Transcript chapter-6

3/29/2016
Norah Ali Al Moneef
king Saud university
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W = F. Δx
units: N.m or joule, j
• W = F Δx cos q
• Work = the product of force and displacement,
times the cosine of the angle between the
force and the direction of the displacement.
• How much work is done pulling with a 15 N
force applied at 20o over a distance of 12 m?
• W = FΔx cos q
• W = 15 N (12 m) (cos 20) = 169J
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20o
2
Sign of work:
+ve work: Force acts in the same direction as the displacement.
 object gains energy
ve work:
Force acts in the opposite direction as the displacement.
 object loses energy
When force F is at right angles to displacement s (F ⊥s):
 cos 90 = 0
 F Δx cos q = 0
 No work is done on the load
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No work is done if:
F=0
or
Δx = 0
or
q = 90o
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Example: When Work is Zero
•
•
•
•
•
A man carries a bucket of water horizontally at constant velocity.
The force does no work on the bucket
Displacement is horizontal
Force is vertical
cos 90° = 0
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example
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example
• An Eskimo returning pulls a sled as shown. The total mass of
the sled is 50.0 kg, and he exerts a force of 1.20 × 102 N on
the sled by pulling on the rope. How much work does he do on
the sled if θ = 30° and he pulls the sled 5.0 m ?
W  ( F cos q ) x
 (1.20  10 2 N )(cos 30 )(5.0m)
 5.2  10 2 J
• Suppose µk = 0.200, How much work done on the sled by friction,
and the net work if θ = 30° and he pulls the sled 5.0 m ?
Fnet , y  N  mg  F sin q  0
W fric  ( f k cos180 )x   f k x
N  mg  F sin q
   k Nx    k (mg  F sin q )x
 (0.200)(50.0kg  9.8m / s 2
Wnet  WF  W fric  WN  Wg
 5.2 10 2 J  4.3 10 2 J  0  0
 90.0 J
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 1.2 10 2 N sin 30 )(5.0m)
 4.3 10 2 J
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Example :
A horizontal force F pulls a 10 kg carton across the floor
at constant speed. If the coefficient of sliding friction
between the carton and the floor is 0.30, how much work
is done by F in moving the carton by 5m?
The carton moves with constant speed.
Thus, the carton is in horizontal
equilibrium.
F = f = μk N = μk mg.
Thus F = 0.3 x 10 x 9.8
= 29.4 N
Therefore work done W = FS
=(29.4 Cos 0o)
=147 J
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6-2 Kinetic Energy
1
KE  mv 2
2
Same units as work
Remember the Eq. of motion
v 2f
vi2

 ax
2
2
1
1
Multiply both sides by m,
2
mv f  mvi2  max
2
2
KE f  KEi  Fx
W = F x
KE f  KE i  W net Work-Energy Theorem
W = K
When work is done on a system and the only change in the system is
in its speed, the work done by the net force equals the change in
kinetic energy of the system.
The speed of
the system increases if the work done on it is positive
( the kinetic energy of the particle increases).
The speed of the system decreases if the net work done on it is
negative( the kinetic energy of the object decreases).
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An object’s kinetic energy depends on its mass and its speed.
The faster something is moving and the heavier it is, the more
work it can do, so …
If ‘v’ tripled
then KE will increase by 9 times.
If ‘v’ quadruples,
then KE will increase by 16 times
example
M= 4 kg
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• Do changes in velocity and mass have the same effect
on kinetic energy?
• No—changing the velocity of an object will have a
greater effect on its kinetic energy than changing its
mass.
• This is because velocity is squared in the kinetic energy
equation.
• For instance, doubling the mass of an object will
double its kinetic energy.
• But doubling its velocity will quadruple its kinetic
energy.
