What is the work done by the two x
Transcription
What is the work done by the two x
What is the work done by the two forces together in moving the object a distance Δx to the right as shown in the diagram? The magnitude of each force is F and the directions of the two forces are 90o apart. A) 2FΔx B) FΔx C) 2FΔx Fnet F F Δx x ( ) D) 1+ 2 FΔx E) None of these. Force = constant, Δx is in a straight path: Work = component of the net force in the direction of motion times displacement Δx: Wnet = ( Fnet )x Δx = FΔx L21 M 10/13/14 a+er lecture 1 Assignments For this week and next: • You should have read up through Ch. 6 and Ch. 7 of Wolfson and Prof. Dubson’s notes. • Now do: CAPA 7 and HW 7 this week. • Remember: Midterm 2 will be Thursday of next week. Today: • Continue: Energy, Work, Work-KE Principle, Conservative Forces. • Introduce: Potential Energy, Conservation of Energy. L21 M 10/13/14 a+er lecture 2 Last Time: Work, Kinetic Energy 1. Definition: Energy is the scalar conserved quantity that obeys the first law of thermodynamics: W (+Q) = ΔU (Unit = joule or J) 2. Definition: Work done by a force F : f WF = ∫ F ⋅dr (in some circumstances: F ⋅ Δr ) i 3. Theorem: Work-KE Principle: Wnet = WFnet = ΔKE (point-like object, no internal moving parts) 4. Definition: A conservative force is such that the Work done by the force is independent of the path taken in applying the force. In this case, a “Potential Energy” for the force can be defined. L21 M 10/13/14 a+er lecture 3 Consider a force that varies with position according to this graph: WF = Work done by the force = area under the F vs x graph 1 = F0 x for a spring this is 2 1 1 2 = (kx)x = kx 2 2 What force do we know has a magnitude proportional to displacement? Hooke’s Law: Fspring = kx ⇒ Wspring 1 2 = kx 2 We’ll see this is related to elastic potential energy. L21 M 10/13/14 a+er lecture 4 A hockey puck sliding on an ice rink is moving at 1 m/s when it slides onto a carpet that someone left on the ice. The puck comes to rest after moving 1m on the carpet. How far along the carpet would the puck go, if its initial speed was 2m/s? A) 1.5 m B) 2 m C) 3 m D) 4 m E) Impossible to determine N f = µk N W-KE prinl: mg L21 M 10/13/14 a+er lecture Δx = 4m Wnet = W friction = − f Δx = − µ k mgΔx 1 2 = ΔKE = KE f − KEi = 0 − mvi 2 1 2 vi2 µ k mgΔx = mvi ⇒ Δx = 2 2 µk g 5 Potential Energy PE is a kind of stored energy associated with the position or geometry of a physical system. It is typically defined for a particular force. Several varieties: gravitational, elastic, …. Definition: PE is the amount of work done on a system by an external force when KE does not change and no heat flows (no friction): ΔPE = Wext when ΔKE = 0 Units: J L21 M 10/13/14 a+er lecture 6 Potential Energy ΔPE = Wext when ΔKE = 0 Form of Gravitational Potential Energy ΔPEgrav Lift mass m at a constant speed a height Δy 0 0 = ma = Fext – mg Fext = mg L21 M 10/13/14 a+er lecture ΔPEgrav = Wext = +Fext Δy = mg Δy PEgrav = mgy If we decide that PEgrav = 0 when y = 0. PE is always defined relative to a “reference level” where it is zero. 7 Potential Energy ΔPE = Wext when ΔKE = 0 Elastic Potential Energy PEelastic of a spring: PEelastic = Wext 1 2 = kx 2 If we decide that PEelastic = 0 when x = 0 (spring is relaxed). PE is always defined relative to a “reference level” where it is zero. L21 M 10/13/14 a+er lecture 8 Conservation of “Mechanical” Energy KE = (1/2) mv2 PEgrav = mgy Emechanical = KE + PE = constant (isolated system – i.e., no work done on the system by outside force or by the system outside of it, no dissipation – i.e., no heat loss) Conservation of Mechanical Energy: KE can change into PE and PE into KE, but the total is constant for an isolated sytem with no dissipation. Investigate swing: http://phet.colorado.edu/sims/pendulum-lab/pendulum-lab_en.html L21 M 10/13/14 a+er lecture 9 Conservation of “Mechanical” Energy KE = (1/2) mv2 PEgrav = mgy Emechanical = KE + PE = constant (isolated system, no dissipation) 1: A swing consists of a mass m attached to a massless string of Example length L. The swing is released from rest at a height h above the lowest point. What is the speed v of the swing when it reaches height h/2? (No friction) (1) Choose a reference level where PE = 0. (2) Ei = E f → KEi + PEi = KE f + PE f PE = 0 L21 M 10/13/14 a+er lecture 0 1 2 h + mgh = mv + mg 2 2 v = gh 10 Conservation of “Mechanical” Energy KE = (1/2) mv2 PEgrav = mgy Emechanical = KE + PE = constant (isolated system, no dissipation) 1: A swing consists of a mass m attached to a massless string of Example length L. The swing is released from rest at a height h above the lowest point. What is the speed v of the swing when it reaches height h/2? (No friction) (1) Choose a reference level where PE = 0. PE = 0 (2) Ei = E f → KEi + PEi = KE f + PE f 0 L21 M 10/13/14 a+er lecture 1 2 h + 0 = mv − mg 2 2 v = gh 11 KE = (1/2) mv2 PEgrav = mgy Conservation of “Mechanical” Energy Emechanical = KE + PE = constant (isolated system, no dissipation) A swing consists of a mass m attached to a massless string of length L. The pendulum is released from rest at a height h above the lowest point. What is the speed v of the pendulum when it reaches its lowest point, h = 0? (No friction) A) gh B) 2 gh PE = 0 L21 M 10/13/14 a+er lecture C) g h D) 2gh KEi + PEi = KE f + PE f 0 1 2 + mgh = mv + 0 2 v = 2gh 12 General Statement of Conservation of Energy If there is heat transfer, but the system is isolated. KE + PE + Etherm = constant (isolated system) thermal energy Note: Thermal energy is the negative of the work done by friction, which is normally negative because it points opposite to the direction of motion: Ethermal = -Wfrict L21 M 10/13/14 a+er lecture 13 A projectile is fired with an initial speed vo at an angle θ from the horizontal. What is the KE of the projectile when it is on the way down at a height h above the ground? (Assume no air resistance.) PE = 0 Ei = E f A) (1/2)mvo2 + mgh KEi + PEi = KE f + PE f B) mgh 1 2 mv0 + 0 = KE f + mgh 2 1 2 KE f = mv0 − mgh 2 C) (1/2)mvo2 – mgh D) Impossible to tell. L21 M 10/13/14 a+er lecture 14 A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. How is the initial height h related to the speed at the bottom of the ramp? PE = 0 v2 A) h = 2g B) h = 2gv C) h = g / v v2 D) h = g L21 M 10/13/14 a+er lecture Ei = E f KEi + PEi = KE f + PE f 1 2 0 + mgh = mv + 0 2 v2 h= 2g 15