# Conservative and Nonconservative Forces

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```Conservative and Nonconservative Forces
Work done by gravity on a closed path is zero:
1
Conservative and Nonconservative Forces
Work done by friction on a closed path is not
zero:
2
Vacuum
Ball is dropped from rest at a height h and experiences only a gravitational force. Gravity
does positive work transforming Potential Energy into Kinetic Energy
vi = 0
mg
h
h/2 mg
h/4
mg
3
Canola Oil
4
The oil produces a constant non-conservative force FNC against
the ball opposite to the direction of motion
vi = 0
mg
FNC
h
FNC
mg
h/2
h/4
mg
If FNC=mg, the forces balances and acceleration ceases,
and the ball moves downward at constant speed
Change in potential energy implies motion as conservative
forces depend only on position
∆U
Work Done=Energy Flow
∆KE
∆U
Work Done=Energy Flow
∆KE
WNC
5
While traveling on Delaware Avenue at 25 mph during a snowy day, you approach an intersection with a red light. It is known the coefficient of friction for your car tires is 0.20 in these conditions. What is the minimum distance to stop the car? Flat road d
f
distance to stop the car
frictional force between tires and snowy road
6
Two blocks are connected by a string that
passes over a pulley . The block of mass m1
lies on a horizontal surface and is connected to
a spring with a force constant k. The system is
released from rest when the spring is
unstretched.
y
h
If the hanging block of m2 falls a distance h
before coming to rest, calculate the coefficient
of kinetic friction μk between the block of m1
and the surface.
Initial and final configuration is at rest therefore Two forms of potential energy: gravitation and the spring.
Initial condition for the spring: unstretched at x=0
Initial PE for the spring is zero Final condition: block m2 falls a distance h and consequently the spring stretched to x=h. x
KEi = KEf=0
Ui = Ugi + Usi= m2gh + 0
Uf = Ugf + Usf= 0 + &frac12;kh2
7
N
KEi = KEf=0
Ui = Ugi + Usi= m2gh + 0= m2gh
Uf = Ugf + Usf= 0 + &frac12;kh2= &frac12;kh2
y
Total initial mechanical energy
x
Total final mechanical energy
Frictional force for the block on the flat surface is the product of the normal force m1g and the coefficient of friction μk
Friction does negative work on the system
Substituting for Wnc
Solve for μk
8
Clicker
You see a leaf falling to the ground with constant speed. When you first notice it, the leaf has initial total energy PEi + KEi. You watch the leaf until just before it hits the ground, at which point it has final total energy PEf + KEf. How do these total energies compare? a) PEi + KEi &gt; PEf + KEf
b) PEi + KEi = PEf + KEf
c) PEi + KEi &lt; PEf + KEf
d) impossible to tell from the information provided
As the leaf falls, air resistance exerts a force on it opposite to its
direction of motion. This force does negative work, which prevents
the leaf from accelerating. This frictional force is a
nonconservative force, so the leaf loses mechanical energy as it
falls, and its final total mechnical energy is less than its initial total
energy.
9
Potential Energy Curves and Equipotentials
The curve of a hill or a roller coaster is itself
essentially a plot of the gravitational potential
energy:
The potential energy U and kinetic energy K add to the total
energy E0 (dashed line) at all x values. K vanishes at A and D,
which are the turning points of the motion.
11
Contour maps are a 2D form of gravitation potential energy curves:
450
350
250
150
50
Contour maps are lines of equal elevation or potential energy drawn on a map. Each line passes through a potential energy of a given value.
1) Lines may never cross or touch.
2) Energy spacing is the same for adjacent lines If an circular object is placed somewhere on the contour map at rest, it will move toward a lower potential energy line Recall the problem from chapter 6
However what can we say about the
motion after the two separate.?
Consider the following problem:
From Newton’s 3rd law, we know the force that the bow acting exerts on the arrow is
matched by a force in the opposite direction on the bow and the archer. This force causes
the archer to slide backwards with the speed requested by the above problem.
The acceleration, forces or bow energy are not given so we are unable to solve this
problem using models we have learned so far.
In an isolated system of two particles, the only force on one particle is that from the
other particle.
Newton’s third law action-reaction pair
Newton’s third law action-reaction pair
Newton’s 2nd law states there
must be an acceleration
associated with the forces
Replace the acceleration
with time rate of change of
the velocity
If mA and mm are constant, we can then rewrite the equation The time rate change of the sum =0, therefore it must be constant
= constant
What is this quantity??
Linear momentum of a particle or an object of mass m with
a velocity is defined as the product of the mass and velocity
Like velocity, momentum is a vector quantity, possessing
a direction as well as a magnitude.
We can now solve this problem
Final and Initial total momentum of archer and arrow must equal Pi=Pf
Initial conditions: Vmi =0 and VAi =0 Solve for the archer’s final velocity
The net force acting on an object is the
rate of change of its momentum:
The General form of Newton’s 2nd law
If the net force is zero, the total momentum does not change: Newton’s 1st Law
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