Physics 170 Week 10, Lecture 1
http://www.phas.ubc.ca/∼gordonws/170
Physics 170 Week 10, Lecture 1
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Textbook Chapter 14:
Section 14.1-3
Physics 170 Week 10, Lecture 1
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Learning Goals:
• We will define the concept of work due to a force.
• We will use a number of examples to illustrate the concept of
work and to show how it can be useful in solving problems.
• We will study the law of work and energy for a system with
more than one particle
• We will illustrate the law of work and energy with an example
• By the end of this lecture, the student should know how to
compute the work due to a force during a motion by doing a
line integral.
• The student should also be able to apply the principle of work
and energy to compute the change in kinetic energy of a
particle that is moving under the influence of a force.
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Work due to a force
• Consider a particle which moves from position ~r to position
~r0 = ~r + d~r so that the displacement is infinitesimal d~r.
• Assume that a force F~ is acting on the particle as it moves.
• Assume that d~r is “small enough” so that the force F~ is a
constant throughout the motion and the displacement is in a
straight line.
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Work due to a force cont’d.
• The work done by the force on the particle during the motion
is defined as
dU = F~ · d~r
• Units are force time distance – Newton-meters = Joules or
foot-pounds.
• This is the usual definition of work as force times distance the
dot product projects F~ on d~r, U = F~ · d~r = |F~ ||d~r| cos θ
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Work due to a force cont’d.
• The work done by the force on the particle during the motion
in an infinitesimal dispalcement is dU = F~ · d~r.
• To get the work for a finite, rather than infinitesimal
displacement, we have to integrate the “line integral”.
Z ~r2
U1−2 =
F~ · d~r
~
r1
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How to do the line integral
• The work is the line integral U1−2 =
R ~r2
~
r1
F~ · d~r
• If you know the equation of the curve over which the
displacement occurs, ~r(s) and you know the value of the force
F~ (~r(s)) when the particle is at a position along the curve, then,
if ~r1 = ~r(s1 ) and ~r2 = ~r(s2 ), the integral is
Z s2
Z ~r2
d
F~ (~r(s)) · ~r(s) ds
F~ · d~r =
U1−2 =
ds
s1
~
r1
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Work due to a constant force field:
• Consider a force which is constant, for example, gravity near
the surface of the earth action on a particle with mass m,
F~ = −mg k̂. Consider a curve ~r(s) = x(s) î + y(s) ĵ + z(s) k̂.
The work done in moving between two points is
Z s2
Z s2
¢
¡
¢ d ¡
d
~·
F
U1−2 =
s1
ds
Z
~r(s)ds =
s1
~
s2
U1−2 =
(−mg)
~
s1
−mg k̂ ·
ds
x(s)î + y(s)ĵ + z(s)k̂ ds
d
z(s) ds = −mg (z(s2 ) − z(s1 ))
ds
• Work U1−2 = −mg(z2 − z1 ) is −mg times the change in height.
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Example: The 10 kg block shown in the figure rests on the smooth
incline. If the spring is originally stretched by 0.5 m, determine the
total work done by all the forces acting on the block when a
horizontal force P = 400 N pushes the block up the plane s = 2 m.
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Example cont’d:
Forces:
~ = −mg ĵ
Gravity: W
P: P~ = −400 î N
~ = N (sin 30 î + cos 30 ĵ)
Reaction: N
³
´
Spring: F~s = −(30 N/m)(s + (0.5 m)) − cos 30 î + sin 30 ĵ
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Example cont’d:
Displacement: We
³ put the origin at the
´ point where the motion
starts. ~r(s) = s − cos 30 î + sin 30 ĵ
³
´
d~r(s) = − cos 30 î + sin 30 ĵ ds , s ∈ [0, 2 m]
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Example cont’d:
~
Increments
of work: find dU
³
´ = F · d~r where
d~r = − cos 30 î + sin 30 ĵ ds and
~ = −mg ĵ,
W
P~ = −400 î N ,
~ = N (sin 30 î + cos 30 ĵ),
N
³
´
F~s = −(30 N/m)(s + (0.5 m)) − cos 30 î + sin 30 ĵ
³
´
Gravity: dUW = −mg ĵ · − cos 30 î + sin 30 ĵ ds
³
´
P: dUP = −400 N î · − cos 30 î + sin 30 ĵ ds
Reaction:
³
´
dUN = N (sin 30 î + cos 30 ĵ) · − cos 30 î + sin 30 ĵ ds = 0
³
´
Spring: dUs = −(30 N/m)(s + (0.5 m)) − cos 30 î + sin 30 ĵ ·
³
´
− cos 30 î + sin 30 ĵ ds Now integrate from s = 0 to s = 2m.