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example
• A 1500 kg car moves down the freeway at 30 m/s
K.E. = ½(1500 kg)(30 m/s)2= 675,000 kg·m2/s2 = 675 kJ
• A 2 kg fish jumps out of the water with a speed of 1 m/s
K.E. = ½(2 kg)(1 m/s)2= 1 kg·m2/s2 = 1 J
•Calculate the work done by a child of weight 300N who
climbs up a set of stairs consisting of 12 steps each of height
20cm.
work = force x distance
the child must exert an upward force equal to its weight
the distance moved upwards equals (12 x 20cm) = 2.4m
work = 300 N x 2.4 m= 720 J
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Kinetic Energy – the energy of motion
• KE = ½ m v 2
•
units: kg (m/s) 2 = (kgm/s2) m
•
= Nm = joule
• Work Energy Theorem: W =  KE
• Work = the change in kinetic energy of a system.
Note: v = speed NOT velocity.
The direction of motion has no relevance to kinetic energy.
Example : A 6 kg bowling ball moves at 4 m/s.
a) How much kinetic energy does the bowling ball have?
b) How fast must a 2.5 kg tennis ball move in order to have the same
kinetic energy as the bowling ball?
K = ½ mb vb²
KE = 48 J
KE = ½ mt vt²
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K = ½ (6 kg) (4 m/s)²
v = √2 K/m = 6.20 m/s
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M=6.00 kg
F  ma 
a
F 12

 2m / s 2
m 6
2
2
v  v0  2ax
v   2ax 
1/ 2
Another solution
 3.46m / s
F  ma 
a
F 12

 2m / s 2
m 6
2
2
v  v0  2ax
v   2ax 
1/ 2
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 3.46m / s
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Example :
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.
W =F d cos 0 =12x 3x1 = 36J
Δk = w
0.5m V2 =W =36 J
1
 6(v ) 2  36 J
2
v  12  3.46 m / s
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Example
What acceleration is required to stop a 1000kg car traveling 28
m/s in a distance of 100 meters?
1
mv 2  max
2
1
 ( 28) 2  a  100
2
784
a 
 3.92m / s 2
200
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example
example
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example
Calculate the speed of a car of mass 1200kg if its kinetic
energy is 15 000J
K = ½ m v2
15 000J = ½ x 1200kg x v2
15 000 = 600 x v2
15 000 ÷ 600 = v2
25 = v2
v = 25
speed = 5.0 ms-1
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example
Calculate the braking
distance a car of mass 900 kg
travelling at an initial speed
of 20 ms-1 if its brakes exert a
constant force of 3 kN.
k.E. of car = ½ m v2
= ½ x 900kg x (20ms-1)2
= ½ x 900 x 400
= 450 x 400
k = 180 000 J
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The work done by the brakes
will be equal to this kinetic
energy.
W = F Δx
180 000 J = 3 kN x Δx
180 000 = 3000 x Δx
Δx = 180 000 / 3000
braking distance = 60 m
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• An object does not have to be moving to have energy.
• Some objects have stored energy as a result of their positions or
shapes.
• When you lift a book up to your desk from the floor or compress a
spring to wind a toy, you transfer energy to it.
• The energy you transfer is stored, or held in readiness.
• It might be used later when the book falls to the floor
• Stored energy that results from the position or shape of an object
is called potential energy.
• This type of energy has the potential to do work.
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Gravitational Potential Energy
• Potential energy related to an object’s height is called
gravitational potential energy.
• The gravitational potential energy of an object is equal to the
work done to lift it.
• Remember that Work = Force × Distance.
• The force you use to lift the object is equal to its weight.
• The distance you move the object is its height.
• You can calculate an object’s gravitational potential energy
using this formula:
• Gravitational Potential Energy = Weight x Height
change in Gravitational potential energy .
= mass x gravitational field strength x change in height
ΔU = m g Δh
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king Saud university
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Gravitational potential energy:
Ug  m g h
-Potential energy only depends on y (height) and not on x
(lateral distance)
-MUST pick a point where potential energy is considered zero!
