Transcript Work and Energy - Nutley Schools

Work and Energy
Potential Energy
Work is the measure of energy transfer .
Objectives: After completing this unit,
you should be able to:
• Define kinetic energy and potential energy,
along with the appropriate units.
• Describe the relationship between work and
kinetic energy, and apply the WORK-ENERGY
THEOREM.
• Define and apply the concept of POWER.
Energy
Energy is anything that can be converted
into work; i.e., anything that can exert a
force through a distance. The SI unit for
energy is the joule (J)
Energy is the capability for doing work.
Potential Energy
Potential Energy: Ability to do work by
virtue of position or condition.
The SI unit for energy is the joule (J)
 An elevated object. U  mgh
h  elevation above an arbitrary point
1 2
 A stretched or compressed spring. U  kx
2
Potential Energy
Potential energy is measured as a
difference from an arbitrarily chosen
reference point
Example Problem: What is the potential energy of
a 50-kg person in a skyscraper if he is 480 m
above the street below?
Gravitational Potential Energy
What is the P.E. of a 50-kg
person at a height of 480 m?
U = mgh = (50 kg)(9.8 m/s2)(480 m)
U = 235 kJ
Kinetic Energy
Kinetic Energy: Ability to do work by
virtue of motion. (Mass with velocity)
Examples –
- A speeding car
- A bullet
For an object with mass m and moving at velocity v
1 2
Kinetic energy = K = mv
2
Examples of Kinetic Energy
What is the kinetic energy of a 5-g bullet
traveling at 200 m/s?
5g
200 m/s
K  mv 1 (0.005
kg)(200
m/s)
m

1
2
2
K
1
2
2
2
(0.005kg )  200   1.00 x102 J
2
s

K  12 mv 2  12 (1000 kg)(14.1 m/s) 2
What is the kinetic energy of a 1000-kg
car traveling at 14.1 m/s?
2
1
m

K  (1000kg ) 14.1   9.94 x104 J
2
s

Work and Kinetic Energy
A force acting on an object changes its
velocity, and does work on that object.
vi
m
vf
d
F
F
m
a
W ork  Fx  ( ma ) x ;
Work  mv  mv
1
2
2
f
1
2
v v
2
i
2
f
2
0
2x
The Work-Energy Theorem
Work is equal
to the change
in ½mv2
Work  mv  mv
1
2
2
f
1
2
2
i
If we define kinetic energy as ½mv2 then we
can state a very important physical principle:
The Work-Energy Theorem: The work
done by a force is equal to the change in
kinetic energy that it produces.
Example 1: A 20 g projectile strikes a mud bank,
penetrating a distance of 6 cm before stopping. Find
the stopping force F if the entrance velocity is 80 m/s.
Work = ½
0
mvf2 -
6 cm
80 m/s
½
F d= - ½ mvi2
d
mvi2
F=?
F (0.06 m) cos 1800 = - ½ (0.02 kg)(80 m/s)2
F (0.06 m)(-1) = -64 J
F = 1067 N
Work to stop bullet = change in K.E. for bullet
Example 2: A bus slams on brakes to avoid an
accident. The tread marks of the tires are 25m long. If
mk = 0.7, what was the speed before applying brakes?
Work = DK
Work = F(cos q) d
F= mk.FN= mk mg
Work = - mk mg d
-½ mvi 2 = -mk mgd
d=25 m
f
DK = ½ mvf 2 - ½ mvi 2
vi=
vi = 2(0.7)(9.8 m/s2)(25 m)
2mkgd
vi = 18.52 m/s
Summary
Potential Energy: Ability to do work by virtue
1
U

mgh
U

kx
of position or condition.
2
2
Kinetic Energy: Ability to do work by
K  12 mv 2
virtue of motion. (Mass with velocity)
The Work-Energy Theorem: The work done by
a force is equal to the change in kinetic energy
that it produces.
Work = ½ mvf2 - ½ mvi2
Conservation of Energy
Law of conservation of energy – the total
energy of a system remains unchanged
unless an outside force does work on the
system, or the system does work on
something outside the system.
Conservation of Energy
The energy of a system can interconvert
between potential and kinetic energy, but
the sum of the two will always remain
constant.
Conservation of Energy
U  mgh
E U  K
1 2
K  mv
2
Conservation of Energy
If the ball falls straight down or slides down an incline without friction
E total  E  mgh before the fall or the slide begins. Where h is the initial height of the ball.
1
2
E total  E  mgh ' + m  v ' 
2
Where the primed quantities represent the height and speed of the ball
at any point during the fall or slide.
v
max
 2 gh Where v max is the velocity of the ball immediately before it hits zero height.
Conservation of Energy
Sample problem1 : A 5.0 kg ball is dropped from a
height of 10.0 meters.
a. What is its velocity when it is 8.0 meters high?
b. What is its velocity immediately prior to hitting
the ground?
Conservation of Energy
Solution:
a. Before it is dropped the ball only has
gravitational potential energy since there is no motion.
Therefore:
E=U+K=U+0=mgh=(10)(5)(9.81)= 490.5J=U
Conservation of Energy
Solution (continued):
a. (continued). At 8.0 meters the ball’s potential energy
is U=(8)(5)(9.81)=392.4J
Since its total energy is conserved,
E=U+K and K=E-U
K=490.5-392.4=98.1J
1 2
2K
but K  mv so v 

