Aim: How do we explain conservation of energy in systems under the influence of non-conservative forces?

Do Now: The energies indicated on your vertical axis are evenly space; that is E3 – E2 = E2 – E1 . The energy E1 is equal to U(x1 ) and the energy E3 is equal to U(x3 ). a) Determine the numerical values of x1 and x3 . b) Describe the motion of the particle if its total energy is E2 . c) What is the particle’s speed at x = x1 if its total energy E =58 J? d) The particle is released from rest at x =1/2x. Find its speed as it passes through x = x1

Recall Extended Definition of Work and Energy Wtot = Wc + WNC + Wext

When friction is present.. Energy is not conserved. Ei + Wkinetic friction =Ef

Example 1 A child of mass 25 kg slides down a slide of height 3 m. When she reaches the bottom of the slide her speed is 5 m/s. How much internal energy was produced? Mgh-Wf=1/2 Mv2 25(10)(3)-Wf=1/2(25)(5)2 Wf =-437.5 J So 437.5 J of internal energy are produced

Example 2 The particle shown below has a mass of 0.1 kg is released from rest at A and the surface of the bowl is rough. The speed of the particle at B is 1.50 m/s. Assume the radius of the bowl is 0.2m. What is its kinetic energy at B? KE=1/2mv2 KE=1/2(.1) (1.5)2 =0.1125 J How much mechanical energy is transformed due to friction as the particle moves from A to B? Ui -Wf = KE Mgh – Wf =.1125 J .1(10)(.2)-Wf =0.1125 Wf =-0.0875 J so 0.0875 J are lost due to friction Is it possible to determine the coefficient of friction from these results in any simple manner? No

Problem 2

Example 3 A 10 kg block is released from point A. The track is frictionless except for the portion between points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2,250 N/m, and compresses the spring 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C. (Solution on next page)

Problem 3 MgH – Wf =1/2kx2 10(10)(3) – Wf =1/2(2250)(.3)2 Wf = -198.75 J So Wf =Ff d 198.75= Ff (6) Ff =33.125 N And Ff = µFN =µmg 33.125=µ(100) µ=0.33

Example 4 A sled of mass m is given a kick on a frozen pond. The kick imparts to it an initial speed of v m/s. The coefficient of kinetic friction between the sled and ice is µ. Use energy considerations to find the distance (d) the sled moves before it stops. ½ mv2 = Wf 1/2mv2 = Ff d 1/2mv2 = µFN d 1/2mv2 =µmgd d= v2 /(2µg)