U g  U  U o  mg ( h  ho )
U g  U  U o   mg ( h  ho )
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example
Calculate the change in Gravitational potential energy (g.p.e). when a
mass of 200 g is lifted upwards by 30 cm.
(g = 9.8 Nkg-1)
ΔU = m g Δh
= 200 g x 9.8 Nkg-1 x 30 cm = 0.200 kg x 9.8 Nkg-1 x 0.30 m = 0.59 J
change in g.p.e. = 0.59 J
example
What is the potential energy of a 12 kg mass raised from the ground
to a to a height of 25 m?
Potential Energy = weight x height change
Weight = 12 x 10 = 120 N
Height change = height at end - height at start = 25 - 0 = 25 m
Potential energy = 120 x 25 = 3000 J
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king Saud university
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example
Calculate the gravitational potential energy in the
following systems
a. a car with a mass of 1200 kg at the top of a 42 m
high hill
(1200 kg)( 9.8m/s/s)(42 m) = 4.9 x 105
b. a 65 kg climber on top of Mt. Everest (8800 m
high)
(65 kg) (9.8m/s/s) (8800 m) = 5.6 x 106 J
c. a 0.52 kg bird flying at an altitude of 550 m
(.52 kg) (9.8m/s/s)(550) = 2.8 x 103 J
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example
Who is going faster at the bottom?
• Assume no friction
• Assume both have the same
speed pushing off at the top
same
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Conservation of Energy
• Law of conservation of energy
- energy cannot be created or destroyed
closed system:
- all energy remains in the system
- nothing can enter or leave
open system:
- energy present at the beginning of the
system may not be at present at the end
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Conservation of Energy
• When we say that something is conserved it means that
it remains constant, it doesn’t mean that the quantity can
not change form during that time, but it will always have
the same amount.
• Conservation of Mechanical Energy:
MEi = MEf
• initial mechanical energy = final mechanical energy
If the only force acting on an object is the force of gravity:
Ko + Uo = K + U
½ mvo² + mgho = ½ mv² + mgh
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example
• At a construction site a 1.50 kg brick is dropped from rest
and hit the ground at a speed of 26.0 m/s. Assuming air
resistance can be ignored, calculate the gravitational
potential energy of the brick before it was dropped?
K+U = Ko+Uo
½ mv2+mgh=½ mv2o+ mgho
½ mv2+0 = =o+ mgho
U0 = mgho = ½ (1.50 kg) (26.0m/s)2 = 507 J
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example
A child of mass 40 kg climbs
up a wall of height 2.0 m and
then steps off. Assuming no
significant air resistance
calculate the maximum:
(a) gravitational potential
energy (gpe) of the child
(b) speed of the child
g = 9.8 Nkg-1
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(a) maximum gravitational
potential energy ( max gpe)
occurs when the child is on the
wall
Δ U = mgΔh
= 40 x 9.8 x 2.0
max gpe = 784 J
(b) max speed occurs when the
child reaches the ground
½ m v2 = m g Δh
½ m v2 = 784 J
v2 = (2 x 784) / 40
v2 = 39.2
v = 39.2
max speed = 6.3 ms-1
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example
A 10 kg shopping cart is pushed from rest by a 250 N
force against a 50 N friction force over 10 m distance.
m = 10.0 kg
vi = 0
Fp = 250.0 N
Fk = 50.0 N
Δx = 10.0 m
A. How much work is done by each force
on the cart?
Wg = 0
WN = 0
Wp = Fp Δx cos q= (250.0 N)(10.0 m)cos 0
FN
Fk
Fp
Fg
Wp = 2500 J
Wk = Fk Δx cos q= (50.0 N)(10.0 m)cos 180= -500 J
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B. How much kinetic energy has the cart gained?
