Transcript Work = Fd cos(q)

Applications of Work
and Energy Equations
James Prescott Joule
AP Physics B
Lecture Notes
Problem #1
A student holds her physics textbook, which has a mass of 1.50 kg from
the second floor which is 3.0 m from the ground. She then drops it.
a) How much work is done on the book by the student in simply holding it?
b) How much work will have been done by the force of gravity during
the time the book falls 3.0 m?
1) A known mass held, then dropped;
determine a) Wst and b) Wgr
PHYSICS
3.0 m
2) Given: m = 1.50 kg, and h = 3.0 m;
Find: Wst and Wgr
3) Work / gravity
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Problem #1
4-5) a) Net work on the book by student:
Since the book does not move,
there is no displacement and
therefore, no work done.
Work = Fd cos(q)
Wst = F(0) cos(q) =
0.0 J
b) Net work on the book by force of gravity:
While the book is falling, the
Net force is equal to the weight (mg).
PHYSICS
-3.0 m
Work = Fd cos(q)
mg
Wgr = (1.50kg)(-9.81m/s2)(-3.0m)cos(0) =
+44 J
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Problem #2
If a person pushes on the lawn mower with a constant force of 90.0 N at
an angle of 40.0 to the horizontal:
How much work does he do in pushing it a horizontal distance of 7.50 m?
1) A constant force at angle
thru distance; determine Wh
90.0 N
2) Given: F = 90.0 N, and q = 40.0;
d = 7.50 m
Find: Wh
40.0
3) Work / horizontal
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Problem #2
4-5) Net work on the mower by person:
The force vector and the
displacement vector must be
parallel for work to be done.
Work = Fd cos(q)
90.0 N
40.0
F cos(q) = Fh
7.50 m
40.0
Fh = (90.0N)(cos 40.0)
Fh = 68.94 N
Wh = (68.94N)(7.50m) =
+517 J
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Problem #3
A 0.75 kg block slides down a 20.0 inclined plane with a uniform velocity.
The horizontal length of the incline is 1.20 m.
a) How much work is done by the force of friction on the block as it slides
the total length of the incline?
b) What is the net work done on the block?
y
N
Ff
mg sin(q)
mg cos(q)
q
x
mg
1.20 m
1) Box slides at constant velocity
against friction;
determine Wf , and Wnet
2) Given: m = 0.75 kg, and
q = 20.0;L = 1.20 m
Find: Wf and Wnet
q
3) Work / incline-friction
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Problem #3
4-5) a) Net forces parallel to the motion:
y
Ff
N
mg sin(q)
mg cos(q)
x
Only two forces do work
because there are only two
forces parallel to the motion.
The force of kinetic friction Ff
and the weight component down
the incline mg sin(q).
Work done by the frictional force:
Ff = mN
W = Ffd cos(q) = Ffd cos(180) = Ffd (-1)
W = mNd(-1)
Ff = mg sin (q)
from uniform motion
N = mg cos(q)
from diagram
m
Ff
N
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Problem #3
N = mg cos(q) from diagram
mg sinq
m

 tanq
N mg cosq
Ff
d
q
W = mNd(-1)
Therefore:
L
d
cosθ
L
from right triangle incline
L
W = -tan(q)mgcos(q)(
)
cosθ
W = -tan(20.0)mgL = -(.3639)(0.75kg)(9.81m/s2)(1.20m)
W=
-3.21 J
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Problem #3
4-5) b) Net forces parallel to the motion:
y
Ff
mg sin(q)
Since the force of kinetic friction Ff
is equal to weight component down
the incline mg sin(q), the net force
is equal to zero.
x
The Net work done:
Wnet = Fnetd cos(q)
Wnet = (0)d cos(q)
Wnet =
0.0 J
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Problem #3
A car of mass m is on an icy driveway inclined at an angle of 20.0.
Determine the acceleration of the car (assume the incline is frictionless).
N
1) Unknown mass down frictionless incline;
determine acceleration.
mg cos(q)
y
2) Given: q = 20.0;
Find: am
Ff
mg sin(q)
3) Newton’s 2nd Law / inclined frictionless
x
mg
q
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Work Done by a Constant Force
Work = Fd cos(q)
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Work Equation
F
q
Fcos(q)
d
Work = Fd cos(q)
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Units of Work
System
Unit of Work
SI
newton-meter
Joule (J)
cgs
dyne-centimeter
erg
British
foot-pound
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Work-Kinetic Energy Theorem
vf
vi
F
m
x
The net work done on an object by a net force acting on
it is equal to the change in kinetic energy of the object
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Work-Kinetic Energy Theroem
A net force acts on a moving object
vi
vf
F
Kinetic
Energy
m
F
a
m
x
2
2
v f2  v i2  2ax
mv f mv i
Fx 
2
2
F
2
2
v f  v i  2  x
Work  KEf - KEi

m
m v f2 - v i2
Fx 
2

Work NET  Δ Kinetic Energy
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Work Done Against Friction
Friction acts on moving object
m
Ff
m
d
Ff  μN
Wf  Ff d
Wf  mNd
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The Force to Stretch a Spring
F  -kx
Hooke’s Law
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x
F  kx
F
F
Work
kx
kx 2
2
0
x
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Work Done by Gravity
Object falls in a gravitational field
m
mg
h
m
Wg = mgh
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Work and Energy Equations
Work done by force (F) =
Fd cos( q)
Kinetic energy =
mv 2
2
Work done by gravity =
mgh
Work done by a spring =
Work done against friction =
kx
2
2
mNd
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Gravity and Kinetic Energy
Ball drooped from rest
Find the final velocity of the ball:
mv 2
2
mgh
Wg  KE
h
mv 2
mgh 
2
v  2gh
v
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Spring and Kinetic Energy
Work done by a spring
Find the final velocity of the block:
v
k
m
m
x
kx 2
2
Ws  KE
kx 2 mv 2

2
2
mv 2
2
kx 2
v
m
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Spring and Friction
Work done by friction
k
Find the displacement of the block:
v=0
m
m
x
kx 2
2
Ws  Wf
d=?
mNd
2
kx
 mNd
2
2
mg
kx 2
d
2mmg
kx
 mmgd
2
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Kinetic Energy and the Incline Plane
Find the final velocity of the block:
vo = 0
d
h
q
mv 2
mgd sinq  
2
mv 2
2
mgh
Wg  KE
V=?
mv 2
mgh 
2
h = d sin(q)
v  2gd sinq 
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Gravitational Potential Energy
Work Done due to Gravity
PE  mghf - mghi
m
PEi  mgh
PE  0 - mgh
PE  -mgh
Wg  mgh
h
Wg  -PE
m
PEf  0
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Elastic Potential Energy
kx 2
PE 
2
kx 2
PE  0 2
k
x
k
kx 2
Ws 
2
PE  0
kx 2
PE  2
Ws  - PE
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Conservation of Mechanical Energy
The mechanical energy (Emec) of a system is
Emec  K  U
When no external forces act on a system
Emec  0
Emec  K  U  0
K  U  0
KEf - KEi   PEf - PEi   0
KEf  PEf  KEi  PEi
Conservation of Mechanical Energy
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Potential and Kinetic Energy
Conservation of Mechanical Energy
U
K
U
U
U
K
U
K
K
K
Power
Energy can be transferred from one form to another
through the process of work. The rate at which work
is done is called power
Average power
Pavg
ΔE

 F vavg
Δt
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Work and
Energy
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