Physics 207: Lecture 14, Pg 1 Lecture 14Goals: Assignment: l l HW6 due Wednesday, Mar. 11 l l For Tuesday: Read Chapter 12, Sections 1-3, 5 & 6 ” do not concern yourself with the integration process in regards to “center of mass” or “moment of inertia” Chapter 10 Chapter 10 Understand spring potential energies & use energy diagrams Understand spring potential energies & use energy diagrams Chapter 11 Chapter 11   Understand the relationship between force, displacement and work   Recognize transformations between kinetic, potential, and thermal energies   Define work and use the work-kinetic energy theorem   Use the concept of power (i.e., energy per time)

Physics 207: Lecture 14, Pg 2 Energy for a Hooke’s Law spring l Associate ½ kx 2 with the “potential energy” of the spring m

Physics 207: Lecture 14, Pg 3 Energy for a Hooke’s Law spring m l l Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

Physics 207: Lecture 14, Pg 4 Energy diagrams l In general: Energy K y U E mech Energy K x U E mech Spring/Mass system Ball falling

Physics 207: Lecture 14, Pg 5 Equilibrium l Example  Spring: F x = 0 => dU / dx = 0 for x=x eq The spring is in equilibrium position l In general: dU / dx = 0  for ANY function establishes equilibrium stable equilibrium unstable equilibrium U U

Physics 207: Lecture 14, Pg 6 Comment on Energy Conservation l We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved.  Mechanical energy is lost:  Heat (friction)  Bending of metal and deformation l Kinetic energy is not conserved by these non-conservative forces occurring during the collision ! l Momentum along a specific direction is conserved when there are no external forces acting in this direction.  In general, easier to satisfy conservation of momentum than energy conservation.

Physics 207: Lecture 14, Pg 7 Mechanical Energy l Potential Energy (U) l Kinetic Energy (K) l If “conservative” forces (e.g, gravity, spring) then E mech = constant = K + U During  U spring +K 1 +K 2 = constant = E mech l Mechanical Energy conserved Before During After 1 2

Physics 207: Lecture 14, Pg 8 Energy (with spring & gravity) l E mech = constant (only conservative forces) l At 1: y 1 = h ; v 1y = 0 At 2: y 2 = 0 ; v 2y = ? At 3: y 3 = -x ; v 3 = 0 l E m1 = U g1 + U s1 + K 1 = mgh l E m2 = U g2 + U s2 + K 2 = ½ mv 2 l E m3 = U g3 + U s3 + K 3 = -mgx + ½ kx l Given m, g, h & k, how much does the spring compress? l E m1 = E m3 = mgh = -mgx + ½ kx 2  Solve ½ kx 2 – mgx +mgh = h 0 -x mass: m

Physics 207: Lecture 14, Pg 9 Energy (with spring & gravity) l When is the child’s speed greatest? (A) At y 1 (top of jump) (B) Between y 1 & y 2 (C) At y 2 (child first contacts spring) (D) Between y 2 & y 3 (E) At y 3 (maximum spring compression) h 0 -x mass: m

Physics 207: Lecture 14, Pg 10 Energy (with spring & gravity) l When is the child’s speed greatest? l A: Calculus soln. Find v vs. spring displacement then maximize (i.e., take derivative and then set to zero) l B: Physics: As long as F gravity > F spring then speed is increasing Find where F gravity - F spring = 0  -mg = kx Vmax or x Vmax = -mg / k So mgh = U g23 + U s23 + K 23 = mg (-mg/k) + ½ k(-mg/k) 2 + ½ mv 2  2gh = 2(-mg 2 /k) + mg 2 /k + v 2  2gh + mg 2 /k = v max h 0 -x mg kx

Physics 207: Lecture 14, Pg 11 Inelastic Processes l If non-conservative” forces (e.g, deformation, friction) then E mech is NOT constant l After  K 1+2 < E mech (before) l Accounting for this loss we introduce l Thermal Energy (E th, new) where E sys = E mech + E th = K + U + E th Before During After 1 2

Physics 207: Lecture 14, Pg 12 Energy & Work l Impulse (Force vs time) gives us momentum transfer l Work (Force vs distance) tracks energy transfer l Any process which changes the potential or kinetic energy of a system is said to have done work W on that system  E sys = W W can be positive or negative depending on the direction of energy transfer l Net work reflects changes in the kinetic energy W net =  K This is called the “Net” Work-Kinetic Energy Theorem

Physics 207: Lecture 14, Pg 13 Circular Motion l I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle. l How much work is done after the ball makes one full revolution? v (A) W > 0 (B) W = 0 (C) W < 0 (D) need more info

Physics 207: Lecture 14, Pg 14 Examples of “Net” Work(W net ) Examples of No “Net” Work  K = W net l l Pushing a box on a rough floor at constant speed l l Driving at constant speed in a horizontal circle l l Holding a book at constant height This last statement reflects what we call the “system” ( Dropping a book is more complicated because it involves changes in U and K, U is transferred to K )   K = W net l Pushing a box on a smooth floor with a constant force; there is an increase in the kinetic energy

