Announcements Read Ch. 6 (Work & Energy): All of Chapter 6 CAPA Set #8 due on Friday at 10 pm (shorter assignment) This week in Section, Assignment #5: Work & Energy Print and bring to your Lab Section. Midterm Exam Tuesday October 11 at 7:30 pm Practice exam and solutions versions have a couple differences. Apologies. Refer to solutions version. Office Hours… Nagle – Monday 2-3 pm Gamow F221 Uzdensky – Tuesday 11 am – 12 Help Room Nagle/Kinney – Wed. 1:45-3:45 Help Room Kinney – Thursday 11 am – 12 Help Room *

Exam #2 Tuesday, October 11 evening at 7:30-9:15 PM:

Reminders: Work and Kinetic Energy Work – Force acting and resulting in displacement along direction of force Kinetic Energy – Energy of object in motion

Two marbles, one twice as massive as the other, are dropped to the ground from the roof of a building. (Assume no air resistance.) Just before hitting the ground, the heavier marble has? A)As much kinetic energy as the lighter one. B) Twice as much kinetic energy as the lighter one. C)Half as much kinetic energy as the lighter one. D)Four times as much kinetic energy as the lighter one. E)Impossible to determine. KE = (1/2) mv 2, so if you double the mass you double the KE. Clicker Question Room Frequency BA

A 1 kg mass is moved part way around a square loop as shown. The square is 1 meter on a side and the final position of the mass is 0.5 m below its original position. Assume that g = 10 m/s 2. What is the work done by the force of gravity during this journey? A)10 J B) 5 J C) 0 J D) –10 J E) –5 J -5 J 0 J +10 J 0 J Total work done W = (-5) + (0) + (+10) + (0) = +5 J Clicker Question Room Frequency BA

The work done by the net force on a single object is equal to the change in kinetic energy of that object: Since work is a transfer of energy via a force, this is a statement of conservation of energy. Only applies if energy is not transferred to another form (e.g. a change in potential energy, heat, chemical energy, etc.) Work-Energy Principle W net = W F net = ΔKE = KE f - KE i

Push a book across a frictionless table with constant force F ext Work-Energy Principle #1: F net = F ext (since N and mg cancel) #2: F net = ma  a = F ext / m #3: W = +F ext  x  F ext = W /  x #4: v f 2 –v i 2 = 2a  x v f 2 –v i 2 = 2(F ext /m)  x v f 2 –v i 2 = 2((W/  x)/m)  x ½ m(v f 2 –v i 2 )= W ½ mv f 2 –½mv i 2 =  KE = W

A car of mass m is moving with speed v. The driver applies the brakes and the car skids to a stop. What was the magnitude of the work done by the friction force on the tires? |W fric | = F fric Δx = μ K N Δx = |ΔKE|= ½ mv 2 If the car’s speed were doubled, the distance it would skid before stopping would be: A) Doubled, B) Tripled, C) Quadrupled, D) Halved. Clicker Question Room Frequency BA μ K N Δx = ½ mv 2

PE is a stored energy associated with the position or geometry of a physical system Several varieties: gravitational, elastic, spring, …. Definition: PE is the amount of work done on a system by an external force when Kinetic Energy does not change and no heat energy flows in or out of the system. Potential Energy ΔPE = W ext when ΔKE = 0 Again, effectively a re-statement of conservation of energy.

Demonstration For a moment, they are at rest near the top (KE = 0). Later, they are moving quite fast (large KE). Where did the energy come from since energy is conserved? Gravitational Potential Energy

ΔPE grav = W ext Lift mass m at a constant speed by a height Δy F net = ma = F ext – mg=0 F ext = mg Often define PE grav = 0 when y = 0. PE is always defined relative to a “reference level” where it is zero. Gravitational Potential Energy As height increases, PE grav increases. PE grav = mgy ΔPE grav = +F ext Δy ΔPE grav = +mg Δy

E mechanical = KE + PE Conservation of Mechanical Energy: KE can change into PE and PE into KE, but the total (KE+PE) is constant for an isolated system with no lost energy (dissipation). Mechanical Energy Investigate swing:  E mechanical =  (KE + PE) = 0 E mechanical = (KE + PE) = constant

E mechanical = KE + PE = constant (isolated system, no dissipation) Consider mass m swinging attached to a string of length L. The swing is released from rest at a height h. What is the speed v of the swing when it reaches height h/2? KE = ½ mv 2 PE grav = mgy Conservation of Mechanical Energy 0