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Analyses with Mechanical Energy ENERGY METHODS Learning Objectives After you complete the homework associated with this lecture, you should be able to: • Be able to draw an energy diagram to explain the changes of speed and direction of motion of an object whose mechanical energy is constant. • Use energy methods to analyze motion in systems where work is done by forces other than those that produce potential energies included in mechanical energy. • Describe how total energy is conserved even when dissipative forces act. The Mechanical Energy ( E ) of a system is the sum of its Kinetic (K ) and Potential (U ) energies Emechanical / K + U This gives us a VERY useful tool to analyze Nature ! Ef – Ei = (Wother)i6f The change in the total mechanical energy of a system of particles in going from the initial to the final state equals the work done on the system by forces other than those represented through potential energy. 1 [©2013 RJ Bieniek] 2 [©2013 RJ Bieniek] Conservation of Mechanical Energy Motion in a Potential Energy Well Ef S Ei = (Wother)i6f What if all forces are accounted for in potential energies? Then Wother= 0, and Ei = Ef if (Wother)i6f = 0 Mechanical energy is conserved if no non-conservative forces acts, i.e., if only conservative forces are present (say P Fa , P F b , ... producing potential energies Ua ,Ub , ....) Ei = ½ m1 v1i2 + ½ m2 v2i2 + ··· + Uai + Ubi + ··· = Ef = ½ m1 v1f2 + ½ m2 v2f2 + ··· + Uaf + Ubf + ··· 3 [©2013 RJ Bieniek] Consider a one-dimensional potential well . Fx(x) = S dU/dx Ef = Ei = E = constant K(x) + U(x) = E = constant K(x) = E S U(x) gives variation of K with x. Potential U(x) low Y Kinetic K(x) is high Potential U(x) high Y Kinetic K(x) is low 4 [©2013 RJ Bieniek] Turning points xt are places where object stops and turns around where K(xt) = 0 , i.e., where E = 0 + U(xt) . BALL ON DOUBLE-INCLINED PLANE ' C Little friction Y Ef S Ei = W other Y Ei = Ef C Ei = Ki + Ui = ½ m ' v i2 + mgyi = 0 + mg(0) = 0 ! How high does the ball rise? Since K = ½ m v 2 $ 0, U can only increase until Kf = 0. Thus ball rises to state such that Ef = 0 + mgyf . Ef = Ei = 0 Y yf = yi = 0 Y rises to same height. : Trust-in-the-Physics Pendulum 5 [©2013 RJ Bieniek] 6 [©2013 RJ Bieniek] When Other Forces Act ... Example of Using ΔE = Wother What happens if there are other forces present that are nonconservative or for which you don’t know their potential form? Then just use the good ol’ all-purpose energy analyzing tool: A rocket at rest is ignited on top of a frictionless cliff. It has a constant horizontal thrust force of T. After traveling distance L along the flat top, it leaves the cliff. When it arrives a distance d from the cliff face at point P, it is a distance h below the cliff top. Its velocity is angle θ down from the horizontal. Assume its mass m is constant. What is its speed V at P ? Question to ask yourself: What toolbox do we use? Ef – Ei = (Wother)i6f Wother includes all other forces that you did not account for in potential energies. Thus it might include some conservative forces you did not care to incorporate into potential energy, but always includes all non-conservative forces. 7 [©2013 RJ Bieniek] Ans: Energy 8 [©2013 RJ Bieniek] Want V. Example: A block of mass M is on a level frictionless surface and is used to compress a spring, of force constant k, by a distance L. The block is released from rest while a constant wind blows against it with force magnitude B, down by angle θ from the horizontal. What is its speed V after it has moved distance 3/2 L? 3 forces: PF , PN, PT g ΔE = Ef – Ei = (Wother)i6f [ Kf + Ugf ] S [Ki + Ugi ] = WN + WT = ∫ Nid 0 +P T CD P [ ½mvf2 + mgyf ] S [ ½mvi2 + mgyi ] = |T P| DT (…TDcosθ !) Note: DT = +(L+d) (how much of P D is in the direction of P T) [ ½mV2 + mg(S h) ] S [ ½m(0)2 + mg(0) ] = T [+(L+d)] V2 = 2gh + 2 T (L+d) /m (note: θ is a distractor) 9 [©2013 RJ Bieniek] Example: A block of mass M is on a level frictionless surface and is used to compress a spring, of force constant k, by a distance L. The block is released from rest while a constant wind blows against it with force magnitude B, down by angle θ from the horizontal. What is its speed V after it has moved distance 3/2 L? NON-MECHANICAL ENERGY ΔE = Ef S Ei = (Wother)i6f where E = K + U = mechanical energy Ef S Ei = (Wother)i6f PF g &P F S conservative Y Ug & US, but treat P N and P B as “other” [½MVf2 + Mgyf + ½ksf2] – [½MVi2 +Mgyi + ½ksi2] = WN + WB ½MV2 + MgY + ½k(0)2 S [0 + 10 [©2013 RJ Bieniek] MgY + ½k( S L)2] = P N •D P + PB •DP ½MV2 S ½ k L2 = 0 + BD D ½MV2 S ½ k L2 = ( S B cosθ) (3/2 L) We can change this into an even more powerful form by noting that (Wother)i6f is an energy term, and is a means of generating non-mechanical energy. A wonderful example is friction. If you rub your hands together, you will note two things: C you can hear it (sound energy) C they warm up (heat energy) and then solve for V. 11 [©2013 RJ Bieniek] 12 [©2013 RJ Bieniek] CONSERVATION OF TOTAL ENERGY We can just lump the non-mechanical energies (e.g., heat energy, internal chemical energies, etc.) into a generic “other energy” term gother (or equivalently, internal ginternal) that a system has. We define a quantity called TOTAL ENERGY õ to be the sum of MECHANICAL ENERGY E and OTHER ENERGY g: Example: A small block is on an undulating track that is frictionless except for two rough patches. It is initially pressed against a pivoted spring of force constant k that has been compressed distance d. It is released from rest at height H above the bottom of the track. The block comes to rest in the midst of the second rough patch at height h = aH. What is change in internal energy Δg (mainly heating)? õ= E+g = K+U +g The total energy of a “closed system” is conserved, giving THE LAW: õi = õf 13 [©2013 RJ Bieniek] õf = õi Ef + gf = Ei + gi gf S gi = Ei S Ef Δ g = Ei S Ef (NOTE : Δ g = S ΔE = S (Wfric)i6f ) Δ g = [½ m' v i2 + m g yi + ½ k si2 ] – [½ m' v f2 + m g yf + ½ k ' s f2 ] = [ m g (+ H) + ½ k(S d)2 ] S [ m g (+aH) ] Y Δ g = + b m g H + ½ k d2 = S (Wfric)i6f Increase in internal energy equals the loss of mechanical energy due to the work of friction in a closed system. 15 [©2013 RJ Bieniek] 14 [©2013 RJ Bieniek]