2
m
 2  98.1  6.26 m
5.0
s
Conservation of Energy
Solution:
b. Right before the ball hits, h=0, U=0,
and E=K+0=490.5J
2K
v

m
 2  490.5  14.0 m 
5.0
s
2 gh
Conservation of Energy
Pendulum – the energy of a pendulum is also
a result of work done against gravity. The pendulum
is at equilibrium when the bob hangs straight down.
As the bob is moved from equilibrium it rises and
acquires potential energy.
Conservation of Energy
Conservation of Energy
Pendulum – the potential energy of a pendulum
displaced at an angle q from equilibrium is
U=mgh = mgl(1-cos q
Where l is the length of the pendulum
The energy of the system E is
E=mgl(1-cos qmax
Where qmax is the angle at the maximum displacement
from equilibrium.
Conservation of Energy
As the pendulum swings it loses potential energy
and gains kinetic energy. At the bottom of the swing
U=0 and K=E
1 2
U  mgl (1  cos )
K  mv
2
E  U  K  mgl (1  cos max )
Conservation of Energy
Example:
A pendulum is 2.0m long, and has a bob that has a
mass of 2.0 kg. The bob is pulled from the
equilibrium position, so that the angle with the
vertical axis θ=10o, and released.
a. What is the potential energy of the system before
the bob is released?
b. What is the speed of the bob as it goes
through the equilibrium position?
c. What is the speed of the bob when the angle
with the vertical is 5.0o?
Conservation of Energy
Example:
a. What is the potential energy of the system before
the bob is released?
U=mgl(1-cos qmaxE29.8121cos10o)=0.59J
Conservation of Energy
Example:
b. What is the speed of the bob as it goes
through the equilibrium position?
At equilibrium, U=0, and E=K
1 2
2K
2(0.59)
m
K  mv 
v
 0.768
2
m
2
s
Conservation of Energy
Example:
c. What is the speed of the bob when the angle
with the vertical is 5.0o?
K=E-U
U=mgl(1-cos q29.8121cos5.0o)=0.149J
K=E-U=0.59J-0.149J=.441J
2(0.441)
m
v
 0.664
2
s
Conservation of Energy
Energy of a spring-mass system:
The potential energy of a spring- mass system is equal to the work done on
1 2
the system U= kx
2
where k is the force constant, and x is the displacement from equilibrium
Conservation of Energy
After the mass is released, the spring stretches or
compresses the velocity of the mass changes
and the energy of the system interconverts
between potential and kinetic energy.
The maximum displacement from the
equilibrium position is called the amplitude A.
Conservation of energy
Conservation of Energy
Conservation of Energy
Conservation of Energy
1 2
U  kx
2
1 2
K  mv
2
1 2
E  U  K  kA
2
Conservation of Energy
Example:
A system consists of a spring attached to a 1.0kg mass
that slides on a frictionless horizontal surface.
The spring has a force constant of 200N/m.
a. If the spring is stretched to a distance of 0.20m from
the equilibrium position, what is the potential energy of
the system?
b. What is the speed of the mass when it is going through
the equilibrium position?
c. What is the speed of the mass when it is at a
distance of 0.1 m from the equilibrium position?
Conservation of Energy
Example:
a. If the spring is stretched to a distance of 0.20m from
the equilibrium position, what is the potential energy of
the system?
Potential energy = the work done on the system
1 2 1
2
U  kA  (200) .20   4.0 J  E
2
2
Conservation of Energy
Example:
b. What is the speed of the mass when it is going through
the equilibrium position?
At the equilibrium position, U=0 and K=E
2K
v

m
 2  4   2.83 m
1
s
Conservation of Energy
Example:
c. What is the speed of the mass when it is at a
distance of 0.1 m from the equilibrium position?
At 0.1m from equilibrium E=U+K
1 2 1
2
U  kx   200  0.10   1J
2
2
K  E  U  4  1  3J
2K
v

m
 2  3  2.45 m
1
s
Conservation of Energy
During an elastic collision – when two moving
objects collide, the kinetic energy of the system is the
same before and after the collision.
During an inelastic collision – when two moving
objects collide, the kinetic energy of the system is not
the same before and after the collision.