Math Review Night: Work and the Dot Product

Math Review Night: Work and the Dot Product

Dot Product A scalar quantity Magnitude: A B = A B cosθ 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 A B = A (cos θ ) B = A B A B = A (cos θ ) B = A B

Dot Product Properties A B = B A ca B = c( A B) ( A + B) C = A C + B C

Dot Product in Cartesian Coordinates With unit vectors ˆi, ˆj and kˆ ˆi ˆi = ˆj ˆj = kˆ kˆ = 1 ˆi ˆj = ˆi k ˆ = ˆj k ˆ = 0 ˆi ˆi = ˆi ˆi cos(0) = 1 ˆi ˆj = ˆi ˆj cos( π /2) = 0 Example: A = A ˆ ˆ ˆ ˆ ˆ ˆ xi + Ay j+ A zk, B = Bxi + Byj+ Bzk A B = A B + A B + A B x x y y z z

Kinetic Energy Velocity v = v x î + v y ĵ + v z ˆk Kinetic Energy: K = 1 2 m( v v) = 1 2 m(v 2 + v 2 x y + v 2 z ) 0 Change in kinetic energy: ΔK = 1 2 mv 2 1 f 2 mv 2 = 1 0 2 m( v f v f ) 1 2 m( v 0 v 0 ) 1 2 2 m(v x, f 2 + v y, f 2 + v z, f ) 1 2 m(v 2 x,0 2 + v y,0 + v 2 z,0 )

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: W = F Δ r = F Δ r cosθ = F cosθ Δ r = F Δr F 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

Work as a Dot Product Let the force exerted on an object be F = F ˆi + F ˆj F y = F sin β Displacement: Δ r = Δx ˆi W x F x = F cosβ y = F Δ r = FΔx cos β = ( F ˆi + F ˆj ) ( Δ x ˆi ) = F Δx x y x

Work Done Along an Arbitrary Path Δ W = F Δr i i i W i= N f = lim F Δ r = F dr i i N Δr 0 i i= 1 0

Work in Three Dimensions Let the force acting on an object be given by F = F ˆi + F ˆj + F kˆ x y z The displacement vector for an infinitesimal displacement is dr = dx ˆi + dy ˆj + dy kˆ The work done by the force for this infinitesimal displacement is dw = F dr = ( F ˆi + F ˆj + F kˆ ) ( dx ˆi + dy ˆj + dy kˆ ) x y z dw = F x dx + F y dy + F z dz Integrate to find the total work f f f f W = F dr = F dx + F dy + F dz x y z 0 0 0 0

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 F = mg ĵ, along a parabolic curve. The body begins at the point (x 0, y 0 ) and ends at the point (x f, y f ). What is the work done by the gravitation force on the body?

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 m s ). Initially the object is at a distance r 0 from the center of the sun. The object moves to a final distance r f 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 l 0 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 m s ). Initially the object is at a distance r 0 from the center of the sun. The object moves to a final distance r f 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 F = ma, F = ma, F = ma x x y y z z r= rf r= rf r= rf r= rf W = F dr = F dx + F dy + F dz x y z r= r r= r r= r r= r 0 0 0 0 r= r r= r r= r f f f W = ma dx + ma dy + ma dz x y z r= r r= r r= r 0 0 0

y f y 0 Recall Work-Energy Theorem in Three-Dimensions x f x ma x dx = m dv f x x 0 dt dx x = m dx f x 0 dt dv v x, f = mv x x x dv x 0 v x,0 Repeat argument for y- and z-direction ma y dx = 1 2 mv 2 y, f 1 2 mv 2 y,0 Adding these three results z f W = ( ma dx + ma dy + ma dz) dx z 0 x y z z f z 0 ma z dx = 1 2 mv 2 x, f = 1 2 mv 2 z, f 1 2 mv 2 z,0 1 2 mv 2 x,0 1 1 1 1 W = m( v + v + v ) m( v + v + v ) = mv mv 2 2 2 2 2 2 2 2 2 2 2 2 x, f y, f z, f x,0 y,0 z,0 f 0 W = ΔK