Transcript Ch 5 Powerpoint - Mr. Ward`s Physics Page

5.2 Kinetic Energy and the Work-Kinetic
Energy Theorem
d
F
Vi
F
Vf = V
Given Wnet we can solve for an object’s
change in velocity
d
F
vi
Wnet = F d
Wnet = (m a) d
Wnet = m (v2 - vi2) / 2
Wnet = 1/2 m v2 - 1/2 m vi2
F
vf = v
v2 = vi2 + 2 a d
a d = (v2 - vi2) / 2
Kinetic Energy = KE = (1/2) m v2
Wnet = KEf – KEi = delta KE
5.2 Kinetic Energy and the Work-Kinetic
Energy Theorem
Objects with KE can do work on other objects
The more KE object A has, the more work it
can do on object B
Remember, b/c Newton’s 3rd law, when object
A does work on object B, object B also does
work on object A
Checking for comprehension
Two football players are going to sack the QB.
Player 1 has a mass of m and has a velocity of v.
Player 2 has a mass of ½ m and has a velocity of
2 v. Who is more likely to cause serious
injuries to the QB? Why?
Ans: the less massive guy b/c…. He has more KE
Example Problem:
A 1400 kg car has a net forward force of 4500
N applied to it. The car starts from rest and
travels down a horizontal highway. What are
its kinetic energy and speed after it has
traveled 100 m? (Ignore the losses in KE
because of friction and air resistance.)
Solve using Work-KE theorem and kinematic
equations
5.3 Potential Energy
Objects with KE can do work on another object
What about objects sitting still?
They have the potential to do work
Therefore they have potential energy
5.3 Potential Energy
Gravitational potential energy: the energy that
an object has as a result of its position
PE = mgh
Wg = mghi mgh
Wg =f PEi – PEf = - delta PE
1 kg
1m
1m
Mr Gener: “What is
the PE of the
block?”
You: “Depends…
Where does h = 0?”
Because location of h = 0 point is arbitrary, we
need only to concern ourselves with the
difference in PE between two locations
Example: A 60.0 kg round
guy is at the top of a slope.
At the initial point A the
round guy is 10.0 m
vertically above point B.
a) Setting h = 0 at B, solve for the PE of the
guy at A then B and then find the difference
b) Setting h = 0 at A, solve for the PE of the guy
at A then B and then find the difference
Ans: delta PE = 5880 J.. Regardless of where h = 0
Homework:
Sections 5.2 – 5.3
Page 142, Problems 9, 11, 14, 21
5.4 Conservative and Nonconservative Forces
Conservative force: the work it does on a object
moving between two points is independent of the
path the object takes between the two points
Example: gravity
A force is nonconservative if it leads to the
dissipation of KE or PE (mechanical energy)
Example: friction
5.5 Conservation of Mechanical Energy
The total mechanical energy in any isolated
system remains constant if the objects interact
only through conservative forces
Wnet = Wc = Wg = PEi - PEf
Wnet = KEf - KEi
KEf - KEi = PEi - PEf
KEi + PEi = KEf + PEf
½ mvi2 + mghi = ½ mvf2 + mghf
Checking for comprehension:
Block 1
2vi
Block 2 v
i
h
h
Two blocks, of the same mass, are 2 h
above the ground. Block 1 has an initial
velocity of 2vi downward and block 2 has
an initial velocity of vi upward
a) Which block has more mechanical energy? 1
b) How would your answer change if block 1
started off only h above the ground? Not enough info
Example 5.5: A sled and its rider together weight
800 N. They move down a frictionless hill through
a distance of 10.0 m. Assuming the initial speed of
the rider-sled system is 5.00 m/s down the hill,
solve for the speed of the rider-sled system at the
bottom of the hill?
vi = 5.00 m/s
10 m
Homework:
Section 5.4 – 5.5
Page 142-143, Problems 23, 25, 27, 29
5.6 Nonconservative Forces and the Work-KE
Theorem
Remember
Wnet = KEf - KEi
Wnet = Wc+Wnc
Wc+Wnc = KEf - KEi
Wc = PEi - PEf
Therefore: Wnc = KEf – KEi – (PEi – PEf)
Wnc = (KEf + PEf) – (KEi + PEi)
5.6 Nonconservative Forces and the Work-KE
Theorem
Wnc = (KEf + PEf) – (KEi + PEi)
The work done by all nonconservative forces
equals the change in the mechanical energy of the
system
The mechanical energy lost from the system is
transformed into another form of energy such as
sound, light or heat
Example Problem:
A 3.00 kg create slides down a ramp. The ramp
is 1.00 m long and inclined at an angle of 30.0o.
The crate starts from rest at the top of the ramp,
experiences a constant frictional force of
magnitude 5.00 N, and continues to move a short
distance on the flat floor. Use energy methods to
determine the speed of the crate when it reaches
the bottom of the ramp.
What happened to
the energy lost due
1m
to friction?
30o
Things to remember:
ΣW = (F d) if F and d are parallel
KE = ½ mv2
PE = mgh
For conservative forces: KEi + PEi = KEf + PEf
Wnet = Wc + Wnc = ½ mvf2 - ½ mvi2
If mechanical energy is lost, the lost energy
must have been transformed into another form
of energy such as heat, sound or light
5.8 Power
Power is defined as the time rate of energy transfer
Powerave = W / (Δt)
= (F Δx) / Δt
= F (Δx / Δt)
=Fv
Units for Power is J/s or watts
Example problem:
An elevator of mass M carries a max load of mass m.
A constant frictional force of Ff retards its motions
upwards. What is the minimum power that the motor
must deliver to lift the fully loaded elevator at a
constant speed of v?
a = 0 m/s2
Ft – Ff – (M+m)g = 0
Ft
v
mg
Ff
Ft = Ff + (M+m)g
P = F v = Ft v
= [Ff + (M+m)g]v
Homework:
Section 5.6 – 5.8
Page 144-145, Problems 33, 37, 39, 41, 43