Page 1Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 1Physics 207, Lecture 14, Oct. 20Goals:Goals:Assignment: Assignment: HW6 due Wednesday, Oct. 22, HW7 available today For Wednesday: Read Chapter 12, Sections 1-3, 5 & 6do not concern yourself with the integration process in regards to “center of mass” or “moment of inertia””••Chapter 11Chapter 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)Physics 207: Lecture 14, Pg 2Mechanical Energy Potential Energy (U) Kinetic Energy (K) If “conservative” forces (e.g, gravity, spring) thenEmech= constant = K + UDuring Uspring+K1+K2 = constant = Emech Mechanical Energy conservedBeforeDuringAfter12Page 2Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 3Energy (with spring & gravity) 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+ ½kx2 Solve ½kx2 –mgx+mgh= 0132h0-xmass: mPhysics 207: Lecture 14, Pg 4Energy (with spring & gravity) When is the child’s speed greatest?(A) At y1(B) Between y1& y2(C) At y2(D) Between y2& y3(E) At y3132h0-xmass: mPage 3Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 5Energy (with spring & gravity) 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> Fspringthen speed is increasingFind where Fgravity- Fspring= 0 -mg = kxVmaxor xVmax= -mg / kSo mgh = Ug23+ Us23 + K23 = mg (-mg/k) + ½ k(-mg/k)2 + ½ mv2 2gh = 2(-mg2/k) + mg2/k + v2 2gh + mg2/k = vmax2132h0-xmgkxPhysics 207: Lecture 14, Pg 6Inelastic Processes If non-conservative” forces (e.g, deformation, friction) thenEmechis NOT constant After K1+2 < Emech(before) Accounting for this loss we introduce Thermal Energy (Eth, new)where Esys= Emech+ Eth= K + U + EthBeforeDuringAfter12Page 4Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 7Energy & 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 ∆Esys= WW can be positive or negative depending on the direction of energy transfer Net work reflects changes in the kinetic energyWnet= ∆KThis is called the “Net” Work-Kinetic Energy Theorem Physics 207: Lecture 14, Pg 8Circular 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(B) W = 0(C) W < 0(D) need more infoPage 5Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 9Examples of “Net” Work (Wnet)Examples of No “Net” Work∆K = Wnet Pushing a box on a rough floor at constant speed Driving at constant speed in a horizontal circle Holding a book at constant heightThis 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 = Wnet Pushing a box on a smooth floor with a constant force; there is an increase in the kinetic energyPhysics 207: Lecture 14, Pg 10Changes in K with a constant F In 1-dimension, F = ma = m dv/dt = m dv/dx dx/dt = m dv/dx vby the chain rule so that F dx = mv dv∫∫=xfxifivvxxxxxdvmvdxF If F is constant∫∫=xfxifivvxxxxxdvmvdxF KmvmvxFxxFxixfxifx∆=−=∆=−221221)(Page 6Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 11Net Work: 1-D Example (constant force) Net Work is F ∆∆∆∆x= 10 x 5 N m = 50 J= 10 x 5 N m = 50 J 1 Nm ≡1 Joule and this is a unit of energy Work reflects energy transfer∆∆∆∆x A force F = 10 N pushes a box across a frictionless floorfor a distance ∆∆∆∆x = 5 m.Fθ = 0°StartFinishPhysics 207: Lecture 14, Pg 12Units:N-m (Joule)Dyne-cm (erg)= 10-7 JBTU = 1054 Jcalorie= 4.184 Jfoot-lb = 1.356 JeV = 1.6x10-19Jcgs OthermksForce x Distance = WorkNewton x[M][L] / [T]2 Meter = Joule[L] [M][L]2/ [T]2Page 7Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 13Net Work: 1-D 2ndExample (constant force) Net Work is F ∆∆∆∆x= = --10 x 5 N m = 10 x 5 N m = --50 J50 J Work reflects energy transfer∆∆∆∆xF A force F = 10 N is opposite the motion of a box across a frictionless floor for a distance ∆x = 5 m.θ = 180°StartFinishPhysics 207: Lecture 14, Pg 14Work in 3D….221221)(zizfzifzmvmvzFzzF −=∆=− x, y and z with constant F:221221)(yiyfyifymvmvyFyyF −=∆=−221221)(xixfxifxmvmvxFxxF −=∆=−2222221221 with zyxifzyxvvvvKmvmvzFyFxF++=∆=−=+∆+∆+∆Page 8Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 15Work: “2-D” Example (constant force) (Net) Work is Fx∆∆∆∆x= F cos(= F cos(-45°) ∆∆∆∆x= 50 x 0.71 Nm = 35 J= 50 x 0.71 Nm = 35 J Work reflects energy transfer∆∆∆∆xF A force F = 10 N pushes a box across a frictionless floor for a distance ∆x = 5 m and ∆y = 0 m θ = -45°StartFinishFxPhysics 207: Lecture 14, Pg 16 Useful for performing projections.A •••• î = Axî •••• î = 1î •••• j = 0îAAxAyθA • B = (Ax)(Bx) + (Ay)(By) + (Az)(Bz) 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 θYou choose the way that works best for you!A · B |A| |B| cos(θ)Page 9Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 17Scalar Product (or Dot Product)Compare:A • B = (Ax)(Bx) + (Ay)(By) + (Az)(Bz)with A as force F, B as displacement ∆r and apply the Work-Kinetic Energy theoremNotice:F • ∆r = (Fx)(∆x) + (Fy)(∆z ) + (Fz)(∆z)Fx∆x +Fy∆y + Fz∆z = ∆KSo hereF • ∆r = ∆K = WnetMore generally a Force acting over a Distance does Work Physics 207: Lecture 14, Pg
View Full Document
Unlocking...