Download Ch 7 Kinetic Energy and Work

Ch 7
Kinetic Energy and Work
Question: 7
Problems: 3, 7, 11, 17, 23, 27, 35, 37,
41, 43
Technical definition of energy – a scalar quantity
that is associated with that state of one or more
objects
The state of an object describes its position and its
motion.
motion requires energy:
flying a plane requires energy (from the fuel)
a thrown ball gets its energy from the thrower
One definition of energy is the capacity to do work.
This chapter covers kinetic energy, which is just one
form of energy, and then work.
The work-energy theorem relates the two quantities.
SI unit of energy is the Joule
1 J = 1 kg m2/s2
another unit of energy is the calorie
Kinetic energy – energy associated with motion.
Anything that moves has kinetic energy.
kinetic energy: K = ½ mv2
Kinetic energy is related to the mass and the velocity.
Example. A 1 kg ball is thrown with v = 20 m/s
K = ½ (1kg)(20m/s)2 = 200 J
Work
When you apply a force to an object, you can either
accelerate or decelerate the object.
Since the velocity changes, the kinetic energy
changes.
Work is the transfer of energy via a force.
Symbol for work is W. Easily confused with weight.
Work is a scalar.
The unit of work is also the Joule.
Relation of work and kinetic energy:
Let a force act in the x-direction on a particle with 1-D
motion: Fx=max
v
2
2
0
v
2a x d
Multiply equation by ½ m:
substituting
Yields:
1
1
2
mv
mv02
2
2
Fx max
1
mv 2
2
1
mv02
2
max d
Fx d
Fxd is the work
Work
• W=Fd
Work depends on displacement
If there is no displacement, there is no work.
- If you are holding a book still in the air,
you are doing no work
- If you are sitting still in a chair, gravity is
doing no work.
- If you lean on a wall and the wall does not move,
you are doing no work.
Only components of forces parallel to the
displacement do work.
F
F sin
F cos
d
W = (F cos ) d
W = F d cos
The vertical component does no work.
Only the component of the force that is along the
direction of the displacement, does any work.
W = F d cos
Work is the dot product
the force and the

 between
displacement. W F d
In this definition: W = F d cos
F and d are magnitudes, are always positive
The work will be positive or negative based on the cos .
Work – kinetic energy theorem
W= K
or
Kf = Ki + W
Positive work increases the kinetic energy.
Speed up the object
Negative work decreases the kinetic energy.
Slows down the object
Work done by gravitational force
The gravitational force is never turned off, so
whenever an object has a change in elevation,
there is work done by the gravitational force..
Wg = mg d cos
Since Fg is always down, whenever the object is
lowered, Fg does positive work.
When the object is raised, Fg does negative
work.
Wg = mg d cos
Since Fg is always down, whenever the object
is lowered vertically, d is down and = 00.
Fg does positive work.
When the object is raised straight up, d is up
and = 1800.
Fg does negative work.
When raising or lowering an object you have to
consider the work done by the applied force and the
work done by gravity.
K = Wa + Wg
If the initial and final velocities of the object being
moved are equal then:
Wa = -Wg
It does not matter if the Kf and Ki are zero or not, as
long as they are equal.
You can lift something at constant velocity, or you can
pick up a stationary object and then hold it still.
If you have a non-constant force that picked up
an object, calculating your work directly may be
tricky.
However you can calculate the work done by
gravity and then take the opposite of Wg.
Example of Olympic snatch. The lift requires two
separate pulls requiring different muscle groups.
Instead of trying to determine how much force each
pull exerted and over what distance, you can instead
find the work done by gravity and then take: Wa = -Wg.
Example
You lower a bucket down a well, with constant
velocity a distance of 8 meters. You exert a
constant force of 19.6 N. The bucket has a
mass of 2 kg. Find the work done by you, and
the work done by gravity.
W = mg
W = (2kg) g
W = 19.6 N
19.6 N
W
The displacement is down the well.
W = (F cos ) x
Your work:
Work by gravity:
Wy = (19.6 N cos 180) 8m
Wy = - 156.8 J
Wg = (19.6 N cos 0) 8m
Wg = 156.8 J
Notice that the work done by gravity was mgh,
where h is the change in height.
The net work is zero, the bucket is lowered at a
constant velocity.
Example
A 500 N force directed 30 degrees above the
horizontal is used to pull a 50 kg sled across
the ground. The coefficient of friction
between the sled and the ground is 0.3. The
sled is pulled 5 meters.
a) What is the work done by the pulling force
b) work done by friction.
c) work done by the normal force
d) work done by gravity
= 300
P = 500 N,
m = 50 kg
= 0.3
x=5m
FN
P
Ff
W = mg = 490 N
W
FN + P y = W
FN = W – Py = mg – 500 N (sin 30)
FN = (50 kg)g – 250 N = 240 N
Ff = FN = 0.3*240 N = 72 N
W = (F cos ) x
Work by pulling force:
Wp = (500 N cos 30) 5 m = 2165 J
Work by friction:
Wf = (72 N cos 180) 5 m = - 360 J
Work by normal force:
WN = (240 N cos 90) 5 m = 0 J
Work by gravity:
Wg = (490 N cos 270) 5 m = 0 J
Net work is then 2165 J – 360 J = 1805
Work by a general variable force
When the force is variable we cannot use:
W = (F cos ) x.
This is because the work done over each interval,
x is different.
Instead we have to integrate the force over the total
displacement to find the work.
W
xf
xi
F ( x)dx
The area under the Force vs. displacement
cure is equal to the work. (see fig. 7-13)

