy Transfer due to
Force
• Force to lift trunk at constant
speed
– Case a Ta – mg = 0 T = mg
– Case b 2Tb - mg =0 or T = ½ mg
• But in case b, trunk only moves ½
distance you pull rope.
• F * distance is same in both!
Ta
mg
Tb Tb
mg
Loading with a Crane
A 3,000 kg truck is to be loaded onto a ship by a
crane that exerts an upward force of 31 kN on
the truck. This force, which is large enough to
overcome the gravitational force and keep the
truck moving upward, is applied over a distance of
2.0 m.
(a) Find the work done on the truck by the crane.
(b) Find the work done on the truck by gravity.
(c) Find the net work done on the truck.
Wapp = Fapp y Dy = (31 kN)(2.0 m) = 62 kJ
Wg = mg y Dy = (3000 kg)(-9.81 m/s2 )(2.0 m) = -58.9 kJ
Wnet = Wapp  Wg = (62.0 kJ)  (-58.9 kJ) = 3.1 kJ
Preflight 1
FN
You are towing a car up a hill with constant velocity.
The work done on the car by the normal force is: T
38% 1. positive
13% 2. negative
50% 3. zero
correct
V
W
“work done by normal force is 0 because q is 90
and cos 90 = 0”
28
Preflight 2
FN
You are towing a car up a hill with constant velocity.
The work done on the car by the gravitational forceT
is:
0% 1. positive
63% 2. negative
37% 3. zero
correct
V
W
The work done on the car by the gravitational force is
negative because it hinders motion up the hill
Preflight 3
FN
You are towing a car up a hill with constant velocity.
The work done on the car by the tension force is:
T
63% 1. positive
25% 2. negative
12% 3. zero
V
correct
W
“work done by tension is positive because the angle of
the force is between 0 and 90”
KE ACTS
Car 1 has twice the mass of Car 2, but
they both have the same kinetic energy.
If the speed of Car 1 is v, approximately
what is the speed of Car 2?
a) 0.50 v
b) 0.707 v
1
1
2
m1v1 = m2 v22
2
2
m1 = 2m2
2m2 v12 = m2 v22
2v12 = v22
v2 = 2  v1
c) v
d) 1.414 v
e) 2.00 v
Preflight 4
FN
You are towing a car up a hill with constant velocity.
The total work done on the car by all forces is:
T
63% 1. positive
12% 2. negative
25% 3. zero
correct
V
W
The total work done is positive because the car is
moving up the hill. (Not quite!)
Total work done on the car is zero because the forces
cancel each other out.
Block w/ friction
A block is sliding on a surface with an initial speed of 5
m/s. If the coefficent of kinetic friction between the
block and table is 0.4, how far does the block travel
before stopping?
y
Y direction: SF=ma
N-mg = 0
N = mg
Work
WN = 0
Wmg = 0
Wf = f Dx cos(180)
= -mmg Dx
5 m/s
N
f
x
mg
W=DK
-mmg Dx = ½ m (vf2 – v02)
-mg Dx = ½ (0 – v02)
mg Dx = ½ v02
Dx = ½ v02 / mg
= 3.1 meters
Summary
• Energy is Conserved
• Work = transfer of energy using
force
– Can be positive, negative or zero
– W = F d cos(q)
• Kinetic Energy (Motion)
– K = ½ m v2
• Work = Change in Kinetic Energy
 S W = DK