1. Energy I Week 6

2. Contents Overview Overview 3 Laws of Thermodynamics 3 Laws of Thermodynamics Energy Energy Work Work Power Power Review Review

3. 3 Laws of Thermodynamics 1. Energy is conserved. 2. Entropy rises with time. 3. At absolute zero, entropy is a minimum.

4. Energy Energy is a conserved quantity. Energy is a conserved quantity. It remains constant throughout a reaction. It remains constant throughout a reaction. However, it tells you nothing about what happens during the reaction. However, it tells you nothing about what happens during the reaction.

5. Work When you do work on an object, you supply energy to it. When you do work on an object, you supply energy to it. W = F Δx W = F Δx Measured in J. Measured in J. F ΔxΔxΔxΔx

6. Work When you do work on an object, you supply energy to it. When you do work on an object, you supply energy to it. W = F Δx cos ө W = F Δx cos ө = F Δx = F Δx ө F ΔxΔxΔxΔx F cos ө

7. Work (cont.) W = F Δx cos ө W = F Δx cos ө When F and Δx are in the same direction, work is positive. When F and Δx are in the same direction, work is positive. When F and Δx are in opposite directions, work is negative. When F and Δx are in opposite directions, work is negative.

8. Potential Energy F = mg F = mg W = F Δh W = F Δh = mg Δh mg Δh is called the gravitational potential energy. mg

9. Kinetic Energy F = ma F = ma v 2 - v 0 2 = 2a Δx v 2 - v 0 2 = 2a Δx a = (v 2 - v 0 2 )/2 Δx W = FΔx W = FΔx = ma Δx = m (v 2 - v 0 2 )/2 = ½mv 2 - ½mv 0 2 = Δ(½mv 2 ) = Δ(kinetic energy)

10. Summary P.E. = mg Δh P.E. = mg Δh K.E. = ½mv 2 K.E. = ½mv 2 Total Energy = P.E. + K.E. Total Energy = P.E. + K.E. By the first law of thermodynamics, this quantity should be constant. By the first law of thermodynamics, this quantity should be constant.

11. Free Fall a = -g a = -g v 2 - v 0 2 = -2g Δh v 2 - v 0 2 = -2g Δh Multiply both sides by ½m. This yields: Multiply both sides by ½m. This yields: ½mv 2 - ½mv 0 2 = -mg Δh ½mv 2 - ½mv 0 2 = -mg Δh ½mv 2 - ½mv 0 2 = -mg (h – h 0 ) ½mv 2 - ½mv 0 2 = -mg (h – h 0 ) ½mv 2 - ½mv 0 2 = mgh 0 – mgh ½mv 2 - ½mv 0 2 = mgh 0 – mgh ½mv 2 + mgh = ½mv 0 2 + mgh 0 ½mv 2 + mgh = ½mv 0 2 + mgh 0 K.E. + P.E. = K.E. 0 + P.E. 0 K.E. + P.E. = K.E. 0 + P.E. 0 Total Energy After = Total Energy Before Total Energy After = Total Energy Before

12. Energy is Conserved! K.E. + P.E. = K.E. 0 + P.E. 0 =)

13. Freefall with Constraints What is the ball’s speed at the bottom? What is the ball’s speed at the bottom? Assume: no friction Assume: no friction K.E. 0 + P.E. 0 = K.E. + P.E. K.E. 0 + P.E. 0 = K.E. + P.E. 0 + mg Δh = ½mv 2 + 0 0 + mg Δh = ½mv 2 + 0 g Δh = ½v 2 g Δh = ½v 2 v = √(2g Δh) v = √(2g Δh) Doesn’t require mass Doesn’t require mass m ΔhΔhΔhΔh

14. Friction The ball moves along a horizontal surface with friction at speed v. The ball moves along a horizontal surface with friction at speed v. How much heat energy is lost after the ball stops? How much heat energy is lost after the ball stops? Heat Energy = ½mv 2 Heat Energy = ½mv 2 The heat energy lost is the work done by friction. The heat energy lost is the work done by friction. m f mg N

15. Pendulum What is the pendulum’s speed at the bottom? What is the pendulum’s speed at the bottom? Δh is the maximum height. Δh is the maximum height. K.E. 0 + P.E. 0 = K.E. + P.E. K.E. 0 + P.E. 0 = K.E. + P.E. mg Δh = ½mv 2 mg Δh = ½mv 2 g Δh = ½v 2 g Δh = ½v 2 v = √(2g Δh) v = √(2g Δh) Δh = L – L cos ө Δh = L – L cos ө = L (1 - cos ө) = L (1 - cos ө) m ө ΔhΔhΔhΔh L L

16. Power Power is the rate at which work is done. Power is the rate at which work is done. P = W/Δt P = W/Δt Measured in J/sec or Watts. Measured in J/sec or Watts.

17. Example A box is at rest. Jimmy pushes the box with 3 Watts of power on a frictionless surface. Then, the box would gain 3 J of kinetic energy per second. A box is at rest. Jimmy pushes the box with 3 Watts of power on a frictionless surface. Then, the box would gain 3 J of kinetic energy per second.

18. Energy-mass Equivalence m v = m 0 /√(1 – v 2 /c 2 ) m v = m 0 /√(1 – v 2 /c 2 ) By the Binomial Theorem, By the Binomial Theorem, m v = m 0 (1 + ½ v 2 /c 2 + 3/8 v 4 /c 4 + …) m v = m 0 (1 + ½ v 2 /c 2 + 3/8 v 4 /c 4 + …) m v = m 0 (1 + ½ v 2 /c 2 ) m v = m 0 (1 + ½ v 2 /c 2 ) m v c 2 = m 0 c 2 (1 + ½ v 2 /c 2 ) m v c 2 = m 0 c 2 (1 + ½ v 2 /c 2 ) m v c 2 = m 0 c 2 + ½ m 0 v 2 m v c 2 = m 0 c 2 + ½ m 0 v 2 m v c 2 = m 0 c 2 + K.E. m v c 2 = m 0 c 2 + K.E. E v = E 0 + K.E. E v = E 0 + K.E. E = mc 2 E = mc 2

19. Review Basic formulae W = F Δx cos ө W = F Δx cos ө P.E. = mg Δh P.E. = mg Δh K.E. = ½mv 2 K.E. = ½mv 2 Conservation of Energy K.E. 0 + P.E. 0 = K.E. + P.E. K.E. 0 + P.E. 0 = K.E. + P.E.

20. Next Week More applications of Energy More applications of Energy