Transcript Work and Kinetic Energy Serway (7.1 – 7.3)

Kinetic Energy and Work
Physics 1D03 - Lecture 19
Kinetic Energy
Definition: for a particle moving with speed v, the kinetic
energy is
K = ½ mv2
(a SCALAR quantity)
Then the Work-Energy Theorem says:
The total work done by all external forces acting on a
particle is equal to the increase in its kinetic energy.
W = ΔK
= Kf – Ki
Physics 1D03 - Lecture 19
• Kinetic Energy is measured in joules (1J=1N·m).
• Kinetic energy is a scalar; the work-energy theorem is a
scalar relation.
• This theorem is equivalent to Newton’s Second Law. In
principle, either method can be used for any problem in
particle dynamics.
Physics 1D03 - Lecture 19
How to deal with friction
If there is friction in the system, then:
ΔK=Wf
= -ffd
Since
ΔK = Kf - Ki = -ffd
Therefore Kf = Ki - ffd
Physics 1D03 - Lecture 19
Example 1
A bartender slides a 1-kg glass 3 m along the bar to a
customer. The glass is moving at 4 m/s when the
bartender lets go, and at 2 m/s when the customer
catches it.
a) Find the work done by friction
b) Calculate the force of friction.
Physics 1D03 - Lecture 19
Example 2
A 6.0 kg block initially at rest is pulled to the right for
3.0m with a force of 12N over a surface.
Determine its final velocity if:
a) the surface has no friction
b) the surface has a coefficient of kinetic friction of 0.15
How else could we solve this problem ????
Try it !!!
Physics 1D03 - Lecture 19
Solution
Physics 1D03 - Lecture 19
Quiz
A mass is attached to a horizontal spring and rests on
a frictionless table. Starting from the unstretched
position at x=0, the spring is displaced by x=A.
When does the mass have the highest speed?
a)
b)
c)
d)
when you let it go at x=A
when it goes through x=0
when it gets to x=-A, on the other side of x=0
it always has the same speed
Physics 1D03 - Lecture 19
Example 3
A block of mass 1.6kg resting on a frictionless surface is
attached to a horizontal spring with a spring constant
k=1.0x103 N/m (for a spring, E= ½ kx2). The spring is
compressed to 2.0cm and released from rest.
a) Calculate the speed of the block as it passes the x=0
point.
b) Calculate the block’s speed at the x=1.0 cm point.
c) Calculate the block’s speed the first time is passes
though the x=0cm point if there is a constant frictional
force of 4.0 N.
Physics 1D03 - Lecture 19
Solution
Physics 1D03 - Lecture 19
Example 4
You drop a rock off the top of the CN Tower
(h=553.33m).
Use the energy-work theorem to
determine the rock’s speed as it hits the ground below.
Physics 1D03 - Lecture 19
Quiz
Your friends at the International Space Station (orbiting
at 350 km above the Earth’s surface) were tired of you,
and pushed you out of an air lock. Assuming negligible
initial speed, if the Earth did not have an atmosphere,
how fast would you hit the ground:
A) 86 m/s
B) 2620 m/s
C) 6860 m/s
D) depends on the direction you take (straight down, or
at an angle)
Physics 1D03 - Lecture 19
10 min rest
Physics 1D03 - Lecture 19
Potential Energy
•
•
•
Work and potential energy
Conservative and non-conservative forces
Gravitational and elastic potential energy
Physics 1D03 - Lecture 19
mg
Gravitational Work
To lift the block to a height y
requires work (by FP :)
FP = mg
WP = FPy
= mgy
When the block is lowered,
gravity does work:
y
mg
Wg1 = mg.s1 = mgy
or, taking a different route:
y
Wg2 = mg.s2 = mgy
s1
s2
Physics 1D03 - Lecture 19
Work done (against gravity) to lift the box is “stored” as
gravitational potential energy Ug:
Ug =(weight) x (height) = mgy
(uniform g)
When a block moves up, work done by gravity is
negative (decrease speed)
When a block moves down, work done by gravity is
positive (increase speed)
• The position where Ug = 0 is arbitrary.