Physics 207: Lecture 14, Pg 15 Changes in K with a constant F l l If F is constant

Physics 207: Lecture 14, Pg 16 Net Work: 1-D Example (constant force) = 10 x 5 N m = 50 J Net Work is F  x = 10 x 5 N m = 50 J l 1 Nm ≡ 1 Joule and this is a unit of energy l Work reflects energy transfer xx A force F = 10 N pushes a box across a frictionless floor for a distance  x = 5 m. F  = 0° Start Finish

Physics 207: Lecture 14, Pg 17 Units: N-m (Joule) Dyne-cm (erg) = J BTU= 1054 J calorie= J foot-lb= J eV= 1.6x J cgsOthermks Force x Distance = Work Newton x [M][L] / [T] 2 Meter = Joule [L] [M][L] 2 / [T] 2

Physics 207: Lecture 14, Pg 18 Net Work: 1-D 2 nd Example (constant force) = -10 x 5 N m = -50 J Net Work is F  x = -10 x 5 N m = -50 J l Work reflects energy transfer xx F A force F = 10 N is opposite the motion of a box across a frictionless floor for a distance  x = 5 m.  = 180° Start Finish

Physics 207: Lecture 14, Pg 19 Work in 3D…. l x, y and z with constant F:

Physics 207: Lecture 14, Pg 20 Work: “2-D” Example (constant force) = F cos( = 50 x 0.71 Nm = 35 J ( Net) Work is F x  x = F cos( -45°)  x = 50 x 0.71 Nm = 35 J l Work reflects energy transfer xx F A force F = 10 N pushes a box across a frictionless floor for a distance  x = 5 m and  y = 0 m  = -45° Start Finish FxFx

Physics 207: Lecture 14, Pg 21 l Useful for performing projections. A  î = A x î  î = 1 î  j = 0 î A AxAx AyAy  A  B = (A x )(B x ) + (A y )(B y ) + (A z )(B z ) l l Calculation can be made in terms of components. Calculation also in terms of magnitudes and relative angles. Scalar Product (or Dot Product) A  B ≡ | A | | B | cos  You choose the way that works best for you! A · B ≡ |A| |B| cos(  )

Physics 207: Lecture 14, Pg 22 Scalar Product (or Dot Product) Compare: A  B = (A x )(B x ) + (A y )(B y ) + (A z )(B z ) with A as force F, B as displacement  r and apply the Work-Kinetic Energy theorem Notice: F   r = (F x )(  x) + (F y )(  z ) + (F z )(  z) F x  x +F y  y + F z  z =  K So here F   r =  K = W net More generally a Force acting over a Distance does Work

Physics 207: Lecture 14, Pg 23 Definition of Work, The basics Ingredients: Force ( F ), displacement (  r ) “Scalar or Dot Product”   r displacement F Work, W, of a constant force F acts through a displacement  r : (Work is a scalar) W = F ·  r (Work is a scalar) If we know the angle the force makes with the path, the dot product gives us F cos  and  r If the path is curvedat each point and

Physics 207: Lecture 14, Pg 24 Remember that a real trajectory implies forces acting on an object l Only tangential forces yield work!  The distance over which F Tang is applied: Work a v path and time a a a = 0 Two possible options: Change in the magnitude of v Change in the direction of v a = 0 a a a a = + a radial a tang a = + l F radial F tang F = +

Physics 207: Lecture 14, Pg 25 Definition of Work, The basics Ingredients: Force ( F ), displacement (  r )  r rr r displacement F Work, W, of a constant force F acts through a displacement  r : (Work is a scalar) W = F ·  r (Work is a scalar) Work tells you something about what happened on the path! Did something do work on you? Did you do work on something? If only one force acting: Did your speed change?

Physics 207: Lecture 14, Pg 26 Exercise Work in the presence of friction and non-contact forces A. 2 B. 3 C. 4 D. 5 A box is pulled up a rough (  > 0) incline by a rope-pulley- weight arrangement as shown below.   How many forces (including non-contact ones) are doing work on the box ?   Of these which are positive and which are negative?   Use a Free Body Diagram   Compare force and path v

Physics 207: Lecture 14, Pg 27 Home Exercise Work Done by Gravity l An frictionless track is at an angle of 30° with respect to the horizontal. A cart (mass 1 kg) is released from rest. It slides 1 meter downwards along the track bounces and then slides upwards to its original position. l How much total work is done by gravity on the cart when it reaches its original position? (g = 10 m/s 2 ) 1 meter 30° (A) 5 J(B) 10 J(C) 20 J(D) 0 J h = 1 m sin 30° = 0.5 m

Physics 207: Lecture 14, Pg 28 Lecture 14 Assignment: l HW6 due Wednesday, March 11 l For Tuesday: Read Chapter 12, Sections 1-3, 5 & 6 do not concern yourself with the integration process