PHYSICS 231 Chapter 5: Energy & work! Remco Zegers 1
WORK Work: Transfer of energy Quantitatively: The work W done by a constant force on an object is the product of the force along the direction of displacement and the magnitude of displacement. W=(Fcosθ)Δx Units: =Nm=Joule 2
n An example n T θ=45 o T f k F g Δx F g The work done by the person on the suitcase: W=(Tcos45 o )Δx The work done by F g on the suitcase: W=(F g cos270 o )Δx=0 The work done by n on the suitcase: W=(F g cos90 o )Δx=0 The work done by friction on the suitcase: W=(f k cos180 o )Δx=-u k nδx The work done by the suitcase on the person: W=(Tcos225 o )Δx opposite 3
Non-constant force W=(Fcosθ)Δx: what if Fcosθ not constant while covering Δx? Example: what if θ changes while dragging the suitcase? Fcosθ Area=ΔA=(Fcosθ)Δx Fcosθ W=Σ(ΔA) The work done is the same as the area under the graph of Fcosθ versus x 4
Power: The rate of energy transfer Work (the amount of energy transfer) is independent of time. W=(Fcosθ)Δx no time in here! To measure how fast we transfer the energy we define: Power(P)=W/Δt (J/s=Watt) (think about horsepower etc). P =(Fcosθ)Δx/Δt=(Fcosθ)v average 5
Example A toy-rocket of 5.0 kg, after the initial acceleration stage, travels 100 m in 2 seconds. What is the work done by the engine? What is the power of the engine? h=100m 6
Potential Energy Potential energy (PE): energy associated with the position of an object within some system. Gravitational potential energy: Consider the work done by the gravity in case of the rocket: W gravity =F g cos(180 o )Δh=-mgΔh=-(mgh f -mgh i )=mgh i -mgh f =PE i -PE f The system is the gravitational field of the earth. PE=mgh Since we are usually interested in the change in gravitational potential energy, we can choose the ground level (h=0) in a convenient way. 7
Another rocket A toy rocket (5kg) is launched from rest and reaches a height of 100 m within 2 seconds. What is the work done by the engine during acceleration? h=100m 8
Kinetic energy Consider object that changes speed only Δt=2s Δx=100m V=0 a) W=FΔx=(ma)Δx used Newton s second law b) v=v 0 +at so t=(v-v 0 )/a c) x=x 0 +v 0 t+0.5at 2 so x-x 0 =Δx=v 0 t+0.5at 2 Combine b) & c) d) aδx=(v 2 -v 02 )/2 Combine a) & d) Rocket: W=½5(100 2-0 2 ) =25000 J!! That was missing! W=½m(v 2 -v 02 ) Kinetic energy: KE=½mv 2 When work is done on an object and the only change is its speed: The work done is equal to the change in KE: W=KE final -KE initial 9
Conservation of mechanical energy Mechanical energy = potential energy + kinetic energy In a closed system, mechanical energy is conserved * V=100 ME=mgh+½mv 2 =constant 5 kg What about the accelerating rocket? h=100m At launch: ME=5*9.81*0+½5*0 2 =0 At 100 m height: ME=5*9.81*100+½5*100 2 =29905 We did not consider a closed system! (Fuel burning) * There is an additional condition, see next lecture V=0 10
question A energy In the absence of friction, which energy-time diagram is correct? potential energy total energy kinetic energy B C energy energy time time time 11
Example of closed system A snowball is launched horizontally from the top of a building at v=12.7 m/s. The building is 35m high. The mass is 0.2 kg. Is mechanical energy conserved? V 0 =12.7 m/s h=35m d=34m 12
Previously Work: W=Fcos(θ)Δx Energy transfer Power: P=W/Δt Rate of energy transfer Potential energy (PE) Energy associated with position. Gravitational PE: mgh Energy associated with position in grav. field. Kinetic energy KE: ½mv 2 Energy associated with motion NEXT: Conservative force: Work done does not depend on path Non-conservative force: Work done does depend on path Mechanical energy ME: ME=KE+PE Conserved if only conservative forces are present KE i +PE i =KE f +PE f Not conserved in the presence of non-conservative forces (KE i +PE i )-(KE f +PE f )=W nc 13
Work and energy WORK POTENTIAL ENERGY KINETIC ENERGY 14
Mechanical Energy Mechanical energy Gravitational Potential Energy (mgh) Kinetic Energy ½mv 2 15
A running person While running, a person dissipates about 0.60 J of mechanical energy per step per kg of body mass. If a 60 kg person develops a power of 70 Watt during a race, how fast is she running (1 step=1.5 m long) What is the force the person exerts on the road? W=FΔx P=W/Δt=Fv Work per step: 0.60 J/kg * 60 kg=36 J Work during race: 36*(racelength(L)/steplength)=24L Power= W/Δt=24L/Δt=24v average= 70 so v average =2.9 m/s F=P/v so F=24 N 16
