MSU Physics 231 Fall Physics 231 Topic 5: Energy and Work Alex Brown October 2, 2015

MSU Physics 231 Fall What’s up? (Friday Sept 26) 1) The correction exam is now open. The exam grades will be sent out after that on Wednesday Oct 7. 2) Homework 04 is due Tuesday Oct 13 th and covers Chapters 5 and 6. It is a little longer that usual so you may want to start early.

MSU Physics 231 Fall

4 Key Concepts: Work and Energy Work and Energy Work and it’s association with forces  Constant forces (driving up a hill)  Variable forces (stretching a spring) Kinetic Energy  Work-Energy Theorem  Relationship to velocity Potential Energy Conservation of Mechanical Energy Power Covers chapter 5 in Rex & Wolfson

MSU Physics 231 Fall What is Energy? Motion – kinetic energy Ability to produce motion – potential energy How to we transfer energy?

MSU Physics 231 Fall Work and Energy 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 = (Fcos  )  x = F x  x Units: N·m = Joule 1 calorie = J 1 Calorie = 4184 J = 1 kcal  F Fcos  xx

MSU Physics 231 Fall Non-constant force/angle W=(Fcos  )  x: what if Fcos  is not constant while covering  x? Example: what if  or F changes while pulling the block? Area=  A= (Fcos  )  x  xx xx Fcos  x The work done is the area under the graph of Fcos  vs x W=  (  A)=total area Fcos  x F

MSU Physics 231 Fall A person drags a block over a floor with a force parallel to the floor. After 4 meters, the floor turns rough and instead of a force of 2N a force of 4N must be applied. The force-distance diagram shows the situation. How much work did the person do over 8 meter? a) 0 Jb) 16 Jc) 20 J d) 24 Je) 32 J 48 m 2N 4N 0 Force distance Work: area under F-x diagram: 4x2 + 4x4 = 24 J Example 4 m

MSU Physics 231 Fall Work and Energy Positive work is done when the angle is less than 90 degrees, energy goes into the object 1) energy can be stored (potential energy increases) 2) motion can be created (kinetic energy increases)

MSU Physics 231 Fall Work and Energy Negative work – when the angle between the force and the displacement is more the 90 degrees, energy is removed: 1) stored energy can be decreased (potential energy decreases) 2) motion can be reduced (kinetic energy decreases) Special case for friction force, the angle is 180 degrees; potential or kinetic energy is removed and heat is created No work – when the angle between the force and the displacement is equal to 90 degrees

MSU Physics 231 Fall Work and Energy Sled is pulled across a surface at constant speed. Where does the energy go in this case? Answer: two forces are acting; the energy goes into friction (the ground/sled heat up!)

MSU Physics 231 Fall v T No work is done because the force acts in a perpendicular direction to the displacement. Using the definition of work W = F (Δs) cos  because  = 90 º, then W = 0. Clicker Question: Tension and Work a) tension does no work at all b) tension does negative work c) tension does positive work A ball tied to a string is being whirled around in a circle. What can you say about the work done by tension?

MSU Physics 231 Fall Power: The rate of energy transfer Work (the amount of energy transfer) is independent of time. W=(Fcos  )  x … total over all time To measure how fast we transfer the energy we define: Power = P = (Work/time) = (W/  t) (J/s=Watt) P = (Fcos  ) ·  x/  t = (Fcos  ) · v average 1 Watt = horsepower

MSU Physics 231 Fall While running, a person dissipates about 0.60 J of mechanical energy per step per kg of body mass. A) If a 60 kg person develops a power of 70 Watt during a race (distance = L), how fast is she running (1 step=1.5 m)? B) What is the force the person exerts on the road? W=F  x P=W/  t=Fv A) Work per step:(0.60 J/kg)  (60 kg) = 36 J Work during race: (36 J) [( L ) / ( step-length )] = (36 J)  ( L) / (1.5 m) = (24L) J/m Power = W/  t = 24 L/  t = 24 v = 70 Watts v =2.9 m/s B) F=P/v so F=24 N A Runner

MSU Physics 231 Fall Net Work When many forces are acting the net work done by all of them is the sum of each term W net = (F x1 + F x2 …..)  x = = F x,net  x = (F net cos  )  x

MSU Physics 231 Fall Question A worker pushes a wheelbarrow 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)0 b)100 J c)200 J d)300 J e)500 J W net = F x,net  x = (50-30) (10) = 200 J

MSU Physics 231 Fall Example A toy-rocket of 5.0 kg, after the initial acceleration stage, the speed is constant and travels 100 m in 2 seconds. A) What is the work done by the engine? B) What is the power of the engine? A) W = (Fcos  )  h = m rocket g  h = (5.0 kg)  (9.81 m/s 2 )  (100 m) = 4905 J Force by engine must balance gravity! h=100m B) P=W/  t = 4905/2=2453 Watt (=3.3 horsepower) or P=(Fcos  )v = mg(h/t) =5.0  9.81  100/2=2453 Watt

