1. Ch. 8: Conservation of Energy

2. Energy Review Kinetic Energy: K = (½)mv 2 –Associated with movement of members of a system Potential Energy –Determined by the configuration of the system (location of the masses in space). –Gravitational PE: U g = mgy –Elastic PE (ideal spring): U e = (½)kx 2 Internal Energy –Related to the temperature of the system

3. Types of Systems Nonisolated Systems –Energy can cross the system boundary in a variety of ways. –The total energy of the system changes Isolated Systems –Energy does not cross the boundary of the system So, the total energy of the system is a constant (CONSERVED!)

4. Ways to Transfer Energy Into or Out of A System Work – transfers energy by applying a force & causing a displacement of the point of application of the force Mechanical Waves – allows a disturbance to propagate through a medium Heat – is driven by a temperature difference between two regions in space

5. Matter Transfer – matter physically crosses the boundary of the system, carrying energy with it Electrical Transmission – transfer by electric current Electromagnetic Radiation – energy is transferred by electromagnetic waves More Ways to Transfer Energy Into or Out of A System

6. Examples of Energy Transfer a) Work b) Mechanical Waves c) Heat d) Matter Transfer e) Electrical Transmission f) Electromagnetic Radiation 

7. TOTAL Energy is conserved “Total” means the sum of all possible kinds of energy. “Conserved” means that it remains constant in any process. In other words, Total Energy can be neither created nor destroyed, but only can be transformed from one form to another or transferred across a system boundary.  If the total amount of energy in a system changes, it can only be due to the fact that energy has crossed the boundary of the system by some method of energy transfer Conservation of Energy

8.  The total change in the energy of a system = the total energy transferred across a system boundary.  E system =  E system = total energy of the system T = energy transferred across the system boundary Established symbols: T work = W & T heat = Q Others just use subscripts The Work-Kinetic Energy Theorem is a special case of Conservation of Energy A full expansion of the above equation gives:  K +  U +  E int = W + Q + T MW + T MT + T ET + T ER

9. For an isolated system, ΔE mech = 0 –Remember E mech = K + U –This is conservation of energy for an isolated system with no nonconservative forces acting If nonconservative forces are acting, some energy is transformed into internal energy Conservation of Energy becomes:  E system = 0 E system is all kinetic, potential, & internal energies The most general statement of the isolated system model Isolated System

10. For an isolated system, the changes in energy can be written out and rearranged K f + U f = K i + U i This applies only to a system in which only conservative forces act!

11. Example 8.1 – Free Fall Calculate the speed of the ball at a distance y above the ground Use energy System is isolated so the only force is gravitational which is conservative So, we can use conservation of mechanical energy!

12. Conservation of Mechanical Energy K f + U gf = K i + U gi K i = 0, the ball is dropped Solve for v f : The equation for v f is consistent with the results obtained from kinematics

13. An actor, mass m actor = 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 m bag = 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 be θ. Calculate the maximum value θ can have such that sandbag lifts off floor. Example 8.2 – Grand Entrance

14. Free Body Diagrams  Actor Sandbag at bottom Step 1: To find actor’s speed at bottom, let y i = initial height above floor & use Conservationof Mechanical Energy K i + U i = K f + U f or 0 + m actor gy i = (½)m actor (v f ) 2 + 0 (1) mass cancels. From diagram, y i = R(1 – cos θ) So, (1) becomes: (v f ) 2 = 2gR(1 – cosθ) (2) Step 2: Use N’s 2 nd Law for actor at bottom of path (T = cable tension). Actor: ∑F y = T – m actor g = m actor [(v f ) 2 /R] or T = m actor g + m actor [(v f ) 2 /R] (3) Step 3: Want sandbag to not move.  N’s 2 nd Law for sandbag: ∑F y = T – m bag g = 0 or T = m bag g (4) Combine (2), (3), (4) : m bag g = m actor g + m actor [2g(1 – cosθ)]. Solve for θ: cosθ = [(3m actor - m bag )/(2 m actor )] = 0.5 or, θ = 60 °