Physics 1710 Chapter 8—Potential Energy

1. 0 Physics 1710 Chapter 8—Potential Energy Power = dW/dt = (Fdx)/dt = F dx/dt = F v (for F constant) = (20.0 N )(36 x 103 m/ 3600 sec) = 200. N m/s = 200. W (#4) Solution:

2. R 0 Physics 1710 Chapter 8—Potential Energy h What is the minimum height from which a small rolling ball must be started from rest so that it will complete a loop-the-loop? Review

3. v d R 0 Physics 1710 Chapter 8—Potential Energy v2/R = g K = U - W ½ mv 2 + ½ (2/5 mv 2)= mgh v 2 = Rg = 10/7 hg h = 0.7 R h = 7/10(22.0 cm) = 15.4 cmd = 2R+h = 69.4 cm Physics Works! (When you include all relevant effects) h What is the minimum height from which a small rolling ball must be started from rest so that it will complete a loop-the-loop?

4. 0 Physics 1710 Chapter 8—Potential Energy 1′ Lecture Potential Energy is U = -∫ F•d r The sum of all energy, potential and kinetic, of a system is conserved, in the absence of dissipation: E = U + K – W F = - ∇U = negative gradient of U.

5. 0 Physics 1710 Chapter 8—Potential Energy Potential Energy: W = ∫ F•d r U = -W = -∫ F•d r Potential Energy is the negative of the work required to put the system in the current state.

6. F 0.50 m x 0 Physics 1710 Chapter 8—Potential Energy What is the potential energy of a 0.100 kg ball placed up a 45 o ramp 0.50 above the table?

7. F - mg 0.50 m x 0 Physics 1710 Chapter 8—Potential Energy What is the potential energy of a 0.100 kg ball placed up a 45 o ramp 0.50 above the table? U = - F‧x = - (-mg)h= mg h

8. 0 Physics 1710 Chapter 8—Potential Energy Example: Elevated Mass F = -mg Potential Energy: Ug = -∫0hFdy = -∫0h(- mg) dy Ug = mg∫0h dy = mgh Thus, the potential energy stored in an elevated mass is proportional to the height h and the weight of the mass.

9. 0 Physics 1710 Chapter 8—Potential Energy Relationship Between F and U: U = -∫ F•d r So U = -∫ [ Fx dx + Fy dy + Fz dz] Then Fx =-dU/dx ; Fy =-dU/dy; Fz =-dU/dz F = -∇U F= -gradient of U

10. 0 Physics 1710 Chapter 8—Potential Energy The Force is equal to the negative gradient of the potential energy: F = -∇U Fx = -∂U/∂x Fy = -∂U/∂y Fz = -∂U/∂z

11. 0 Physics 1710 Chapter 8—Potential Energy  L Example: Pendulum U = mg h h = L(1- cos  ) U = mg L(1- cos  ) s= L  FS= - (1/L)dU/d  = - mg sin  s

12. 0 Physics 1710 Chapter 8—Potential Energy Example: Ball on a slope h = ax + by U = mgh Fx = -∂U/∂x = -∂(mgh)/∂x = -mg∂h/∂x Similarly: Fy = -∂U/∂y = -mg b Thus, F = -mg( ai + b j )

13. 0 Physics 1710 Chapter 8—Potential Energy Example: Mass on a Spring Potential Energy: U = ½ k x 2 F =dU/dx F= -½ k dx2/dx F= -k x Thus, the force is equal to the negative of the gradient of the potential energy.

14. 0 Physics 1710 Chapter 8—Potential Energy Force: z = ar2 U = mgz Fr = -∂U/∂r = -∂(mgz)/∂r = -mg∂z/∂r = - 2amgr = - k r Like a mass on a spring!

15. 0 Physics 1710 Chapter 8—Potential Energy Conservation of Energy: The sum of all energy in a system is conserved, i.e. remains the same. E = U + K

16. 0 Physics 1710 Chapter 8—Potential Energy Dissipative (non-conservative) Forces: W = ∫ F•d r =∫ (C vx2 )dx =∫ (C vx2 )(dx /dt) dt =∫ (C vx3 )dt E = U + K -W

17. 0 Physics 1710 Chapter 8—Potential Energy Summary: The Potential Energy is equal to the negative of the work done on the system to put it in its present state. U = -∫ F•d r F = - ∇U The sum of all energy, potential and kinetic, of a system is conserved, in the absence of dissipation. E = U + K – W

18. -F h F Physics 1710 Chapter 8—Potential Energy U = m g h P = dU/dt = mg dh/dt mg =(100. kg)(9.8N/kg) = 98.0 N dh/dt = 10 m/10 s = 1 m/s P = 98. W Potential Energy: