Transcript Lecture 14.KineticEn..

Kinetic Energy
Lecturer:
Professor Stephen T. Thornton
Reading Quiz:
Two marbles, one twice as heavy as the other, are
dropped to the ground from the roof of a building.
Just before hitting the ground, the heavier marble
has
A) the same kinetic energy as the lighter one.
B) half as much kinetic energy as the lighter one.
C) twice as much kinetic energy as the lighter one.
D) four times as much kinetic energy as the
lighter one.
Answer: C
The velocities will be the same in this
case, so the only difference in the
kinetic energy is due to the mass.
Because the mass is twice as much,
the kinetic energy is twice as much.
K  1 mv2
2
Last Time
Discussed the concept of work
Today
Discuss kinetic energy
Work-energy theorem
Conceptual Quiz
In a baseball game, the
catcher stops a 90-mph
A) catcher has done positive work
pitch. What can you say
B) catcher has done negative work
about the work done by
C) catcher has done zero work
the catcher on the ball?
Conceptual Quiz
In a baseball game, the
catcher stops a 90-mph
A) catcher has done positive work
pitch. What can you say
B) catcher has done negative work
about the work done by
C) catcher has done zero work
the catcher on the ball?
The force exerted by the catcher is opposite in direction to the
displacement of the ball, so the work is negative. Or using the
definition of work (W = F d cos q ), because q = 180º, then W < 0.
Note that because the work done on the ball is negative, its
speed decreases.
Follow-up: What about the work done by the ball on the catcher?
Previous kinematic equation:
v  v  2ad
2
f
2
i
or rearranging,
2ad  v  v
2
f
2
i
 Fnet 
2
2
2
 d  v f  vi ,
 m 
Fnet d 
1
1
multiply by m / 2
mv  mv
2
2
1 2 1 2
Wnet  Wtotal  mv f  mvi
2
2
2
f
2
i
Definition of kinetic energy
1 2
K  mv
2
Unit of kinetic energy is the joule, J.
Q. Why do we define this?
A. It turns out to make our
calculations easier!
Work-Energy Theorem
Wtotal  K  K f  K i 
The total work done on an
object is equal to the change in
its kinetic energy:
This is a general result,
even for a force not
constant in magnitude and
direction. Do spring
demo.
1
2
mv 
2
f
1
2
2
i
mv
Kinetic Energy and
the Work-Energy Principle
Energy was traditionally defined as
the ability to do work. We now know
that not all forces are able to do work;
however, we are dealing in these
chapters with mechanical energy,
which does follow this definition.
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Because work and kinetic energy can be equated,
they must have the same units: kinetic energy is
measured in joules. Energy can be considered as
the ability to do work:
To find work, we have to be sure about what
force is exerting the effort. Here we might ask
about the work done by friction, gravity, air
resistance, or the engine. Not a good figure.
Forces are not accurate.
We need to make sure we can find the
work done by every force. (sloppy diagram)
Wengine  Fd (don't like)
Wfriction  Ffriction  d   Ffriction d
Wgravity  Fgravity  d   mgd sin q
Wair resis  Fair resis  d   Fair resis d
Doing Work Against Gravity
Energy is reclaimed in this case.
Doing Work Against Friction
Energy is not reclaimed in this case.
Does the Earth do work on the moon?
The Moon revolves around
the Earth in a nearly circular
orbit, with approximately
constant tangential speed,
kept there by the
gravitational force exerted
by the Earth. Gravity does
no work, because the
displacement and force are
perpendicular.
Cable Tension, Work. A 265-kg load is
lifted 23.0 m vertically with an
acceleration a = 0.150g by a single
cable. Determine (a) the tension in the
cable; (b) the net work done on the
load; (c) the work done by the cable on
the load; (d) the work done by gravity
on the load; (e) the final speed of the
load assuming it started from rest.
Crate Friction. A 46.0-kg crate,
starting from rest, is pulled across a
floor with a constant horizontal
force of 225 N. For the first 11.0 m
the floor is frictionless, and for the
next 10.0 m the coefficient of
friction is 0.20. What is the final
speed of the crate after being pulled
these 21.0 m?
Paintball. In the game of paintball, players
use guns powered by pressurized gas to
propel 33-g gel capsules filled with paint at
the opposing team. Game rules dictate that a
paintball cannot leave the barrel of a gun with
a speed greater than 85 m/s. Model the shot
by assuming the pressurized gas applies a
constant force F to a 33-g capsule over the
length of the 32-cm barrel. Determine F (a)
using the work-energy principle, and (b) using
the kinematic equations and Newton’s second
law.
Conceptual Quiz
A child on a skateboard is
moving at a speed of 2 m/s.
After a force acts on the
child, her speed is 3 m/s.
What can you say about
the work done by the
external force on the child?
A) positive work was
done
B) negative work was
done
C) zero work was done
Conceptual Quiz
A child on a skateboard is
moving at a speed of 2 m/s.
After a force acts on the
child, her speed is 3 m/s.
What can you say about
the work done by the
external force on the child?
A) positive work was
done
B) negative work was
done
C) zero work was done
The kinetic energy of the child increased because her
speed increased. This increase in KE was the result of
positive work being done. Or, from the definition of
work, because W = KE = KEf – KEi and we know that
KEf > KEi in this case, then the work W must be
positive.
Follow-up: What does it mean for negative work to be done on the child?
Conceptual Quiz
If a car traveling 60 km/hr
can brake to a stop within
20 m, what is its stopping
distance if it is traveling
120 km/hr? Assume that
the braking force is the
same in both cases.
A)
B)
C)
D)
E)
20 m
30 m
40 m
60 m
80 m
Conceptual Quiz
If a car traveling 60 km/hr
can brake to a stop within
20 m, what is its stopping
distance if it is traveling
120 km/hr? Assume that
the braking force is the
same in both cases.