MSU Physics 231 Fall Clicker Question: Force and Work N T mg displacement Any force not perpendicular to the motion will do work: N does no work T does positive work f k does negative work mg sin  does negative work a) one force b) two forces c) three forces d) four forces e) no forces are doing work A box is being pulled up a rough incline by a rope connected to a pulley. How many forces are doing non-zero work on the box? Mg sin   W = (Fcos  )·  x fkfk

MSU Physics 231 Fall 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 a falling object: W gravity = F g cos(0 o )  h = mg  h = mg h i – mg h 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. W = (Fcos  )·  x

MSU Physics 231 Fall (A) (B) (C) (D) (E)

MSU Physics 231 Fall Kinetic energy: Consider an object that changes speed only W = Fcos   x = (ma)  x … used Newton’s second law  x=100m t=0 V O  t=2s VfVf (E)  x = (v f 2 -v 0 2 )/2a 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 = KE f - KE i W=½m(v f 2 -v 0 2 ) = ½mv f 2 - ½mv 0 2

MSU Physics 231 Fall Conservation of energy Mechanical energy = Potential Energy + Kinetic energy ME = PE + KE Mechanical energy is conserved if: the system is closed (no energy can enter or leave) the forces are ‘conservative’ (see soon) Heat, chemical energy (e.g battery or fuel in an engine) Are sources or sinks of internal energy (not mechanical). We’re not talking about this!

MSU Physics 231 Fall Example of closed system At launch: ME = mgh + ½mv  9.81  = J At ground: ME = mgh + ½mv 2 = 0 + ½  0.2  v 2 = 0.1 v 2 J t=0 s h=35 m v=0 m/s t>0 s v at h=0? An object (0.2 kg) is dropped from a height of 35 m. Assuming no friction, what is the velocity when it reaches the ground? Conservation of ME: J = 0.1 v 2 J v = 26.2 m/s

MSU Physics 231 Fall Clicker Quiz! time energy In the absence of friction, which energy-time diagram is correct? potential energy total energy kinetic energy A time energy time B C energy

MSU Physics 231 Fall t=0 t=3t=2t=1 t=5  Where is the kinetic energy… 1)highest? 2)lowest ? AE DC B Parabolic Motion

MSU Physics 231 Fall t=0 t=3t=2t=1 t=5  Where is the potential energy… 1)highest? 2)lowest ? A EDC B Parabolic Motion

MSU Physics 231 Fall A swing h 30 o L=5m If released from rest, what is the velocity of the ball at the lowest point? (PE+KE) = constant PE release =mgh (h=5-5cos(30 o )) =6.57m J KE release =0 PE bottom =0 KE bottom =½mv 2 ½mv 2 =6.57m so v=3.6 m/s

MSU Physics 231 Fall Ball on a track A B In which case has the ball the highest velocity at the end? A) Case AB) Case BC) Same speed In which case does it take the longest time to get to the end? A) Case AB) Case BC) Same time h h end

MSU Physics 231 Fall Conservation of mechanical energy Mechanical energy = potential energy + kinetic energy In a closed system, mechanical energy is conserved ME = PE+KE = mgh + ½mv 2 = constant h=100m What about the accelerating rocket? V O =0 V=100 m/s At launch: ME = 5*9.81*0 + ½5*0 2 = 0 J At 100 m height: ME = 5*9.81*100 + ½5*100 2 = J M=5 kg We did not consider Fuel burning -Another source of energy that is not mechanical energy *

MSU Physics 231 Fall

MSU Physics 231 Fall Roller coaster KE PE TME NC With friction

MSU Physics 231 Fall 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: work done by gravitational force h=10m Using the stairs: W g = mgh f -mgh i = mg(h f -h i ) 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! Conservative Forces

MSU Physics 231 Fall 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. (In this example difference forces are applied so that the KE is the same at the end.) object on rough surface top view Non-Conservative Forces

MSU Physics 231 Fall 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 =E nc if some forces are nonconservative. E nc =positive energy dissipated by non-conservative forces (in the books notation E nc = -W f ) Example: throwing a snowball from a building taking into account air resistance

MSU Physics 231 Fall Overview Newton’s second Law F=ma Work W=(Fcos  )  x Equations of kinematics x(t)=Xo + Vo t + ½at 2 v(t)=Vo + at Work-energy Theorem E nc =ME i - ME f Conservation of mechanical energy ME i = ME f E nc =0Closed system

MSU Physics 231 Fall Clicker Quiz! time energy When there is friction, which mechanical energy-time diagram is correct? potential energy total energy kinetic energy A time energy time B C energy

MSU Physics 231 Fall 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 At ground level: ME= ½mv 2 + mgh = ½mv = ½mv 2 KE i + PE i = KE f + PE f KE = ½ mv 2 PE = mgh Conservation of energy: ½mv 2 = 392m  v 2 = 2x392 so v=28 m/s At highest point: ME= ½mv 2 + mgh = 0 + m*9.81*40 = 392m Step 1: Step 2: Step 3: Step 4:

MSU Physics 231 Fall Conservation of mechanical energy What is the speed of m 1 and m 2 when they pass each other? At time of release: PE 1i = m 1 gh 1i = 5.00*9.81*4.00= J PE 2i = m 2 gh 2i = 3.00*9.81*0.00= 0.00 J KE 1i = ½m 1 v i 2 = 0.5*5.00*(0.) 2 = 0.00 J KE 2i = ½m 1 v i 2 =0.5*3.00*(0.) 2 = 0.00 J Total= J = v 2 v = 3.13 m/s M2 3 kg 5 kg M1 4.0 m At time of passing: PE 1f = m 1 gh 1f = 5.00*9.81*2.00 = 98.0 J PE 2f = m 2 gh 2f = 3.00*9.81*2.00 = 58.8 J KE 1f = ½m 1 v 2 = 0.5*5.00*(v) 2 = 2.5v 2 J KE 2f = ½m 2 v 2 = 0.5*3.00*(v) 2 = 1.5v 2 J Total= v 2 J ME = (PE 1 +PE 2 +KE 1 +KE 2 )=constant

MSU Physics 231 Fall Friction (non-conservative) The pulley is now not frictionless. The friction force equals 5 N. What is the speed of the objects when they pass? E nc = f friction  x = 5.00*2.00 = 10.0 J v=2.7 m/s M2 3 kg 5 kg M1 4.0 m ME i = ME f + E nc = v 2 Without Friction: v = 3.13 m/s

MSU Physics 231 Fall Question! In the absence of friction, when m 1 starts to move down: 1)potential energy is transferred from m 1 to m 2 2)potential energy is transformed into kinetic energy 3)m 1 and m 2 have the same kinetic energy M2 3 kg 5 kg M1 4.0 m TRUE FALSE

MSU Physics 231 Fall Question A ball rolls down a slope as shown in the figure. The starting velocity is 0 m/s. There is some friction between the ball and the slope. Which of the following is true? h a)The kinetic energy of the ball at the bottom of the slope equals the potential energy at the top of the slope b)The kinetic energy of the ball at the bottom of the slope is smaller than the potential energy at the top of the slope c)The kinetic energy of the ball at the bottom of the slope is larger than the potential energy at the top of the slope

MSU Physics 231 Fall Quiz A ball rolls down a slope and back up a more shallow slope as shown in the figure. The starting velocity is 0 m/s. There is some friction between the ball and the slope. Which is true? h2 a)The maximum height reached on the right (h2) is the same as the height the ball started at on the left (h1) b)The maximum height reached on the right (h2) is smaller than the height the ball started at on the left (h1) c)The maximum height reached on the right (h2) is larger than the height the ball started at on the left (h1) h1

MSU Physics 231 Fall 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., the velocity is 0 m/s at that point). How much energy is dissipated by friction? a)0. b)8.2 J c)9.8 J d)18 J e)27.8 J Initial:ME = ½mv 2 (kinetic only) = ½x1x6 2 = 18 J Final: ME = mgh (potential only) = 1x9.8x1 = 9.8 J E nc = = 8.2 J kinetic energy: ½mv 2 potential energy: mgh g=9.81 m/s 2 ME i – ME f = E nc (KE i +PE i ) - (KE f +PE f )= E nc

MSU Physics 231 Fall Question An outfielder who is 2M tall 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 Two components of velocity at start: v ox = v o cos(30 o ) = 34.6 m/s v oy = v o sin(30 o ) = 20 m/s At highest point: only horizontal velocity v x = v ox = 34.6 m/s v y = 0 m/s kinetic energy: ½mv 2 = ½(0.15)(34.6) 2 = 90 J

MSU Physics 231 Fall Question An outfielder who is 2m tall throws a baseball of 0.15 kg at a speed of 40 m/s and angle of 30 degrees with the field. How high does the ball go at its highest point, ignoring friction? Initial: KE + PE = ½mv 2 + mgh = ½ (0.15)(40) 2 + (0.15)(9.81)(2) = At highest point: KE + PE = 90 + mgh = h = h h = 22.4 m ie, the ball travels 20.4 m higher than the player’s height

MSU Physics 231 Fall Work and Energy Work: W = Fcos(  )  x Energy transfer The work done is the same as the area under the graph of Fcos  versus x Power: P = W/  t 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 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 ME i = ME f Not conserved in the presence of non-conservative forces ME i = ME f + E nc

MSU Physics 231 Fall All freely falling objects fall with the same acceleration (g=9.81 m/s 2 ). Because the acceleration is the same for both, and the distance is the same for both, then the final velocities will be the same for both stones. 2 Question: Free Fall 1 a) quarter as much b) half as much c) the same d) twice as much e) four times as much Two stones, one twice the mass of the other, are dropped from a cliff. Just before hitting the ground, what is the VELOCITY of the heavy stone compared to the light one?

MSU Physics 231 Fall Consider the work done by gravity to make the stone fall distance d:  KE = W net = F d cos   KE = mgd Thus, the stone with the greater mass has the greater KE, which is twice as big for the heavy stone. Quiz: Free Fall 2 a) quarter as much b) half as much c) the same d) twice as much e) four times as much Two stones, one twice the mass of the other, are dropped from a cliff. Just before hitting the ground, what is the KINETIC ENERGY of the heavy stone compared to the light one? KE i + PE i = KE f + PE f