PHY PHYSICS 231 Lecture 11+12:work, energy and power Remco Zegers

PHY Previously  Work: W=Fcos(  )  x Energy transfer  Power: P=W/  tRate 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

PHY Work and energy WORK POTENTIAL ENERGY KINETIC ENERGY

PHY Mechanical Energy Mechanical energy Gravitational Potential Energy (mgh) Kinetic Energy ½mv 2

PHY quiz (for credit) A person is dragging a block over a floor, with a force parallel to the floor. After 4 meter, the floor turns rough and instead of a force of 2N and force of 4N must be applied. The force-distance diagram shows to 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

PHY 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=F  x P=W/  t=Fv Work per step: 0.60 J/kg * 60 kg=36 J Work during race: 36*(racelength(L)/steplength)=24L Power= W/  t=24L/  t=24v average= 70 so v average =2.9 m/s F=P/v so F=24 N

PHY 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 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!

PHY 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.

PHY 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

PHY Overview Newton’s second Law F=ma Work W=(Fcos  )  x Equations of kinematics X(t)=X(0)+V(0)t+½at 2 V(t)=V(0)+at Work-energy Theorem W nc =E f -E i Conservation of mechanical energy W nc =0Closed system

PHY Conservation of mechanical energy A) what is the speed of m 1 and m 2 when they pass each other? (PE 1 +PE 2 +KE 1 +KE 2 )=constant At time of release: PE 1 =m 1 gh 1 =5.00*9.81*4.00 =196. J PE 2 =m 2 gh 2 =3.00*9.81*0.00=0.00 J KE 1 =0.5*m 1 *v 2 =0.5*5.00*(0.) 2 =0.00 J KE 2 =0.5*m 1 *v 2 =0.5*3.00*(0.) 2 =0.00 J Total=196. J At time of passing: PE 1 =m 1 gh 1 =5.00*9.81*2.00 =98.0 J PE 2 =m 2 gh 2 =3.00*9.81*2.00=58.8 J KE 1 =0.5*m 1 *v 2 =0.5*5.00*(v) 2 =2.5v 2 J KE 2 =0.5*m 1 *v 2 =0.5*3.00*(v) 2 =1.5v 2 J Total= v 2 196= v 2 so v=3.13 m/s

PHY work How much work is done by the gravitational force when the masses pass each other? W=F  x=m 1 g2.00+m 2 g(-2.00)=39.2 J  Pe start -  Pe passing =( )= 39.2 J The work done by F g is the same as the change in potential energy

PHY 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? (  PE+  KE) start -(  PE+  KE) passing =W nc W nc =F friction  x=5.00*2.00=10.0 J (196.)-(  KE)=10 J  KE=29.2 J=0.5*(m 1 +m 2 )v 2 =4v 2 v=2.7 m/s

PHY A spring F s =-kx +x F s (x=0)=0 N F s (x=-a)=ka F s =(0+ka)/2=ka/2 W s =F s  x=(ka/2)*(a)=ka 2 /2 The energy stored in a spring depends on the location of the endpoint: elastic potential energy. k: spring constant (N/m)

PHY 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? (PE gravity +PE spring +KE ball ) pull =(PE gravity +PE spring +KE ball ) launch mgh pull +½kx pull 2 +½mv pull 2 = mgh launch +½kx launch 2 +½mv launch 2 0.1*9.81*0+½120(0.05) 2 +½0.1(0) 2 = 0.1*9.81*(0.05*sin(10 o ))+½120*(0) 2 +½0.1v pull =8.5E v 2 v=1.7 m/s end

PHY quiz (extra credit) 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 a)1) and 2) and 3) are true b)only 2) and 3) are true c) only 1) and 2) are true d) only 2 is true.

PHY 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

PHY Race track KE PE TME NC With friction

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PHY A swing h 30 o L=5m If relieved 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

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PHY Parabolic motion t=0 t=3t=2t=1 t=5  Where is the kinetic energy… 1)highest? 2)lowest ? Assume height of catapult is negligible to the maximum height of the stone. A E DC B And what about potential energy?

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PHY 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

PHY 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 Horizontal component of velocity at start: v x =v o cos30=34.65 m/s At highest point: only horizontal velocity v x,highest =34.65 m/s kinetic energy: 0.5mv 2 =90 J

PHY 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 W friction =F  x=(50-30)10=200 J

PHY 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: E=0.5mv 2 +mgh= 0.5mv 2 +0=0.5mv 2 At highest point: E=0.5mv 2 +mgh= 0+m*9.8*40=392m Conservation of energy: 0.5mv 2 =392m so v=28 m/s

PHY 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 initial: E=0.5mv 2 (kinetic only)=18J final: E=mgh (potential only)=9.8 J W nc =18-9.8=8.2 J kinetic energy: 0.5mv 2 potential energy: mgh g=9.8 m/s 2