Transcript 7.4) Kinetic Energy andThe Work-Kinetic Energy Theorem • Figure (7.13) - a particle of mass m moving to the right under.

7.4) Kinetic Energy andThe Work-Kinetic Energy Theorem
• Figure (7.13) - a particle of mass m moving to the right under the action of a
constant net force F.
• Because the force is constant - from Newton’s second law, the particle moves
with a constant acceleration a.
• If the particle is displaced a distance d, the net work done by the total force F
is :
(7.12)
 W  Fd  (ma)d
• The relationships below are valid when a particle undergoes constant
acceleration :
d
d  12 ( v i  v f ) t
v  vi
a f
t
m
F
Figure (7.13)
vi
where vi = the speed at t = 0 and vf = the speed at time t.
vf
• Substituting these expressions into Equation (7.12) :
 vf  vi  1
W

m

 2 v i  v f t

 t 
 W  12 mvf2  12 mvi2
(7.13)
2
• The quantity 12 mv represents the energy associated with the motion of the
particle = kinetic energy.
• The net work done on a particle by a constant net force F acting on it equals
the change in kinetic energy of the particle.
• In general, the kinetic energy K of a particle of mass m moving with a speed v
is defined as :
(7.14)
K  12 mv2
• Kinetic energy is a scalar quantity and has the same units as work = Joule (J).
• It is often convenient to write Equation (7.13) in the form :
W  K
• That is :
f
 Ki  K
Ki   W  K f
(7.15)
Work-kinetic energy theorem
Work-kinetic energy theorem
• Must include all of the forces that do work on the particle in the calculation of
the net work done.
• The speed of a particle increases if the net work done on it is positive - because
the final kinetic energy is greater than the initial kinetic energy.
• The speed of a particle decreases if the net work done is negative because the
final kinetic energy is less than the initial kinetic energy.
• Kinetic energy is the work a particle can do in coming to rest, or the amount of
energy stored in the particle.
• We derived the wrok-kinetic energy theorem under the assumption of a constant
net force - but it also is valid when the force varies.
• Suppose the net force acting on a particle in the x direction is Fx.
• Apply Newton’s second law, Fx = max, and use Equation (7.8), the net work
done :
xf
xf
 W    F dx   ma
x
xi
xi
x
dx
• If the resultant force varies with x, the acceleration and speeed also depend on x.
• Acceleration is expressed as :
a
dv dv dx
dv

v
dt dx dt
dx
• Substituting this expression for a into the above equation for W gives :
xf
v
f
dv
 W   mv dx dx   mvdv
xi
vi
2
2
1
1
W

mv

mv
 2 f 2 i
(7.16)
The net work done on a
particle by the net force
acting on it is equal to the
change in its kinetic
energy of the particle
• Situation Involving Kinetic Friction
• Analyze the motion of an object sliding on a horizontal surface - to describe the
kinetic energy lost because of friction.
• Suppose a book moving on a horizontal surface is given an initial horizontal
velocity vi and slides a distance d before reaching a final velocity vf
(Figure (7.15)).
d
vi
Figure (7.15)
vf
fk
• The external force that causes the book to undergo an acceleration in the
negative x direction is the force of kinetic friction fk acting to the left, opposite
the motion.
• The initial kinetic energy of the book is ½mvi2 , and its final kinetic energy is
½mvf2 (Apply the Newton’s second law).
• Because the only force acting on the book in the x direction is the friction force,
Newton’s second law gives -fk = max.
• Multiplying both sides of this expression by d and using Eq. (2.12) in the form
vxf2 - vxi2 = 2axd
for motion under constant acceleration give
-fkd = (max)d = ½mvxf2 - ½mvxi2 or :
(7.17a)
Kfriction  f k d
• The amount by which the force of kinetic friction changes the kinetic energy of
the book is equal to -fkd.
• Part of this lost kinetic energy goes into warming up the book, and the rest
goes into warming up the surface over which the book slides.
• The quantity -fkd is equal to the work done by kinetic friction on the book
plus the work done by kinetic friction on the surface.
• When friction (as welll as other forces) - acts on an object, the work-kinetic
energy theorem reads :
(7.17b)
Ki  Wother  f k d  Kf

• Wother represents the sum of the amounts of work done on the object by
forces other than kinetic friction.
Example (7.7) : A Block Pulled on a Frictionless Surface
A 6.0-kg block initially at rest is pulled to the right along a horizontal,
frictionless surface by a constant horizontal force of 12 N. Find the speed of the
block after it has moved 3.0 m.
n
vf
F
mg
n
fk
d
F
mg
(a)
vf
d
(b)
Figure (7.16)
Example (7.8) : A Block Pulled on a Rough Surface
Find the final speed of the block described in Example (7.7) if the surface is not
frictionless but instead has a coefficient of kinetic friction of 0.15.
Example (7.9) : Does the Ramp Lessen the Work Required?
A man wishes to load a refrigerator onto a truck using a ramp, as shown in
Figure (7.17). He claims that less work would be required to load the truck if
the length L of the ramp were increased. Is his statement valid?
Figure (7.17)
L
Example (7.10) : Useful Physics for Safer Driving
A certain car traveling at an initial speed v slides a distance d to a halt after its
brakes lock. Assuming that the car’s initial speed is instead 2v at the moment
the brakes lock, estimate the distance it slides.
Example (7.11) : A Block-Spring System
A block of mass 1.6 kg is attached to a horizontal spring that has a force constant
of 1.0x103 N/m (Figure (7.10)). The spring is compressed 2.0 cm and is then
released from rest. (a) Calculate the speed of the block as it passes through the
equilibrium position x = 0 if the surface is frictionless.
7.5) Power
• The two cars - has different engines (8 cylinder powerplant and 4 cylinder
engine), same mass.
• Both cars climb a roadway up a hill, but the car with the 8 cylinder engine
takes much less time to reach the top.
• Both cars have done the same amount of work against gravity, but in
different time periods.
• Know (i) the work done by the vehicles, and (ii) the rate at which it is done.
• The time rate of doing work = power.
• If an external force is applied to an object (assume acts as a particle), and if the
work done by this force in the time interval t is W, then the average power
expended during this interval is defined as :

W
t
Average power
• The work done on the object contributes to the increase in the energy of the
object.
• Therefore, a more general definition of power is the time rate of energy
transfer.
• We can define the instantaneous power P as the limiting value of the average
power as t approaches zero :
W dW
  lim

t  0 t
dt
Represent increment of work done by dW
• From Equation (7.2), letting the displacement be ecpressed as ds, that
dW = F·ds.
• Therefore, the instantaneous power can be written :
dW
ds

 F  Fv
dt
dt
(7.18)
v =ds / dt
• The SI unit of power = joules per second (J /s) = watt (W).
1W = 1 J/s = 1 kg ·m2 / s3
• A unit of power in the British engineering system is the horsepower (hp) :
1 hp = 746 W
• A unit of energy (or work) can now be defined in terms of the unit of power.
• One kilowatt hour (kWh) = the energy converted or consumed in 1 h at the
constant rate of 1 kW = 1 000 J /s.
1kWh = (103W)(3600 s) = 3.60 x 106 J
kWh = a unit of energy, not power
Example (7.12) : Power Delivered by an Elevator Motor
An elevator car has a mass of 1 000 kg and is carrying passengers having a
combined mass of 800 kg. A constant frictional force of 4 000 N retards its
motion upward, as shown in Figure (7.18a). (a) What must be the minimum
power delivered by the motor to lift the elevator car at a constant speed of
3.00 m/s?