work dot product presentationk k
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Math Review Night: Work and the Dot Product
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Dot Product A scalar quantity
Magnitude:
The dot product can be positive, zero, or negative
Two types of projections: the dot product is the parallel component of one vector with respect to the second vector times the magnitude of the second vector
cosθ⋅ =A B A B
(cos ) Aθ⋅ = =A B A B B
(cos ) Bθ⋅ = =A B A B A
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Dot Product Properties
( )
( ) C C C
c c
⋅ = ⋅
⋅ = ⋅
+ ⋅ = ⋅ + ⋅
A B B A
A B A B
A B A B
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Dot Product in Cartesian Coordinates
x y z x y z
x x y y z z
ˆ ˆ ˆ ˆˆ ˆA A A , B B B
A B A B A B
= + + = + +
⋅ = + +
A i j k B i j k
A B
ˆ ˆ ˆ ˆ| || | cos(0) 1ˆ ˆ ˆ ˆ| || |cos( /2) 0π
⋅ = =
⋅ = =
i i i i
i j i j
ˆ ˆ ˆWith unit vectors , and i j k
ˆ ˆ ˆ ˆ ˆ ˆ 1ˆ ˆ ˆ ˆˆ ˆ 0
⋅ = ⋅ = ⋅ =
⋅ = ⋅ = ⋅ =
i i j j k k
i j i k j kExample:
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Kinetic Energy • Velocity
• Kinetic Energy: •
• Change in kinetic energy:
v = vx i + vy j+ vzk
K =
12
m(v ⋅ v) =12
m(vx2 + vy
2 + vz2 ) ≥ 0
ΔK =12
mv f2 −
12
mv02 =
12
m(v f ⋅v f ) −
12
m(v0 ⋅v0 )
12
m(vx , f2 + vy , f
2 + vz , f2 ) −
12
m(vx ,02 + vy ,0
2 + vz ,02 )
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Work Done by a Constant Force
Definition: Work
The work done by a constant force on an object is equal to the component of the force in the direction of the displacement times the magnitude of the displacement:
Note that the component of the force in the direction of the displacement can be positive, zero, or negative so the work may be positive, zero, or negative
F
cos cosW Fθ θ= ⋅Δ = Δ = Δ = ΔF r F r F r r
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Work as a Dot Product Let the force exerted on an object be
ˆ ˆx yF F= +F i j
ˆDisplacement: xΔ = Δr i
cosˆ ˆ ˆ( ) ( )x y x
W F x
F F x F x
β= ⋅Δ = Δ
= + ⋅ Δ = Δ
F r
i j i
Fx = F cosβFy = F sinβ
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Work Done Along an Arbitrary Path
i i iWΔ = ⋅ΔF r
010
limi
i N f
i iN iW d
=
→∞ =Δ →
= ⋅Δ = ⋅∑ ∫r
F r F r
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Work in Three Dimensions Let the force acting on an object be given by
The displacement vector for an infinitesimal displacement is
The work done by the force for this infinitesimal displacement is
Integrate to find the total work
ˆ ˆ ˆx y zF F F= + +F i j k
ˆ ˆ ˆ d dx dy dy= + +r i j k
ˆ ˆ ˆ ˆˆ ˆ( ) ( )x y zdW d F F F dx dy dy= ⋅ = + + ⋅ + +F r i j k i j k
dW = Fx dx + Fy dy + Fz dz
f f f f
x y z0 0 0 0
W d F dx F dy F dz= ⋅ = + +∫ ∫ ∫ ∫F r
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Checkpoint Problem: Work done by Constant Gravitational Force
The work done in a uniform gravitation field is a fairly straightforward calculation when the body moves in the direction of the field. Suppose a body is moving under the influence of uniform gravitational force , along a parabolic curve. The body begins at the point and ends at the point . What is the work done by the gravitation force on the body?
F = −mg j
(x0 , y0 )
(x f , y f )
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Checkpoint Problem: Work Constant Forces
An object of mass m, starting from rest, slides down an inclined plane of length s. The plane is inclined by an angle of θ to the ground. The coefficient of kinetic friction is µ.
a) What is the work done by each of the three forces while the object is sliding down the inclined plane?
b) For each force, is the work done by the force positive or negative? c) What is the sum of the work done by the three forces? Is this
positive or negative?
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Checkpoint Problem: Work Done by the Inverse Square
Gravitational Force Consider an object of mass m
moving towards the sun (mass ms). Initially the object is at a distance r0 from the center of the sun. The object moves to a final distance rf from the center of the sun. How much work does the gravitational force between the sun and the object do on the object during this motion?
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Checkpoint Problem: Work Done by a Spring Force
Connect one end of a spring of length l0 with spring constant k to an object resting on a smooth table and fix the other end of the spring to a wall. Stretch the spring until it has length l and release the object. Consider the object-spring as the system. When the spring returns to its equilibrium length what is the work done by the spring force on the body?
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Checkpoint Problem: Work Done by the Inverse Square
Gravitational Force Consider an object of mass m
moving towards the sun (mass ms). Initially the object is at a distance r0 from the center of the sun. The object moves to a final distance rf from the center of the sun. How much work does the gravitational force between the sun and the object do on the object during this motion?
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Work-Energy Theorem in Three-Dimensions
Newton’s Second Law:
Total work:
becomes
, ,x x y y z zF ma F ma F ma= = =
f f f f
0 0 0 0
x y zW d F dx F dy F dz= = = =
= = = =
= ⋅ = + +∫ ∫ ∫ ∫r r r r r r r r
r r r r r r r r
F r
f f f
0 0 0
x y zW ma dx ma dy ma dz= = =
= = =
= + +∫ ∫ ∫r r r r r r
r r r r r r
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Work-Energy Theorem in Three-Dimensions
Recall
Repeat argument for y- and z-direction
Adding these three results
maydx
y0
y f∫ =12
mvy , f2 −
12
mvy ,02
mazdx
z0
z f∫ =12
mvz , f2 −
12
mvz ,02
0
2 2 2 2 2 2 2 2, , , ,0 ,0 ,0 0
( )
1 1 1 1( ) ( )2 2 2 2
fz
x y zz
x f y f z f x y z f
W ma dx ma dy ma dz dx
W m v v v m v v v mv mv
= + +
= + + − + + = −
∫
maxdx
x0
x f∫ = mdvx
dtdx
x0
x f∫ = m dxdt
dvxx0
x f∫ = mvxdvxvx ,0
vx , f∫ =12
mvx , f2 −
12
mvx ,02
W K= Δ