Transcript Work and Energy

Chapter 7.2 and 7.3
Energy
Energy
Energy is anything that can be converted into work; i.e., anything that can
exert a force through a distance.
Mechanic energy = KE + PE
Energy is the capability for doing work.
Energy is conserved. The total energy
remains constant. ME = KE+PE = constant
Potential Energy: Ability to do work
U

mgh
by virtue of position or condition.
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 resultant force is equal to the change in
kinetic energy that it produces.
Work = ½ mvf2 - ½ mvo2
The Work-Energy Theorem
Work is equal
to the change
in ½mv2
Work  mv  mv
1
2
2
f
1
2
2
0
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 resultant force is equal to the
change in kinetic energy that it produces.
Kinetic energy
(energy of motion)
W  F  d  ma  d
 2   0 2  2ad
a
Work-energy
principle
Kinetic energy
  0
2
2
2d
W  m
 2  0 2
2d
W  KE
m 2
KE 
2
m 2 m 0
d 

 KE
2
2
2
Practice quiz (KE)
Q:
The work W accelerate a car
from 0 to 50km/h. How much
work is needed to accelerate
the car from 50 km/h to 150
km/h?
1.
2W
2.
3W
3.
6W
4.
8W
5.
9W
A: Let’s call the two speeds v and
3v, for simplicity. We know that
the work is given by W=KE(2)KE(1)
Potential Energy (depend on position)
Various types of potential energy can be
defined, and each type is associated with a
particular force.
Gravitational potential energy
(the most common example)
A ball held high in the air has potential energy
because of its position relative to the Earth.
W  F  d cos   mg ( y  y0 )  mgh
PE  mgh
Example 2: 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.
6 cm
80
m/s
0
x
2
2
Work = ½ mvf - ½ mvo
F x = - ½ mvo2
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-1
A 1000 kg roller coaster car moves from point 1 to point 2 and then point 3. a)
What is the GPE at 2 and 3 relative to 1? That is, take y=0 at point 1.
(b) What is the change in potential energy when the car goes from point 2 to 3?
Repeat part (a) and (b), but take the reference point (y=0) to be at point 3.
a)
y=0 at point 1,
GPE= mgh(2)= positive
GPE= mgh(3)= negative
point 2
GPE  2.5 105 J
10 m
b) y=0 at point 3,
point1
GPE= mgh(2)= positive
15 m
point 3
GPE= mgh(3)= zero
GPE  2.5 105 J
The change of the potential energy is
negative.
Spring (Hook’s law)
The restoring force of a spring is
Fs  kx
where k is called the spring
constant, and needs to be
measured for each spring.
Fp  kx
Spring energy is one of
the common potential
energy.
It is called also elastic
potential energy (EPE)
EPE
The potential energy stored in a spring is given by
1
2
EPE  k  x
2
which comes from the work done on it by the
average force F, since the force (spring) is not
constant but varies over distance (Note that
potential energy of a spring is always positive.)
Practice quiz (PE)
Q: How does the work required to
stretch a spring 2 cm
compare with the work
required to stretch it 1cm?
1.
Same amount of work
2.
Twice the work
3.
4 times the work
4.
8 times of work
A: 3. EPE depends on the square
of displacement.