1
2
F d = Wnet = KE 1= 0 – mv2,
and thus, |F| d = 2 mv2.
Therefore, if the speed doubles,
the stopping distance gets four
times larger.
A)
B)
C)
D)
E)
20 m
30 m
40 m
60 m
80 m
Work Done by a Constant Force
Solving work problems:
1. Draw a free-body diagram.
2. Choose a coordinate system.
3. Apply Newton’s laws to determine any
unknown forces.
4. Find the work done by a specific force.
5. To find the net work, either
a) find the net force and then find the work it
does, or
b) find the work done by each force and add.
Copyright © 2009 Pearson Education, Inc.
Conceptual Quiz
By what factor does
A) no change at all
the kinetic energy of
B) factor of 3
a car change when
its speed is tripled?
C) factor of 6
D) factor of 9
E) factor of 12
Conceptual Quiz
By what factor does the
A) no change at all
kinetic energy of a car
B) factor of 3
change when its speed
C) factor of 6
is tripled?
D) factor of 9
E) factor of 12
Because the kinetic energy is
1
2
mv2, if the speed increases
by a factor of 3, then the KE will increase by a factor of 9.
Follow-up: How would you achieve a KE increase of a factor of 2?
Conceptual Quiz
Two stones, one twice the
mass of the other, are dropped
from a cliff. Just before hitting
the ground, what is the kinetic
energy of the heavy stone
compared to the light one?
A) quarter as much
B) half as much
C) the same
D) twice as much
E) four times as much
Conceptual Quiz
Two stones, one twice the
mass of the other, are dropped
from a cliff. Just before hitting
the ground, what is the kinetic
energy of the heavy stone
compared to the light one?
A) quarter as much
B) half as much
C) the same
D) twice as much
E) four times as much
Consider the work done by gravity to make the stone
fall distance d:
KE = Wnet = F d cosq
KE = mg d
Thus, the stone with the greater mass has the greater
KE, which is twice as big for the heavy stone.
Follow-up: How do the initial values of gravitational PE compare?
Conceptual Quiz
A car starts from rest and accelerates to 30
mph. Later, it gets on a highway and
A) 0  30 mph
accelerates to 60 mph. Which takes more
B) 30  60 mph
energy, the 0  30 mph, or the 30  60
C) both the same
mph?
Conceptual Quiz
A car starts from rest and accelerates to 30
mph. Later, it gets on a highway and
A) 0  30 mph
accelerates to 60 mph. Which takes more
B) 30  60 mph
energy, the 0  30 mph, or the 30  60
C) both the same
mph?
1
The change in KE ( 2 mv2 ) involves the velocity squared.
So in the first case, we have:
In the second case, we have:
1
2
1
2
1
m (302 − 02) = 2 m (900)
m (602 − 302) = 21 m (2700)
Thus, the bigger energy change occurs in the second case.
Follow-up: How much energy is required to stop the 60-mph car?
Conceptual Quiz
The work W0 accelerates a car from
A) 2 W0
0 to 50 km/hr. How much work is
B) 3 W0
needed to accelerate the car from
C) 6 W0
50 km/hr to 150 km/hr?
D) 8 W0
E) 9 W0
Conceptual Quiz
The work W0 accelerates a car from
A) 2 W0
0 to 50 km/hr. How much work is
B) 3 W0
needed to accelerate the car from
C) 6 W0
50 km/hr to 150 km/hr?
D) 8 W0
E) 9 W0
Let’s call the two speeds v and 3v, for simplicity.
We know that the work is given by W = KE = KEf – Kei.
1
2
Case #1: W0 =
Case #2: W =
1
2
m (v2 – 02) =
1
2
m ((3v)2 – v2) =
m (v2)
1
2
m (9v2 – v2) =
1
2
m (8v2) = 8 W0
Follow-up: How much work is required to stop the 150-km/hr car?
Conceptual Quiz
Two blocks of mass m1 and m2 (m1 > m2)
A) m1
slide on a frictionless floor and have the
B) m2
same kinetic energy when they hit a long
C) they will go the
rough stretch (m > 0), which slows them
same distance
down to a stop. Which one goes farther?
m1
m2
Conceptual Quiz
Two blocks of mass m1 and m2 (m1 > m2)
A) m1
slide on a frictionless floor and have the
B) m2
same kinetic energy when they hit a long
C) they will go the
rough stretch (m > 0), which slows them
same distance
down to a stop. Which one goes farther?
With the same KE, both blocks
m1
must have the same work done
to them by friction. The friction
force is less for m2 so stopping
m2
distance must be greater.
Follow-up: Which block has the greater magnitude of acceleration?
Conceptual Quiz
A golfer making a putt gives the ball an initial
velocity of v0, but he has badly misjudged the
putt, and the ball only travels one-quarter of
the distance to the hole. If the resistance force
due to the grass is constant, what speed
should he have given the ball (from its original
position) in order to make it into the hole?
A) 2 v0
B) 3 v0
C) 4 v0
D) 8 v0
E) 16 v0
Conceptual Quiz
A golfer making a putt gives the ball an initial
velocity of v0, but he has badly misjudged the
putt, and the ball only travels one-quarter of
the distance to the hole. If the resistance force
due to the grass is constant, what speed
should he have given the ball (from its original
position) in order to make it into the hole?
A) 2 v0
B) 3 v0
C) 4 v0
D) 8 v0
E) 16 v0
In traveling four times the distance, the resistive force
will do four times the work. Thus, the ball’s initial KE
must be four times greater in order to just reach the
hole—this requires an increase in the initial speed by a
1
factor of 2, because KE = 2 mv2.