Let: F
Variable force in 3-D

and dr
Fxiˆ Fy ˆj Fzkˆ
dxiˆ dyˆj dzkˆ (the small change in position)
The amount of work produced by the force over the small

displacement interval dr is:
dW
 
F dr
Fx dx Fy dy Fz dz
The work over the total displacement is:
rf
rf 

W
dW
F dr
ri
W
xf
xi
ri
Fx dx
yf
yi
Fydy
zf
zi
Fzdz
Work-Kinetic Energy Theorem with a variable force
The work of a variable force is: W
xf
xi
F ( x)dx
xf
Using Newton’s 2nd law we rewrite as: W
dv
ma dx = m dt dx
Using the chain rule we have:
W
W
xf
xi
(ma)dx m
K
vf
vi
dv
dt
vdv
dv dx
dx dt
1 2
mv f
2
xi
(ma)dx
dv
v
dx
1 2
mvi
2
Spring
A spring force is a particular type of variable force.
Spring forces are important because many forces
behave mathematically like spring forces.
Spring can be used to apply forces
Springs can store energy
These can be done by either compression, stretching,
or torsion.
Springs
Ideal, or linear springs follow a rule called:
Hooke’s Law:
Fs = - k x
Also an ideal spring will have negligible mass.
• k is called the spring constant. This
determines how stiff the spring is.
• x is the distance the spring is deformed
(stretched or compressed) from the
equilibrium length.
• The minus sign tells us that this is a restoring
force.
Restoring force means that the force the spring
exerts, is in the opposite direction of the force
that deforms the spring.
If I pull the box to the right (stretch the spring),
the spring will exert a force to the left.
If I push the box to the left (compress the spring),
the spring will exert a force to the right.
Spring constant
The spring constant, (k), determines how stiff a
spring is.
- High spring constant
Stiff or strong, Hard to stretch or compress
- Low spring constant
Limp or weak, Easy to stretch or compress
- Units for spring constant are force per length:
N/m
Simple spring example
A spring with a spring constant of 250 N/m has a
length of 0.5 meters when un stretched. What
magnitude of force is needed to stretch the
spring so that is 0.75 meters long?
Fs = - k x
Fs = (250 N/m)(0.75m – 0.5m) = 62.5 N
I got rid of the minus sign to show the magnitude.
Note that it would take the same amount of
force to compress the spring by 0.25 meters.
W = (F cos ) x assumes a constant amount of
force. So this will not work for a spring.
The force needed to deform (stretch/compress)
a spring increases as the deformation increases.
xf
Ws
xi
xf
Ws
Ws
xi
(
Fx dx
kxdx
1
k )(x 2f
2
k
xi2 )
xf
xi
xdx
1 2
kxi
2
1 2
kx f
2
If xi=0, the work of a spring force, where x is the
deformation of the spring: Ws = - ½ kx2
When a spring is deformed:
The work done by the spring will be positive if
the spring ends up closer to its relaxed length.
Ws will be negative if the spring ends up
further away from its relaxed length.
Ws is zero if the difference of the spring’s
length to the relaxed length is unchanged.
Similar to the case of an applied force acting
against gravity.
We can find the work of an applied force
acting against a spring using:
K = Wa + Ws
If the ends of the spring are stationary before
and after the displacement.
then: Wa = -Ws
Power
Definition of power is the work done per time.
Pave
W
t
SI unit of power is the watt (W).
1 watt = 1 J/s
(Don’t confuse the W for watt with the W for work)
Instantaneous power: P
dW
dt
U.S. unit of power is horsepower
1 hp = 746 W
For electric power generation/usage, use the
kilowatt-hour. This is the energy transferred in
1hr at the rate of 1kW (1000 J/s).
1kWh = (1000J/s)(3600s) = 3.6x106J
General form for power:
P
P
P
dW F (cos )dx
dt
dt
Fv cos
 
