1. UNIT 4Work, Energy, and Power

2. 1) 2 v1 = v2 2)  2 v1 = v2 3) 4 v1 = v2 4) v1 = v2 5) 8 v1 = v2 ConcepTest 6.5bKinetic Energy II Car #1 has twice the mass of car #2, but they both have the same kinetic energy. How do their speeds compare?

3. 1) 2 v1 = v2 2)  2 v1 = v2 3) 4 v1 = v2 4) v1 = v2 5) 8 v1 = v2 ConcepTest 6.5bKinetic Energy II Car #1 has twice the mass of car #2, but they both have the same kinetic energy. How do their speeds compare? Since the kinetic energy is 1/2 mv2, and the mass of car #1 is greater, then car #2 must be moving faster. If the ratio of m1/m2 is 2, then the ratio of v2 values must also be 2. This means that the ratio of v2/v1 must be the square root of 2.

4. Thursday November 10th CONSERVATION OF MECHANICAL ENERGY

5. TODAY’S AGENDA Thursday, November 10 • Conservation of Mechanical Energy • Hw: Practice D (All) p172 UPCOMING… • Fri: More Conservation of Energy: • Bowling Ball Demo • Power • Mon: Problem Quiz 1 • Tue: Problems @ the Boards

6. Section 3 Conservation ofEnergy Chapter 5 Mechanical Energy • Mechanical energyis the sum of kinetic energy and all forms of potential energy associated with an object or group of objects. ME = KE + ∑PE • Mechanical energy is often conserved. MEi = MEf initial mechanical energy = final mechanical energy (in the absence of friction)

8. Section 3 Conservation ofEnergy Chapter 5 Sample Problem Conservation of Mechanical Energy Starting from rest, a child zooms down a frictionless slide from an initial height of 3.00 m. What is her speed at the bottom of the slide? Assume she has a mass of 25.0 kg.

9. Section 3 Conservation ofEnergy Chapter 5 Sample Problem, continued Conservation of Mechanical Energy 1. Define Given: h = hi = 3.00 m m = 25.0 kg vi = 0.0 m/s hf = 0 m Unknown: vf = ?

10. Section 3 Conservation ofEnergy Chapter 5 Sample Problem, continued Conservation of Mechanical Energy 3. Calculate Substitute values into the equations: PEg,i = (25.0 kg)(9.81 m/s2)(3.00 m) = 736 J KEf = (1/2)(25.0 kg)vf2 Now use the calculated quantities to evaluate the final velocity. MEi = MEf PEi + KEi = PEf + KEf 736 J + 0 J = 0 J + (0.500)(25.0 kg)vf2 vf = 7.67 m/s

11. Section 3 Conservation ofEnergy Chapter 5 Mechanical Energy, continued • Mechanical Energy is not conserved in the presence of friction. • As a sanding block slides on a piece of wood, energy (in the form of heat) is dissipated into the block and surface.

12. Systems and Energy Conservation Ball dropped from rest falls freely from a height h. Find its final speed. h v Energy

13. Systems and Energy Conservation A block of mass m compresses a spring (force constant k) a distance x. When the block is released, find its final speed. v m m x Energy

14. Systems and Energy Conservation m When released from rest, the block slides to a stop. Find the distance the block slides. vf= 0 Friction (m) m k x d Energy

15. END