Math Review Night: Work and the Dot Product

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

Dot Product Properties

( )

( ) C C C

c c

⋅ = ⋅

⋅ = ⋅

+ ⋅ = ⋅ + ⋅

A B B A

A B A B

A B A B

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:

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 )

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

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

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

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

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 )

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?

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?

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?

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?

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

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