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Example cont’d:
Work due to gravity:
³
´
Gravity dUW = −mg ĵ · − cos 30 î + sin 30 ĵ ds
Z
UW = −mg sin 30
2
ds = −mg sin 30 (2m) = −(10)(9.81) sin 30(2) J
0
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Example cont’d:
Total work of each force
Gravity: UW = −(10)(9.81)(sin 30)(2) J
UW = −98.1 J
√
R2
P: UP = 400 cos 30 0 ds = (400)(2) cos 30J = 400 3 J
UP = 693 J
Reaction:
³
´R
2
UN = N (sin 30 î + cos 30 ĵ) · − cos 30 î + sin 30 ĵ 0 ds = 0
UN = 0
R2
Spring: Us = −(30 N/m) 0 ds(s + (0.5 m)) =
−(30 N/m)( 21 s2 + (0.5 m)s)20 = −(30)( 12 (4) + (0.5)(2)) J
Us = −90 J
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Principle of Work and Energy
• Now we consider a displacement where the position of the
particle evolves according to Newton’s second law
d
F~ = m~a = m dt
~v (t) where F~ is all forces acting on the particle.
• Multiply each side by the velocity
d
d
F~ · ~v (t) = m ~v (t) · ~v (t) =
dt
dt
Integrate this expression
Z t2
Z
dt F~ · ~v (t) =
t1
Physics 170 Week 10, Lecture 1
t2
t1
dt
d
dt
µ
µ
1
m~v 2 (t)
2
1
m~v 2 (t)
2
¶
¶
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Principle of Work and Energy cont’d
µ
¶
Z t2
Z t2
d 1
~
m~v 2 (t)
dt F · ~v (t) =
dt
dt 2
t1
t1
The right-hand-side can be integrated
1
1
2
m~v (t2 ) − m~v 2 (t1 ) =
2
2
or
1
1
2
m~v2 − m~v12 =
2
2
Z
Z
t2
dtF~ · ~v (t)
t1
~
r2
F~ · d~r
~
r1
1
2
m~
v
2
is the kinetic energy
The change of kinetic energy of a particle is equal to the work done
on it by all forces.
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Example: The 20 lb crate has a velocity of vA = 12 f t/s when it
is at A. Determine its velocity after it slides s = 6 f t down the
plane. The coefficient of kinetic friction between the crate and the
plane is µk = 0.2.
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Strategy for solving the problem:
• Find the forces acting on the box.
• Find the work done by each of the forces and add them
together to find the total work.
• Equate total work to the change in kinetic energy and use this
to compute the final speed.
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Free-body Diagram
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Forces acting on the box:
This is a two-dimensional problem
Orient x-axis horizontally and y-axis vertically
We find a mathematical expression for force vectors:
~ = −mg ĵ
Gravity: W
³
´
~ = N − 3 î + 4 ĵ
Reaction: N
5
5
³
´
Friction: F~f = νk N 45 î + 35 ĵ
~ . N = 4 mg
Reaction must cancel normal component of W
5
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Increment of work due to forces:
µ
¶
µ
¶
4
3
4
3
~r(s) = −s
î + ĵ
, d~r = −ds
î + ĵ
5
5
5
5
Increment of work is dU = F~ · d~r. We compute this for each force:
~ = −mg ĵ
Gravity: W
~ · d~r = 3 mg ds
dUW = W
5³
´
~ = N − 3 î + 4 ĵ
Reaction: N
5
~ · d~r = 0 → dUN = 0
N
³
Friction: F~f = νk N 45 î +
5
3
5
´
ĵ
dUf = F~f · d~r = −µk N ds
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Work due to forces:
Now we integrate the increments of work to get the total work done
by each force:
~ · d~r = 3 mg ds
Gravity: dUW = W
5
R
R
6
ft
3
~ · d~r = 3 mg
UW = W
ds
=
5
5 (20 lb)(6 f t) = 72.0 f t.lb
0
Reaction: UN = 0
Friction: dUf = F~f · d~r = −µk N ds
R
R 6 ft
~
Uf = Ff · d~r = −µk N 0
ds = −(0.2)(20 lb) 54 (6 f t) =
− 96.0
5 f t lb
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Implement the principle of work and energy
We will now implement the principle that the total work done by
all of the forces that are acting on the crate must equal the change
of its kinetic energy.
Total work due to all forces is is
UW + UN + Uf = 72.0 + 0 −
1
= mv22 −
2
From this equation we find the final
v22
96.0
f t lb
5
1
mv12
2
velocity:
2(32.2 f t/s2 )
96.0
= (12 f t/s) +
(72.0 + 0 −
f t lb)
20lb
5
2
v2 = 17.7 f t/s
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Work due to a force in a system of particles
• The work done on particle with displacement d~ri by the force
F~j dU = F~j · d~ri .
• The work for a finite displacement is the line integral
Z ~r2 X
U1−2 =
F~i · d~rj
~
r1
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i,j
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• Some forces are internal and f~ij = f~ji . In this case, particles
follow trajectories so that
·
¸
Z
d~
r
(t)
d~
r
(t)
j
i
dt f~ij (t) ·
+ f~ji (t) ·
dt
dt
·
¸
Z
d~rj (t) d~ri (t)
= f~ij (t) ·
−
dt
dt
dt
can cancel and we only need to consider external forces.
• This is useful when particles are tied together – can sometimes
be used for pulleys.
