Define Potential Energy of a Hooke’s Law spring Use Energy diagrams Lecture 15 Goals: Chapter 10 Define Potential Energy of a Hooke’s Law spring Use Energy diagrams Chapter 11 (Work) Define Work Employ the dot product Heads up: Exam 2 7:15 PM Thursday, Nov. 3th 1

Energy for a Hooke’s Law spring m Associate ½ ku2 with the “potential energy” of the spring

Energy for a Hooke’s Law spring m Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

Energy diagrams Ball falling Spring/Mass system Emech Emech y In general: Ball falling Spring/Mass system Emech Energy K y U Emech K Energy U u = x - xeq

Energy (with spring & gravity) 1 3 2 h -x mass: m Given m, g, h & k, how much does the spring compress? Emech = constant (only conservative forces) At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0 Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0

Energy (with spring & gravity) 1 3 2 h -x mass: m Given m, g, h & k, how much does the spring compress? Emech = constant (only conservative forces) At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0 Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0 Given m, g, h & k, how much does the spring compress? Em1 = Em3 = mgh = -mgx + ½ kx2 

Energy (with spring & gravity) 1 3 2 h -x mass: m Given m, g, h & k, how much does the spring compress? Emech = constant (only conservative forces) At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0 Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0 Given m, g, h & k, how much does the spring compress? Em1 = Em3 = mgh = -mgx + ½ kx2  Solve ½ kx2 - mgx - mgh = 0

Energy (with spring & gravity) 1 mass: m 2 h 3 -x When is the child’s speed greatest? (A) At y = h (top of jump) (B) Between h & 0 (C) At y = 0 (child first contacts spring) (D) Between 0 & -x (E) At y = -x (maximum spring compression)

Energy (with spring & gravity) 1 2 h mg 3 kx -x When is the child’s speed greatest? (D) Between y2 & y3 A: Calc. soln. Find v vs. spring displacement then maximize (i.e., take derivative and then set to zero) B: Physics: As long as Fgravity > Fspring then speed is increasing Find where Fgravity- Fspring= 0  -mg = kxVmax or xVmax = -mg / k So mgh = Ug23 + Us23 + K23 = mg (-mg/k) + ½ k(-mg/k)2 + ½ mv2  2gh = 2(-mg2/k) + mg2/k + v2  2gh + mg2/k = vmax2

Chapter 11: Energy & Work Impulse (Force over a time) describes momentum transfer Work (Force over a distance) describes energy transfer Any single acting force which changes the potential or kinetic energy of a system is said to have done work W on that system DEsys = W W can be positive or negative depending on the direction of energy transfer Net work reflects changes in the kinetic energy Wnet = DK This is called the “Net” Work-Kinetic Energy Theorem

Work performs energy transfer If only conservative forces present then work either increases or decreases the mechanical energy. It is important to define what constitutes the system

How much work is done after the ball makes one full revolution? Circular Motion I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle. How much work is done after the ball makes one full revolution? v (A) W > 0 Fc (B) W = 0 (C) W < 0 (D) need more info

Examples of “Net” Work (Wnet) DK = Wnet Pushing a box on a smooth floor with a constant force; there is an increase in the kinetic energy Examples of No “Net” Work DK = 0 = Wnet Pushing a box on a rough floor at constant speed Driving at constant speed in a horizontal circle Holding a book at constant height This last statement reflects what we call the “system”

Again a constant F along a line (now x-dir) Recall eliminating t gives Multiply by m/2 And m ax = Fx (which is constant)

Constant force along x, y & z directions Then for each x y z component Fx Fy Fz there is work W = Fx Dx + Fy Dy + Fz Dz Notice that if  is the angle between F and : |F| cos  is the compoent of F parallel to Dr |F| cos  |Dr| = W = Fx Dx + Fy Dz + Fz Dz So here we introduce the “dot” product F  Dr ≡ |F| cos  |Dr| = Fx Dx + Fy Dz + Fz Dz = DK = Wnet A Parallel Force acting Over a Distance does Work F q F|| Δr

Scalar Product (or Dot Product) î A Ax Ay q Useful for finding parallel components A  î = Ax î  î = 1 î  j = 0 Calculation can be made in terms of components. A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz ) Calculation also in terms of magnitudes and relative angles. A  B ≡ | A | | B | cos q You choose the way that works best for you!

Scalar Product (or Dot Product) Compare: A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz ) with A as force F, B as displacement Dr Notice if force is constant: F  Dr = (Fx )(Dx) + (Fy )(Dz ) + (Fz )(Dz) Fx Dx +Fy Dy + Fz Dz = DK So here F  Dr = DK = Wnet Parallel Force acting Over a Distance does Work

Net Work: 1-D Example (constant force) A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m. Start Finish F q = 0° x Net Work is F x = 10 x 5 N m = 50 J 1 Nm ≡ 1 Joule and this is a unit of energy Work reflects energy transfer

For Wednesday: Read Chapter 12, Sections 1-3, 5 Recap Next time: Power (Energy / time) Start Chapter 12 Assignment: HW7 due Tuesday, Nov. 1st For Wednesday: Read Chapter 12, Sections 1-3, 5 do not concern yourself with the integration process 1