Wnet = ∆KE
Wp + Wk = KEf - KEi
2500 J + -500 J = KEf - 0
KEf = 2000J
C. What is the carts final speed?
KE = 1/2 m v2
v = √((2KE)/(m))
v = √((2(2000 J))/(10.0 kg))
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v = 20 m/s
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example
At B
At C
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A truck of mass 3000 kg is to be loaded onto a ship using a crane that exerts a
force of 31 kN over a displacement of 2m.Find the upward speed of truck after
its displacement.
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Nonconservative Forces
• A nonconservative force does not satisfy the conditions of
conservative forces
• Nonconservative forces acting in a system cause a change
in the mechanical energy of the system
• The work done against friction is greater along
the brown path than along the blue path
• Because the work done depends on the path,
friction is a nonconservative force
K  U  K 0  U 0  W friction
K  U  K0  U 0  fk  d
K  U  K 0  U 0  µk N  d
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Work and Energy
• If a force (other than gravity) acts on the system and does work
• Need Work-Energy relation
PEi  KEi  W  PEf  KEf
• or
1
1
2
mghi  mvi  W  mghf  mv 2f
2
2
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example
On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial
speed of 2.2 m/s. How far does it travel if the coefficient of kinetic
friction between the sled and the ice is .10?
W = ∆KE
FΔx= Kf - Ki
m = 10.0 kg
vi = 2.2 m/s
vf = 0 m/s
µk = .10
µk = Fk / FN
Fk = µk (mg)
µk (-mg) Δx = 1/2 m vf2 - 1/2 m vi2
µk (-mg) Δx = - 1/2 m vi2
Δx = (- 1/2 m vi2) / µk (-mg)
Δx = (- 1/2 (10.0 kg) (2.2 m/s)2) / (.10 (-10.0 kg) (9.8 m/s2))
Δx = 2.47 m
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Example
A 20-g projectile strikes a mud bank, penetrating a distance of 6
cm before stopping. Find the stopping force F if the entrance
velocity is 80 m/s.
W =Δ K
F Δ x cosθ = K – K0
W = ½ mv² - ½ mvo²
=0
F Δ x cosθ = ½ mv² - ½ mvo²
F (0.06 m) cos 1800 = 0 - 0 .5 (0.02 kg)(80 m/s)2
F (0.06 m)(-1) = -64 J F = 1067 N
Work to stop bullet = change in K.E. for bullet
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example
A bus slams on brakes to avoid an accident. The tread marks of
the tires are 80 m long. If μk = 0.7, what was the speed before
applying brakes?
Work = F Δx (cos θ)
f = μk.N = μk mg
Work = - μk mg Δx
ΔK = 1/2 mv2 - ½ mvo2
W = ΔK
vo = (2μk g Δ x) ½
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Example
Suppose the initial kinetic and potential energies of a system are
75J and 250J respectively, and that the final kinetic and potential
energies of the same system are 300J and -25J respectively. How
much work was done on the system by non-conservative forces?
1. 0J
2. 50J
correct
3. -50J
4. 225J
5. -225J
Work done by non-conservative forces equals the
difference between final and initial kinetic energies plus
the difference between the final and initial gravitational
potential energies.
W = (300-75) + ((-25) - 250) = 225 - 275 = -50J.
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example
You are towing a car up a hill with constant velocity.
A )The work done on the car by the normal force is:
1. positive
2. negative
3. zero
FN
V
T
correct
The normal force is perpendicular to the displacement,
hence, does no work.
W
b )The work done on the car by the gravitational force is:
1. positive
2. negative
3. zero
correct
With the surface defined as the x-axis, the x
component of gravity is in the opposite direction of the
displacement, therefore work is negative.
C ) The work done on the car by the tension force is:
1. positive
2. negative
3. zero
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correct
Tension is in the same direction as the displacement.
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•
•
•
•
•
Power is the rate at which work is done.
Power is the rate of energy transfer by any method.
The SI unit of power is the watt, W
1 watt = 1 Joule/s
Work
1000 watts = 1 kW
Power 
Work = force · distance
Power = force·distance/time
Power = force·velocity
time
P = F·v
KE Fx
P

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t
t
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example
Calculate the power of a car engine that exerts a force
of 40 kN over a distance of 20 m for 10 seconds.