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Example: A 2-lb block rests on a semi-cylindrical surface. An
elastic cord having a stiffness k=2lb/ft is attached to the block at
B and to the base of the semi-cylinder at C. If the block is released
from rest at A (θ = 0), determine the unstretched length of the
cord so that the block begins to leave the semicylinder at the
instant that θ = 45 degrees. Neglect the size of the block.
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Example:
Cylindrical coordinates
~r = rûr , ~v = rθ̇ûθ , ~a = rθ̈ûθ − rθ̇2 ûr
Tangential-normal coordinates (ût = ûθ , ûn = −ûr )
(rθ̇)2
ûn
~v = vût = rθ̇ût , ~a = rθ̈ut +
r
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Example:
Total forces acting: gravity, normal reaction, spring
F~ = −mg ĵ + N ûr + kr(θ0 − θ)ûθ
ĵ = sin θûr + cos θûθ
F~ = (N − mg sin θ)ûr + [kr(θ0 − θ) − mg cos θ]ûθ
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Example:
Dybnamics: F~ = m~a where
~a = rθ̈ûθ − rθ̇2 ûr
F~ = (N − mg sin θ)ûr + [kr(θ0 − θ) − mg cos θ]ûθ
³
´
(N − mg sin θ)ûr + [kr(θ0 − θ) − mg cos θ]ûθ = m rθ̈ûθ − rθ̇2 ûr
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Example:
³
´
(N − mg sin θ)ûr + [kr(θ0 − θ) − mg cos θ]ûθ = m rθ̈ûθ − rθ̇2 ûr
equations for components:
N − mg sin θ = −mrθ̇2
kr(θ0 − θ) − mg cos θ = mrθ̈
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Example:
Integrate second equation:
kr(θ0 − θ) − mg cos θ = mrθ̈
krθ̇(θ0 − θ) − mg θ̇ cos θ = mrθ̇θ̈
· µ
¶
¸
·
¸
d
1 2
d 1
mrθ̇2
kr θθ0 − θ − mg sin θ =
dt
2
dt 2
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Example:
Z
0
t
¶
¸
· µ
¸ Z t
·
d
d 1
1 2
dt
dt
mrθ̇2
kr θθ0 − θ − mg sin θ =
dt
2
dt 2
0
Impose initial conditions: θ(0) = 0 , θ̇(0) = 0
¶
µ
1
1
kr θθ0 − θ2 − mg sin θ = mrθ̇2
2
2
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Example:
Interpretations of the formula
· µ
·
¶
¸ Z t
¸
Z t
d
1 2
d 1
kr θθ0 − θ − mg sin θ =
mrθ̇2
dt
dt
dt
2
dt 2
0
0
from principle of work and energy: (v = rθ̇) and
Z 2
Z θ
1 2 2
1
mr θ̇ = mv 2 =
F~ · d~r =
Fθ (θ0 )rdθ0
2
2
1
0
µ
¶
Z θ
1 2
0
0
0
2
=
[kr(θ0 − θ ) − mg cos θ ]rdθ = kr θθ0 − θ − mgr sin θ
2
0
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Example:
Interpretations of the formula
· µ
·
¶
¸ Z t
¸
Z t
d
1 2
d 1
kr θθ0 − θ − mg sin θ =
mrθ̇2
dt
dt
dt
2
dt 2
0
0
Conservation of energy
1 2 2 k
kr2
2
(rπ − `0 ) = mr θ̇ + (` − `0 )2
2
2
2
Potential energy stored in a spring is
k
2 (s
− s0 )2
` = r(π − θ) , `0 = r(π − θ0 )
µ
¶
1 2
1 2 2
1
2
kr θθ0 − θ − mgr sin θ = mr θ̇ = mv 2
2
2
2
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Example:
¶
µ
1
1
kr θθ0 − θ2 − mg sin θ = mrθ̇2
2
2
¡
¢
2
2
mrθ̇ (t) = kr 2θ(t)θ0 − θ (t) − 2mg sin θ(t)
Block leaves surface when N = 0
N = mg sin θ − mrθ̇2 = 0 when mrθ̇2 = mg sin θ
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Example:
¡
¢
2
mrθ̇ (t) = kr 2θ(t)θ0 − θ (t) + 2mg sin θ(t)
2
N = mg sin θ − mrθ̇2 = 0 when mrθ̇2 = mg sin θ
¢
¡
2
mg sin θ = kr 2θθ0 − θ − 2mg sin θ
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Example:
¡
mg sin θ = kr 2θθ0 − θ
2
¢
− 2mg sin θ
θ
3mg
θ0 = +
sin θ
2 2θkr
·
¸
θ
3mg
`0 = r(π − θ0 ) = r π − −
sin θ
2 2θkr
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Example:
·
¸
θ
3mg
`0 = r π − −
sin θ
2 2θkr
when θ = π/4,
·
`0 = (1.5 f t) π −
π
3(2 lb)
1
√
−
8
2(π/4)(2 lb/f t)(1.5 f t) 2
¸
`0 = 2.77 f t
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For the next lecture, please read
Textbook Chapter 14:
Section 14.4-6
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