Conservative forces A force is conservative if the work done by the force when Moving an object from A to B does not depend on the path taken from A to B. Example: gravitational force Using the stairs: W g =mgh f -mgh i =mg(h f -h i ) h=10m Using the elevator: W g =mgh f -mgh i =mg(h f -h i ) The path taken (longer or shorter) does not matter: only the displacement does! 17
Non conservative forces A force is non-conservative if the work done by the force when moving an object from A to B depends on the path taken from A to B. Example: Friction You have to perform more work Against friction if you take the long path, compared to the short path. The friction force changes kinetic energy into heat. Heat, chemical energy (e.g battery or fuel in an engine) Are sources or sinks of internal energy. 18
Conservation of mechanical energy only holds if the system is closed AND all forces are conservative ME i -ME f =(PE+KE) i -(PE+KE) f =0 if all forces are conservative Example: throwing a snowball from a building neglecting air resistance ME i -ME f =(PE+KE) i -(PE+KE) f =W nc if some forces are nonconservative. W nc =work done by non-conservative forces. Example: throwing a snowball from a building taking into account air resistance 19
Overview Newton s second Law F=ma Work W=(Fcosθ)Δx Conservation of mechanical energy W nc =0 Closed system Work-energy Theorem W nc =E f -E i Equations of kinematics X(t)=X(0)+V(0)t+½at 2 V(t)=V(0)+at 20
Conservation of mechanical energy A) what is the speed of m 1 and m 2 when they pass each other? 21
work How much work is done by the gravitational force when the masses pass each other? 22
Friction (non-conservative) The pulley is not completely frictionless. The friction force equals 5 N. What is the speed of the objects when they pass? 23
A spring F s =-kx k: spring constant (N/m) +x F s (x=0)=0 N F s (x=-a)=ka F s =(0+ka)/2=ka/2 W s =F s Δx=(ka/2)*(a)=ka 2 /2 The energy stored in a spring depends on the location of the endpoint: elastic potential energy. 24
PINBALL The ball-launcher spring has a constant k=120 N/m. A player pulls the handle 0.05 m. The mass of the ball is 0.1 kg. What is the launching speed? end 25
A B h Ball on a track end h end In which case has the ball the highest velocity at the end? A) Case A B) Case B C) Same speed In which case does it take the longest time to get to the end? A) Case A B) Case B C) Same time 26
Race track With friction KE PE TME NC KE PE TME NC KE PE TME NC KE PE TME NC KE PE TME NC 27
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A swing 30 o L=5m If relieved from rest, what is the velocity of the ball at the lowest point? h 29
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Where is the kinetic energy 1) highest? 2) lowest? Assume height of catapult is negligible to the maximum height of the stone. And what about potential energy? Parabolic motion θ t=0 A t=1 B t=2 C t=3 D E t=5 31
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question An object is lowered into a deep hole in the surface of the earth. What happens to its potential energy? a) increase b) decrease c) remain constant d) cannot tell from information given e) don t know 33
question An outfielder throws a baseball of 0.15 kg at a speed of 40 m/s and angle of 30 degrees with the field. What is the kinetic energy of the baseball at the highest point, ignoring friction? a) 0 J b) 30 J c) 90 J d) 120 J e) don t know 34
question A worker pushes a sled with a force of 50 N over a distance of 10 m. A frictional force acts on the wheelbarrow in the opposite direction, with a magnitude of 30 N. What net work is done on the wheelbarrow? a) don t know b) 100 J c) 200 J d) 300 J e) 500 J 35
question Old faithful geyser in Yellowstone park shoots water hourly to a height of 40 m. With what velocity does the water leave the ground? a) 7.0 m/s b) 14 m/s c) 20 m/s d) 28 m/s e) don t know 36
question A ball of 1 kg rolls up a ramp, with initial velocity of 6 m/s. It reaches a maximum height of 1 m (I.e. velocity 0 at at that point). How much work is done by friction? a) 0. b) 8.2 J c) 9.8 J d) 18 J e) 27.8 J kinetic energy: 0.5mv 2 potential energy: mgh g=9.8 m/s 2 37
h end Consider the above rollercoaster (closed system). The cart starts at a height h. What is its velocity v at the end? Hint: consider the mechanical energy in the beginning and the end. a) v=2gh b) v=gt trip c) v= (2gh) d) v= (2gh/m) e) v=0 m/s kinetic energy: 0.5mv 2 potential energy: mgh 38