Transcript No Slide Title

Chapter (8)
Potential Energy and Conservative Forces
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There are two types of forces:
• conservative (gravity, spring force)
•All microscopic forces are conservative:
•Gravity,
•Electro-Magnetism,
•Weak Nuclear Force,
•Strong Nuclear Force
• nonconservative (friction, tension)
•Macroscopic forces are non-conservative,
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Conservative Forces
A force is conservative if the work it does on an
object moving between two points is independent
of the path taken.
 work done depends only on ri and rf
W PQ (along1) W PQ (along 2)
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 If an object moves in a closed path (ri = rf)
then
total work done by the force is zero.
W PQ (along 1)  W QP (along 2)
W PQ (along 1) W QP (along 2)  0
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Nonconservative Forces
 work done by the force depends on the path
W PQ (along1) W PQ (along 2)
W PQ (along 1)  W QP (along 2)
W PQ (along 1) W QP (along 2)  0
 non-conservative forces dissipate energy
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Work Done by Conservative Forces
Potential Energy: Energy associated with the position of
an object.
For example: When you lift a ball a distance y, gravity
does negative work on the ball. This work can be
recovered as kinetic energy if we let the ball fall. The
energy that was “stored” in the ball is potential energy.
Wc = -DU =-[Ufinal – Uinitial]
Wc = work done by a conservative force
DU = change in potential energy
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Gravitational Potential Energy
Gravitational potential energy
U = mg(y-y0)
U=0 at y=y0
y = height
(e.g. surface of earth).
Work done by gravity:
Wg = mg Dy = mg (y- y0)
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Spring Potential Energy
Uf – Ui =  [Work done by spring on mass]
F=kx
Mass m starts at x=0 (Ui =0) and moves
until spring is stretched to position x.
x
WorkSpring = - ½ kx2
U(x) – 0 =  (1/2 kx2)
USpring(x) = ½ kx2
x = displacement from
equilibrium position
Area in triangle
= kx times increment in x
= Work done by spring
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Conservation of Energy
• Energy is neither created nor destroyed
• The energy of an isolated system of objects
remains constant.
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Mechanical Energy (Conservative Forces)
Mechanical energy E is the sum of the potential
and kinetic energies of an object.
E=U+K
The total mechanical energy in any isolated system
of objects remains constant if the objects interact
only through conservative forces:
E = constant
Ef = Ei  Uf + Kf = Ui+ Ki
DU + DK = DE = 0
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Example: A ball of mass m is dropped from a height h above the
ground, as shown in Figure
(A) Neglecting air resistance, determine the speed of the ball when
it is at a height y above the ground.
Ef  Ei
U f  K f U i  K i
1
1
2
mgy f  mv f  mgy i  mv i2
2
2
1
mgy  mv f2  mgh  0
2
v f  2mg (h  y )
(B) Neglecting air resistance, determine the speed of the ball
when it is at the ground.
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Example: A 0.5 kg block is used to compresses a spring with a
spring constant of 80.0 N/m a distance of 2.0 cm. After the
spring is released, what is the final speed of the block?
Solution
U f  K f U i  K i
1 2 1 2 1 2 1 2
kx f  mv f  kx i  mv i
2
2
2
2
1 2 1 2
0  mv f  kx i  0
2
2
kx
80  (0.02)
vf 

