Physics 207, Lecture 14, Oct. 20 Chapter 11 Goals: 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 Develop a complete statement of the law of conservation of energy Use the concept of power (i.e., energy per time) Assignment: HW6 due Wednesday, Oct. 22, HW7 available today For Wednesday: 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” 1

Mechanical Energy Potential Energy (U) Kinetic Energy (K) If “conservative” forces (e.g, gravity, spring) then Emech = constant = K + U During  Uspring+K1+K2 = constant = Emech Mechanical Energy conserved Before During 1 2 After

Energy (with spring & gravity) 1 2 h 3 mass: m -x 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 y1 (B) Between y1 & y2 (C) At y2 (D) Between y2 & y3 (E) At y3

Energy (with spring & gravity) 1 2 h mg 3 kx -x When is the child’s speed greatest? (D) Between y2 & y3 A: Calculus 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

Inelastic Processes If non-conservative” forces (e.g, deformation, friction) then Emech is NOT constant After  K1+2 < Emech (before) Accounting for this loss we introduce Thermal Energy (Eth , new) where Esys = Emech + Eth = K + U + Eth Before During After 1 2

Energy & Work Impulse (Force vs time) gives us momentum transfer Work (Force vs distance) tracks energy transfer Any process 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

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? (A) W > 0 v (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 = 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” ( Dropping a book is more complicated because it involves changes in U and K, U is transferred to K )

Changes in K with a constant F In 1-dimension, F = ma = m dv/dt = m dv/dx dx/dt = m dv/dx v by the chain rule so that F dx = mv dv If F is constant

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. F q = 0° Start Finish 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

Units: Force x Distance = Work N-m (Joule) Dyne-cm (erg) = 10-7 J Newton x [M][L] / [T]2 Meter = Joule [L] [M][L]2 / [T]2 N-m (Joule) Dyne-cm (erg) = 10-7 J BTU = 1054 J calorie = 4.184 J foot-lb = 1.356 J eV = 1.6x10-19 J cgs Other mks

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

Work in 3D…. x, y and z with constant F:

Work: “2-D” Example (constant force) A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m and y = 0 m Start Finish F q = -45° Fx x (Net) Work is Fx x = F cos(-45°) x = 50 x 0.71 Nm = 35 J Work reflects energy transfer

Scalar Product (or Dot Product) A · B ≡ |A| |B| cos(q) Useful for performing projections. î A Ax Ay q 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 and apply the Work-Kinetic Energy theorem Notice: F  Dr = (Fx )(Dx) + (Fy )(Dz ) + (Fz )(Dz) Fx Dx +Fy Dy + Fz Dz = DK So here F  Dr = DK = Wnet More generally a Force acting over a Distance does Work

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

Remember that a real trajectory implies forces acting on an object v path and time a = + a aradial atang a = + a F Ftang Fradial a = 0 = + Two possible options: Change in the magnitude of v a = 0 Change in the direction of v a = 0 Only tangential forces are important one for work! Direct application of Newton’s Laws gives the kinematics The distance over which FTang is applied: Work

Exercise Work in the presence of friction and non-contact forces A box is pulled up a rough (m > 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 2 3 4 5

Exercise Work in the presence of friction and non-contact forces A box is pulled up a rough (m > 0) incline by a rope-pulley-weight arrangement as shown below. How many forces are doing work on the box ? And which are positive (T) and which are negative (f, mg)? (For mg only the component along the surface is relevant) Use a Free Body Diagram (A) 2 (B) 3 is correct 4 5 v T N f mg

Physics 207, Lecture 14, Oct. 20 Assignment: HW6 due Wednesday, Oct. 22 HW7 available today For Wednesday: Read Chapter 12, Sections 1-3, 5 & 6 do not concern yourself with the integration process Next slides are on Wednesday 1

Work and Varying Forces (1D) Consider a varying force F(x) Area = Fx Dx F is increasing Here W = F · r becomes dW = F dx Fx x Dx Start Finish F F q = 0° Dx Work has units of energy and is a scalar!

Example: Work Kinetic-Energy Theorem with variable force How much will the spring compress (i.e. x) to bring the box to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ? x vo m to F spring compressed spring at an equilibrium position V=0 t Notice that the spring force is opposite the displacement For the mass m, work is negative For the spring, work is positive

Example: Work Kinetic-Energy Theorem with variable force How much will the spring compress (i.e. x = xf - xi) to bring the box to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ? to vo m spring at an equilibrium position x F V=0 t m spring compressed

Conservative Forces & Potential Energy For any conservative force F we can define a potential energy function U in the following way: The work done by a conservative force is equal and opposite to the change in the potential energy function. This can be written as: ò W = F ·dr = - U ri rf Uf Ui U = Uf - Ui = - W = - F • dr ò ri rf

Exercise: Work & Friction Two blocks having mass m1 and m2 where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. m > 0) which slows them down to a stop. Which one will go farther before stopping? Hint: How much work does friction do on each block ? (A) m1 (B) m2 (C) They will go the same distance m1 v1 m2 v2

Exercise: Work & Friction W = F d = - m N d = - m mg d = DK = 0 – ½ mv2 - m m1g d1 = - m m2g d2  d1 / d2 = m2 / m1 (A) m1 (B) m2 (C) They will go the same distance m1 v1 m2 v2

Conservative Forces and Potential Energy So we can also describe work and changes in potential energy (for conservative forces) DU = - W Recalling (if 1D) W = Fx Dx Combining these two, DU = - Fx Dx Letting small quantities go to infinitesimals, dU = - Fx dx Or, Fx = -dU / dx

Non-conservative Forces : If the work done does not depend on the path taken, the force involved is said to be conservative. If the work done does depend on the path taken, the force involved is said to be non-conservative. An example of a non-conservative force is friction: Pushing a box across the floor, the amount of work that is done by friction depends on the path taken. and work done is proportional to the length of the path !

A Non-Conservative Force, Friction Looking down on an air-hockey table with no air flowing (m > 0). Now compare two paths in which the puck starts out with the same speed (Ki path 1 = Ki path 2) . Path 2 Path 1

A Non-Conservative Force Path 2 Path 1 Since path2 distance >path1 distance the puck will be traveling slower at the end of path 2. Work done by a non-conservative force irreversibly removes energy out of the “system”. Here WNC = Efinal - Einitial < 0  and reflects Ethermal

Work & Power: Two cars go up a hill, a Corvette and a ordinary Chevy Malibu. Both cars have the same mass. Assuming identical friction, both engines do the same amount of work to get up the hill. Are the cars essentially the same ? NO. The Corvette can get up the hill quicker It has a more powerful engine.

Work & Power: Power is the rate at which work is done. Average Power is, Instantaneous Power is, If force constant, W= F Dx = F (v0 Dt + ½ aDt2) and P = W / Dt = F (v0 + aDt)

Power is the rate at which work is done. Work & Power: Power is the rate at which work is done. Average Power: Instantaneous Power: Units (SI) are Watts (W): 1 W = 1 J / 1s Example: A person of mass 80.0 kg walks up to 3rd floor (12.0m). If he/she climbs in 20.0 sec what is the average power used. Pavg = F h / t = mgh / t = 80.0 x 9.80 x 12.0 / 20.0 W P = 470. W

Top Middle Bottom Exercise Work & Power Starting from rest, a car drives up a hill at constant acceleration and then suddenly stops at the top. The instantaneous power delivered by the engine during this drive looks like which of the following, Z3 time Power Top Middle Bottom