Sect. 8-3: Mechanical Energy & It’s Conservation

1. Sect. 8-3: Mechanical Energy & It’s Conservation

2. GENERALLY In any process, total energy is neither created nor destroyed. • The total energy of a system can be transformed from one form to another & from one object to another, but the total amount remains constant  The Law of Conservation of Total Energy

3. In Ch. 7, we learned theWork-Energy Principle: The net work W done on an object = the change in the object’s kinetic energy K Wnet = K  The Work-Energy Principle (As we’ve said, this is Newton’s 2nd Law in Work-Energy language!) Note:Wnet =The work done by the net (total) force.

4. If several forces act (both conservative & non-conservative), The total work done is: Wnet =WC + WNC WC = the work done by all conservative forces WNC = the work done by all non-conservative forces • The Work-Energy Principle still holds: Wnet = K • For conservative forces(by thedefinition of potential energy U!): WC = -U  K = -U + WNC or:WNC = K + U

5.  So, in general, WNC = K + U The Total Work done by all Non-conservative forces = The total change in K + the total change in U

6. Sect. 8-3: Mechanical Energy & Its Conservation • In general, we just found that: WNC = K + U • In the (Very!)SPECIAL CASE where there are CONSERVATIVE FORCES ONLY!  WNC = 0 or K + U = 0  the Principle of Conservation of Mechanical Energy • Note:This ISN’T(quite)the same as the Law of Conservation of(total) Energy! It’s aspecial caseof this law(where all forces are conservative)

7. For theSPECIAL CASEof CONSERVATIVE FORCES ONLY! K + U = 0  the Principle of Conservation of Mechanical Energy Note Again! This is NOT(quite) the same as the Law of Conservation of (total) Energy! ThePrinciple of Conservation of Mechanical Energy is a special caseof this law (where all forces are conservative)

8. For CONSERVATIVE FORCES ONLY! K + U = 0  Principle of Conservation of Mechanical Energy Note! This came from the Work-Energy Principle & the definition of Potential Energy for conservative forces! Since the Work-Energy Principle is N’s 2nd Law in Work-Energy language, & since Conservation of Mechanical Energycame from this principle, Conservation of Mechanical Energy is Newton’s 2nd Law in Work-Energy language, in the special case of Conservative Forces only!

9. Conservation of Mechanical Energy So, for conservative forcesONLY!In any process K + U = 0 Conservation of Mechanical Energy • Definitionof Mechanical Energy: E  K + U Conservation of Mechanical Energy  In any process with only conservative forces acting, E = 0 = K + U or E = K + U = Constant So, in any process with only conservative forces acting, the sum K + U is unchanged The mechanical energy changes from U to K or K to U, but the sum remains constant.

10. If & ONLY if there are no nonconservative forces, the sum of the kinetic energy change K& the potential energy change Uin any process is zero. The kinetic energy change K& the potential energy change Uare equal in size but opposite in sign. The Total Mechanical Energy is defined as: Conservation of Mechanical Energyhas the form:

11. Summary: The Principle of Conservation of Mechanical Energy If only conservative forces are acting, the total mechanical energy of a system neither increases nor decreases in any process. It stays constant; it is conserved.

12. Conservation of Mechanical Energy  K + U = 0 or E = K + U = Constant For conservative forces ONLY(gravity, spring, etc.) • Suppose, initially: E = K1 + U1 & finally: E = K2+ U2 E = Constant  K1 + U1= K2+ U2 A very powerful method of calculation!!

13. Conservation of Mechanical Energy  K + U = 0 or E = K + U = Constant • For gravitational U: Ugrav = mgy E = K1 + U1 = K2+ U2  (½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 y1 = Initial height, v1 = Initial velocity y2 = Final height, v2 = Final velocity

14. all U U1 = mgh, K1 = 0 the sum K + U remains constant K+ U = same as at points 1 & 2 half K half U K1 + U1 = K2 + U2 0 + mgh = (½)mv2+ 0 v2 = 2gh U2 = 0, K2 = (½)mv2 all K