 0.25m / s
m
5
2
i
2
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Example: A particle of mass m = 5.00 kg is released from point A
and slides on the frictionless track shown in Figure.
Determine
(a) the particle’s speed at points B and C and
(b) the net work done by the gravitational force in moving the
particle from A to C.
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(a) the particle’s speed at points B
U B  K B U A  K A
1
1
2
2
mgy B  mv B  mgy A  mv A
2
2
1
5  9.8  3.20   5 v B2  5  9.8  5  0
2
245  156.8
vB 
 5.94m / s
2.5
Find the particle’s speed at points C then find the work from
the relation
W  DU
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Problem: find the particle’s speed at points B, where the particle
released from point A and slides on the frictionless track
Problem: Find the distance x.
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Problem: A 0.2-kg pendulum bob is
swinging back and forth. If the speed
of the bob at its lowest point is 0.65
m/s, how high does the bob go above its
minimum height?
Problem: Two objects are connected by a
light string passing over a light frictionless
pulley as shown in Figure. The object of
mass 5.00 kg is released from rest. Using
the principle of conservation of energy, (a)
determine the speed of the 3.00-kg object
just as the 5.00-kg object hits the ground.
(b) Find the maximum height to which the
3.00-kg object rises.
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Problem: An object of mass m starts from rest and slides a distance
d down a frictionless incline of angle . While sliding, it contacts
an unstressed spring of negligible mass as shown in Figure. The
object slides an additional distance x as it is brought momentarily
to rest by compression of the spring (of force constant k). Find the
initial separation d between object and spring.
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Work Done by Nonconservative
Forces
Nonconservative forces change the amount of
mechanical energy in a system.
DE  0
Ef  Ei  0
DE W nc
Wnc = work done by nonconservative force
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Problem: Children and sled with mass of 50 kg slide down a hill
with a height of 0.46 m. If the sled starts from rest and has a speed
of 2.6 m/s at the bottom, how much thermal energy is lost due to
friction (i.e. what is the work that friction does)? If the hill has an
angle of 20° above the horizontal what was the frictional force.
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(U f  K f )  (U i  K i ) W nc
1
1
2
mghf  mv f  mghi  mv i2 Wnc
2
2
Since vi = 0, and hf = 0,
The force done by friction is determined from;
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Example: A child of mass m rides on an irregularly curved slide
of height h = 2.00 m, as shown in Figure. The child starts from
rest at the top.
(A) Determine his speed at the bottom,
assuming no friction is present.
U f  K f U i  K i
1
2
0  mv f  mgh  0
2
v f  2 gh  6.26m / s
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(B) If a force of kinetic friction acts on the child, how much
mechanical energy does the system lose? Assume that vf =3.00
m/s and m = 20.0 kg.
DE  U f  K f  (U i  K i )
1
DE  0  mv f2  (mgh  0)
2
1
DE  mv f2  mgh
2
1
DE   20  9  20  9.8  2  302J
2
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Example: A block having a mass of 0.80kg is given an initial
velocity vA = 1.2m/s to the right and collides with a spring of
negligible mass and force constant k = 50N/m, as shown in Figure.
(A) Assuming the surface to be
frictionless, calculate the maximum
compression of the spring after the
collision.
EC  E A
1 2
1
kx max  mv A2
2
2
x max
mv

k
2
A
 0.15m
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(B) Suppose a constant force of kinetic friction acts between the
block and the surface, with k = 0.50. If the speed of the block at
the moment it collides with the spring is vA = 1.2 m/s, what is the
maximum compression xC in the spring?
DE  E f  E i
1
1
2
DE  kx C  mv A2 ............(1)
2
2
But
DE  f k x C
DE    k mgx C ......................(2)
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then
1 2 1
2
kx C  mv A  k mgx C  0
2
2
1
1
2
2
 50  x C  (0.8)(1.2)  (0.50)(0.8)(9.8)x C  0
2
2
2
25x C  3.92x C  0.576  0
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Problem: A 5.00-kg block is set into motion up an inclined plane
with an initial speed of 8.00 m/s (Fig. P8.33). The block comes
to rest after traveling 3.00m along the plane, which is inclined
at an angle of 30.0° to the horizontal. For this motion
determine
(a) the change in the block’s kinetic energy,
(b) (b) the change in the potential energy of the block–Earth
system,
(c) the friction force exerted on the block (assumed to be
constant).
(d) What is the coefficient of kinetic friction?
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Problem: A 1300-kg car drives up a 17.0-m hill. During the drive,
two nonconservative forces do work on the car:
(i) the force of friction, and
(ii) the force generated by the car’s engine.
The work done by friction is –3.31  105 J;
the work done by the engine is +6.34  105 J.
Find the change in the car’s kinetic energy from the bottom of the
hill to the top of the hill.
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Problem :A skateboard track has the form of a circular arc with a
4.00 m radius, extending to an angle of 90.0° relative to the vertical
on either side of the lowest point, as shown in the Figure. A 57.0 kg
skateboarder starts from rest at the top of the circular arc. What is
the normal force exerted on the skateboarder at the bottom of the
circular arc? What is wrong with this picture?
At bottom, a = v2/r
Which direction?
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Problem : A 1.9-kg block slides down a frictionless ramp, as shown
in the Figure. The top of the ramp is 1.5 m above the ground; the
bottom of the ramp is h = 0.25 m above the ground. The block
leaves the ramp moving horizontally, and lands a horizontal distance
d away. Find the distance d.
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