W = FΔx
= 40 kN x 20 m
= 40 000 x 20 m
= 800 000 J
example
P = ΔW / Δt
= 800 000 J / 10 s
power = 80 000 W
• In 1935, R. Goddard, the America rocket pioneer, launched
several A-series sounding rockets. Given that the engine
had a constant thrust of 890 N, how much power did it
transfer to the rocket while traveling at its maximum
speed of 113.9 m/s.
• P = Fv = 890N x 313.9 m/s = 279 kW.
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example
• Together, two students exert a force of 825 N
in pushing a car a distance of 35 m.
• A. How much work do they do on the car?
• W = F Δx = 825N (35m) =
• B. If the force was doubled, how much work
would they do pushing the car the same
distance?
• W = 2F Δx = 2(825N)(35m) =
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Power Delivered by an Elevator Motor
 A 1000-kg elevator carries a maximum load of 800 kg. A
constant frictional force of 4000 N retards its motion upward.
What minimum power must the motor deliver to lift the fully
loaded elevator at a constant speed of 3 m/s?
Fnet , y  may
Since the speed is constant, so
a=0
T  f  Mg  0
where M = total mass of the system
M= 1000+800=1800 Kg
T  f  Mg  2.16 104 N
P  Fv  (2.16 10 4 N )(3m / s)
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 6.48 10 4 W
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example
OR
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example
In a rifle barrel, a 15.0 g bullet is accelerated from rest to a speed of 780 m/s.
a)Find the work that is done on the bullet.?
Using Work-Kinetic Energy Theorem,
W = ΔKE
W = KEf - KEi
W = 1/2(0.015 kg)(780 m/s)2 - 0
W = 4.56 kJ
b) If the rifle barrel is 72.0 cm long, find the magnitude of the average total force
that acted on the bullet.
Using W = Fcosθd,
F = (4.56 kJ) / (0.72 m)
F = 6.33 kN
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example
A rock climber wears a 7.5 kg backpack while scaling a cliff.
After 30.0 min, the climber is 8.2 m above the starting
point.
a. How much work does the climber do on the backpack?
W = Fd = m g Δx= 7.5kg(9.8m/s2)(8.2m)
b.How much power does the climber expend in this effort?
P = W = 603 J =
t
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0.34 watt
1800 s
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Example : A 200 kg curtain needs to be raised 8m. in as
close to 5 s as possible. You need to decide among three
motors to buy for this, each motor cost a little more the
bigger the power rating. The power rating for the three
motors are listed as 1.0 kw, 3.5 kw and 5.5 kw.
Which motor is the best for the job?
• Given: m = 200 kg
Δx = 8 m
∆t = 5s
• Unknown: Power and work
W = F·Δx
W = m·g·h
W = (200 kg)·(9.81 m/s²)·(8 m) = 15,696 Joules
P = W/∆t
P = 15,696 J / 5 s= 3,139 watts
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example
Calculate the power of a car that maintains a constant speed of
30 ms-1 against air resistance forces of 2 kN
As the car is travelling at a constant speed the car’s engine
must be exerting a force equal to the opposing air
resistance forces.
P=Fv
= 2 kN x 30 ms-1
= 2 000 N x 30 ms-1
power = 60 kW
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example
What power is consumed in lifting a 70-kg robber 1.6 m
in 0.50 s?
example
A 100-kg cheetah moves from rest to 30 m/s in 4 s.
What is the power?
the work is equal to the change in kinetic energy:
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Summary
• If the force is constant and parallel to the displacement, work is force
times distance
• If the force is not parallel to the displacement,
• The total work is the work done by the net force:
• SI unit of work: the joule, J
• Total work is equal to the change in kinetic energy:
where
• Power is the rate at which work is done:
• SI unit of power: the watt,
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Norah
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