15. Sect. 8-4: Using Conservation of Mechanical Energy Example 8-3: Falling Rock U1 = mgy1, K1 = 0 Original height of the rock: y1 = h = 3.0 mS Speed when it falls tol y3 = 1.0 maabove the ground? K1 + U1 = K2 + U2 = K3 + U3 0 + mgh = (½)mv2+ 0= (½)m(v3)2+ mgy3 U3 = mgy3 K3 = (½)m(v3)2 U2 = 0, K2 = (½)m(v2)2

16. Example 8-3 What is it’s speed at y = 1.0 m? Conservation of Mechanical Energy!  (½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 (The mass cancels in the equation!) y1 = 3.0 m, v1 = 0 y2 = 1.0 m, v2 = ? Gives: v2 = 6.3 m/s U only Part U, part K K only

17. Example – Free Fall y1 = h U1 = mgh K1 = 0 • A ball is dropped from height h above the ground. • Calculate the ball’s speed at distance y above the ground. • Neglecting air resistance, the only force is gravity, which is conservative. • So, we can use Conservation of Mechanical Energy y y2 = y U2 = mgy K2 = (½)mv2 v v v y3 = 0, U3 = 0 K3 = (½)m(v3)2

18. Conservation of Mechanical Energy K2 + U2 = K1+ U1 In general, for an initial velocity v1, solve for v: The result for v is, of course, consistent with the results from the kinematics of Ch. 2. For this specific problem, v1 = 0 & K1 = 0, since the ball is dropped. v = (v1)2 + 2g(h – y)

19. Example 8-4: Roller Coaster • Mechanical energy conservation! (Frictionless!)  (½)m(v1)2 + mgy1= (½)m(v2)2 + mgy2 (mass cancels!)Only height differences matter! Horizontal distance doesn’t matter! • Speed at the bottom? y1 = 40 m, v1 = 0 y2 = 0 m, v2 = ? Gives: v2 = 28 m/s • For what y is v3 = 14 m/s? Use: (½)m(v2)2 + 0 = (½)m(v3)2 + mgy3 Gives:y3 = 30 m Height of hill = 40 m. Starts from rest at the top. Calculate: a. The speed of the car at the hill bottom. b. The height at which it has half this speed. Ii Take y = 0 at the bottom of the hill.

20. Conceptual Example 8-5: Speeds on 2 Water Slides frictionless water slides! Two water slides at a pool are shaped differently, but start lllll atthe same height h. Two riders, Paul & Kathleen, start from rest at the same time on different slides. Ignore friction & assume that both slides have the same path length. a. Which rider is traveling FF faster at the bottom? b. Which rider makes it to the FF bottom first? WHY?? Demonstration! both start here!  

21. Example – Grand Entrance • An actor, mass mactor = 65 kg, in a play is to “fly” down to stage during performance. Harness attached by steel cable, over 2 frictionless pulleys, to sandbag, mass mbag = 130 kg, as in figure. Need length R = 3 m of cable between nearest pulley & actor so pulley can be hidden behind stage. For this to work, sandbag can never lift above floor as actor swings to floor. Let initial angle cable makes with vertical beθ. Calculate the maximum value θcan have such that sandbag doesn’t lift off floor.

22. Free Body Diagrams Step 1: To find actor’s speed at bottom, let yi = initial height above floor & use Conservationof Mechanical Energy Ki + Ui = Kf + Uf or 0 + mactorgyi = (½)mactor(vf)2 + 0(1) mass cancels. From diagram, yi = R(1 – cos θ) So, (1) becomes:(vf)2 = 2gR(1 – cosθ) (2) Step 2: Use N’s 2nd Law for actor at bottom of path (T = cable tension).   Actor Sandbag at bottom Actor:∑Fy = T – mactorg = mactor[(vf)2/R] or T = mactorg + mactor[(vf)2/R] (3) Step 3: Want sandbag to not move.  N’s 2nd Law for sandbag: ∑Fy = T – mbagg = 0 or T = mbagg (4) Combine (2), (3), (4): mbagg = mactorg + mactor[2g(1 – cosθ)]. Solve for θ: cosθ = [(3mactor - mbag)/(2 mactor)] = 0.5 or, θ = 60°