• Ug is a function of position only. (It depends only on the
relative positions of the earth and the block.)
• The work Wg depends only on the initial and final heights,
NOT on the path.
Physics 1D03 - Lecture 19
Example
• A rock of mass 1kg is released from rest from a 10m
tall building. What is its speed as it hits the ground ?
• The same rock is thrown with a velocity of 10m/s at
an angle of 45o above the horizontal. What is its
speed as it hits the ground.
Physics 1D03 - Lecture 19
Example
• What minimum speed does a 100g puck need to
make it to the top of a 3.0m long 200 frictionless
ramp?
Physics 1D03 - Lecture 19
Conservative Forces
path 1
A force is called “conservative”
if the work done (in going from A
to B) is the same for all paths
from A to B.
B
A
path 2
W1 = W2
An equivalent definition:
For a conservative force, the
work done on any closed path
is zero.
Total work is zero.
Physics 1D03 - Lecture 19
Quiz
The diagram at right shows a
force which varies with position.
Is this a conservative force?
a) Yes.
b) No.
c) Maybe, maybe not.
Physics 1D03 - Lecture 19
Quiz
The diagram at right shows a
force which varies with position.
Is this a conservative force?
a) Yes.
b) No.
c) Maybe, maybe not.
Physics 1D03 - Lecture 19
For every conservative force, we can define a potential energy
function U so that
WAB = -DU = UA -UB
Note the negative
Since: W = ∆K = - ∆U :
Kinetic ↔ Potential
Examples:
Gravity (uniform g) : Ug = mgy, where y is height
Gravity (exact, for two particles, a distance r apart):
Ug = - GMm/r, where M and m are the masses
Ideal spring: Us = ½ kx2, where x is the stretch
Electrostatic forces (we’ll do this in January)
Physics 1D03 - Lecture 19
Non-conservative forces:
• friction
• drag forces in fluids (e.g., air resistance)
Friction forces are always opposite to v (the direction
of f changes as v changes). Work done to overcome friction is
not stored as potential energy, but converted to thermal energy.
Physics 1D03 - Lecture 19
Conservation of mechanical energy
If only conservative forces do work,
potential energy is converted into kinetic
energy or vice versa, leaving the total
constant. Define the mechanical energy E
as the sum of kinetic and potential energy:
E  K + U = K + Ug + Us + ...
Conservative forces only: W = -DU
Work-energy theorem:
W = DK
So, DK+DU = 0; which means that E
does not change with time:
dE/dt = 0
Physics 1D03 - Lecture 19
Example: Atwood’s Machine
An Atwood's machine supports masses m1=0.205 kg and
m2=0.292 kg. The masses are held at rest beside each
other and then released. Once released the 0.292 kg
mass accelerates downward.
Neglecting friction, what is the
speed of the masses the instant
each has moved through 0.424 m?
Physics 1D03 - Lecture 19
Example: Pendulum
L
The pendulum is released from
rest with the string horizontal.
a) Find the speed at the lowest
point (in terms of the length L
of the string).
vf
Physics 1D03 - Lecture 19
Example: Pendulum
The pendulum is released from
rest at an angle θ to the
vertical.
a) Find the speed at the lowest
point (in terms of the length L
of the string).
θ
vf
Physics 1D03 - Lecture 19
10 min rest
Physics 1D03 - Lecture 19
Energy Examples
Physics 1D03 - Lecture 19
Example
• Tarzan (mass 90kg) swings on a 10m long rope off
the top of BSB to save a student (mass 60kg) from
falling into a construction hole on campus. If Tarzan
starts with the rope in a horizontal position and picks
up the student at the bottom of the swing, determine
how high they will go.
Physics 1D03 - Lecture 19
Example
• In a ballistic pendulum a bullet is shot into a block on
a string. If the block and bullet swing up by a vertical
distance of h, determine the speed of the bullet.
Physics 1D03 - Lecture 19
Example
• A pendulum of length 1m and mass of 100g is released
from an angle 30o. At the bottom of the swing hits a
spring of spring constant k=10N/m. Determine the
maximum compression of the spring.
Physics 1D03 - Lecture 19