F v
dx
F cos
dt
Power is the dot product of the force and velocity vectors.
Only components of forces that are parallel to the velocity
do work.
The power from forces that are perpendicular to the
motion is zero.
Power
Power is related to how fast a force can be applied.
Lifting a heavy weight slowly may not require much power.
Picking up the same weight quickly will require more power.
Weightlifting examples: compare the power required to
perform a bench press and an Olympic snatch.
Assume the weights are moved at constant velocities and
the applied forces are constant.
Bench press: 300 lbs (1335 N),
Range of motion x ~ 0.5 m
  time to raise weight ~3 seconds
P = F v = F x/ t = (1335 N)(0.5 m)/(3s) = 222.5 W
The above assumptions lead us to the power being
constant, (equal to the average power).
Olympic snatch: 100 lbs (445 N)
Range of motion x ~ 1.5 m
time
to
raise
weight
~1
s
 
P = F v = F x/ t = (445 N)(1.5 m)/(1s) = 667.5 W
Compared to the power in the bench press (222.5 W)
Even though the snatch is performed with less weight, it
requires more power because of the larger velocity.
(Could have found the power by finding the work of the applied
force, using Wa = - Wg. Then using power = work/time.)
Bucket Example
You want to lift a 20 kg bucket up a well at a
constant velocity of 0.5 m/s. What power is
needed to do so?
Since the velocity is constant, the upward
force you must pull with is equal to the weight
of the bucket.
 
P = F v = F v = (20kg)g (0.5 m/s) = 98 W
Another bucket example
Again you want to raise the same 20 kg
bucket. It is starting from rest, and you want
to pull on the bucket so that is has a velocity
of 2 m/s. You want to accomplish this over a
time interval of 4 seconds.
Use Power and the work energy theorem.
(net work equals change in kinetic energy.)
first find the change in the kinetic energy.
K = ½ mvf2 – ½ mv02 = ½ (20kg)(2m/s)2
K = 40 J
Work = 40 J
Average power = Work/time = (40 J) /(4s) = 10 W
Shamu example
Calculate the average power needed for the whale to
speed up. The killer whale has mass of 8000 kg
What power is needed to reach speed of 12m/s in a 6
second time interval?
Do work energy theorem
K = ½ mvf2 – ½ mv02 = ½ (8000kg)(12m/s)2
K = 5.76x105J
Power = (5.76x105J)/6 s = 9.6x104 W
This was neglecting drag. The power is actually higher.
9.6x104 W (1hp/746W) = 128 hp
About the same as a car.
Power delivered by elevator motor
A 1000 kg elevator carries a load of 800 kg. A
constant friction force of 4000 N retards it
upward motion. The retarding force behaves
similar to friction. What minimum power in
kilowatts and horsepower must the motor
deliver to lift the fully loaded elevator at a
constant speed of 3 m/s?
Elevator
Since the speed is constant (3m/s), the
acceleration and the sum of all forces equals
zero.
Find the force (T) the motor must pull with to
achieve this motion.
F = ma = 0
T
T – Fr – mg = 0
T = Fr + mg = 4000 N + (1800kg)g
Fr
mg
T = 2.16x104N
Elevator
Now that we know the force the motormust

exert we can find the power using P = F v =F v.
P = F v = (2.16x104N)(3m/s) = 6.48x104W
746 W = 1 hp
So P = (6.48x104W) x (1 hp)/(746 W) = 87 hp
Question 8
Problems: 10, 12, 20, 24, 30, 34, 48