CHAPTER 5 Work and Energy Work: Work:Work done by an agent exerting a constant force is defined as the product of the component of the force in the direction of resulting displacement and the magnitude of the displacement. W = F cos  · d

Dimensional Analysis W = F · d = N · m = kg · m/s 2 · m = kg · m 2 /s 2 = Joule 1 N · m = 1 kg · m 2 /s 2 = 1 Joule

Energy: Energy:The means to do work Energy also has the units of “Joules”. What happened to the energy that caused the force that did the work? Assume: Frictionless Surface F Net,x = F = ma a = F/m Work is transformed into energy of motion. Kinematics: v o = 0m/s v = ? a = F/m t = ?  x = d v 2 = 2 ·F/m · d mv 2 = F · d 2 Energy of Motion = Kinetic Energy = mv 2 2 KE = mv 2 2 d m FaFa

d=h F F g = mg Assume the object is moved at constant velocity upward. W = F · d W = mgh Work is transformed into energy of elevation. Energy of Elevation = Potential Energy = mgh PE = mgh

Work Energy Equation In the absence of friction; work, kinetic energy, and potential energy are conserved. W in + KE 1 + PE 1 = KE 2 + PE 2 W in = Work put into the object or done on the object KE 1 & PE 1 = Kinetic and Potential Energy of the object before the work was performed. KE 2 & PE 2 = Kinetic and Potential Energy of the object at the end of the event. F a · d + ½ mv mgh 1 = ½ mv mgh 2

Example Problem Example Problem (Frictionless Rollercoaster) A motor does work as a 250kg coaster car is raised vertically 85 meters up an incline. How much work is “done on” the car? W = F · d = mgh = 250kg · 9.80m/s 2 · 85m = 210,000 kg ·m 2 /s 2 = 210,000 Joules 210,000J of work resulted in 210,000 of Potential Energy

Example Problem Example Problem (Frictionless Rollercoaster) The 250 kg car is at rest on the top of the incline when it is released and travels to the bottom of the track. What is its velocity at the bottom? W in + KE 1 + PE 1 = KE 2 + PE ,000J = KE ,000J = ½ mv ,000J = ½ (250kg)v 2 2 v 2 = 41m/s Loss of 210,000J of PE results in gain of 210,000J of KE

Potential Energy Stored in a Spring x = spring displacement from the “relaxed” position F = 0N x = 0m F = 98N x =.35m F = 196N x =.70m F s = -kx Hookes Law k = spring constant k = 280N/m in this example k = a property of the spring x=0m x=.35m 10kg 20kg x=.70m

Hookes Law Work = Energy Done on the Spring W = F · d W = Area Under the Curve W = ½ · kx · x W = ½ kx 2 = PE s PE s = ½ kx 2 PE s = Elastic Potential Energy or Spring Potential Energy F (N) kx x x (m) F = kx

Example Problem Example Problem (conversion of PE s ) A spring with a 15kg mass attached to it is stretched horizontally.25m from its relaxed position. This requires a force of 200.N. What is the spring constant? F = kx 200N = k ·.25m k = 800. N/m How much energy is stored in the spring in the stretched position? PE s = ½ kx 2 PE s = ½ · 800N/m · (.25m) 2 PE s = 100. N·m PE s = 100 J Upon its release, the object reaches maximum velocity as it passes through the relaxed position. What is that velocity? PE s1 + KE s1 = PE s2 + KE s2 100.J + 0J = 0J + 100J KE 2 = 100J = ½ mv 2 2 ; m = 15kg v 2 = 3.7m/s

Principle of Conservation of Energy: Energy can never be created or destroyed. Energy can never be created or destroyed. Energy may be transformed from one form into another. Energy may be transformed from one form into another. The total energy in an isolated system remains constant. The total energy in an isolated system remains constant. The total energy of the universe is constant. The total energy of the universe is constant. Mother of All Energy Equations: W in + KE 1 + PE 1g + PE 1s = W out + KE 2 + PE 2g + PE 2s W in = F a · d W out = F f · d W out generally dissipates to heat which also has units of Joules.

Power Power is the rate at which work is done P = W  t Dimensional Analysis P = W = Joules = Watt  t sec 1 Watt = 1 Joule/sec Example Problem Example Problem (Power) A 55kg rock is lifted at constant velocity 15m in 10.sec by a motor. What is the power output of the motor? Work =  PE = mg  h = 8100 Joules  t = 10.sec P = W = 8100J  t 10.sec P = 810J/s P = 810 Watts