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Work, Energy, and Power Need to look at peggy’s notes and add in more look at hers and match them more closely. Add in a bunch of stuff like simple motion and some of her early slides I like. 6.1 Work Work is the exertion of a force in the direction that an object moves Work is tied to motion: No motion, no work! Equal to the magnitude of the force times the magnitude of the displacement Units are Joules (1 J = 1 N-m) For a force in line with the motion: W F d 20 N Moves 20m Ex: A weight lifter is bench pressing a barbell whose weight is 710N. He raises and lowers the barbell 0.65 m at a constant velocity. How much work is done raising the barbell? Lowering it? W F d 710N 0.65m 461.5J The work lowering the weight is -461.5J If force is at an angle, only the portion of the force in the direction of the motion creates work: F θ W ( F cos ) d Ex: How much work is done to pull a kid in a wagon for 22 meters with a force of 40N on a handle that is at a 48 degree angle from the ground? W ( F cos ) d 40 N cos( 48) 22m 588.835 J Force and direction of motion both matter in defining work! There is no work done by a force if it causes no displacement Forces can do positive, negative, or zero work. When a box is pushed on a flat floor for example The normal force and gravity do not work, since they are perpendicular to the direction of motion Cos (90º) = 0 The person pushing the box does positive work, since she is pushing in the direction of motion. Cos (0º) =+1 Friction does negative work, since it points opposite the direction of motion Cos (180º) =-1 Question: If a man holds a 50 kg box at armslength for 2 hours as he stands still, how much work does he do on the box? FBD W =Fcos(θ)d Fapp W =Fcos(θ)·0m = 0J mg Question: If a man holds a 50 kg box at arms length for 2 hours as he walks 1 km forward, how much work does he do on the box? FBD Fapp mg W =Fcos(θ)d cos(90º) = 0 W =Fcos(90º)·1000m = 0J Question: If a man lifts a 50 kg box 2.0 meters, how much work does he do on the box? FBD d = 2m Fapp mg W =Fcos(θ)d W =mgcos(θ)d W =50kg·9.8m/s2·cos(0º)·2m = 980J Work Energy Theorem Work changes mechanical energy! Theorem relates work to this change. Deals only with the work done by the NET FORCE, not individual forces. If an applied force does positive work on a system, it tries to increase mechanical energy. If an applied force does negative work, it tries to decrease mechanical energy. The two forms of mechanical energy are called potential and kinetic energy. v = instantaneous velocity (initial or final) – work is done to accelerate objects and change their velocity Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to a tree house 9.0 meters above the ground. How much work does Jane do when she lifts Tarzan? FBD d = 9.0m Wj =Fcos(θ)d Fapp Wj =mgcos(θ)d mg Wj =70kg·9.8m/s2·cos(0º)·9m = +6174J How much work does gravity do when Jane lifts Tarzan? Wnet = Wj+Wg = 0 Wg = -6174J Joe pushes a 10-kg box and slides it across the floor at constant velocity of 3.0 m/s. The coefficient of kinetic friction between the box and floor is 0.50. How much work does Joe do if he pushes the box for 15 meters? 15m ΣF = 0 FBD W =Fapp·cos(θ)d fk Fapp ΣF = fk – Fapp = 0 fk = Fapp W =fk·cos(θ)d W =μFN·cos(θ)d W =μmg·cos(θ)d W =0.5·10kg·9.8m/s2·cos(0º) ·15m=735J How much work does friction do as Joe pushes the box? Wg = -735J Wnet = Wj + Wg = 0 A father pulls his child in a little red wagon with constant speed. If the father pulls with a force of 16 N for 10.0 m, and the handle of the wagon is inclined at an angle of 60º above the horizontal, how much work does the father do on the wagon? ΣF = 0 θ=60º 10m W =Fapp·cos(θ)d W =16N·cos(60º) ·10m = 80J Kinetic energy Energy due to motion 2 K = ½ m v K: Kinetic Energy m: mass in kg v: speed in m/s Unit: Joules How do we get to the kinetic from the equations we already know? 2 2 KE mv 1 2 Fd m(a t ) 1 2 2 Fd m(at ) 1 2 W Fd v at F ma Fd mad d at 1 2 Fd mv 1 2 2 W KE f KEo 2 Fd ma( at ) 1 2 2 2 A 10.0 g bullet has a speed of 1.2 km/s. What is the kinetic energy of the bullet? m=10.0g =0.010kg v = 1.2km = 1200m K=1/2mv2 K=1/2·0.01Kg·(1200m/s)2= 7200J What is the bullet’s kinetic energy if the speed is halved? K=1/2m(1/2v)2 K=1/2m1/4(v)2 K=1/4(1/2mv2) K=1/4·7200J = 1800J What is the bullet’s kinetic energy if the speed is doubled? Ex: A 58 kg skier is coasting at 3.6 m/s down a 25o slope. The slope suddenly flattens out and he slows down to 1.2 m/s. How much work did the snow do on the skier? W KE f KE0 1 2 mv2f 1 2 mv02 0.5 58kg((1.2 m ) 2 (3.6 m ) 2 ) 334.08 J s s Why did the angle not impact the work done? 6.3 Work done by Gravity Gravitational Potential Energy – energy due to an objects position relative to the earth The object has the potential to do work if it can fall because of gravity W Fd mad Work gravity PE m g h Units: Joules (J) 6.4 Conservation of Mechanical Energy Total mechanical energy (kinetic + potential) of an object remains constant provided the net work done is by gravity KEo PEo KE f PE f Ex: One of the fastest roller coasters in the world is the Magnum at Cedar Point. It includes a drop of 59.4 meters. Neglecting friction and starting with a speed at the top of 0 m/s, how fast is the coaster going at the bottom of the hill? KEo PEo KE f PE f 1/ 2mv2f mghf 1/ 2mv02 mgh0 v f v02 2 g (h0 h f ) (0 m s) 2 2 9.8 m s 2 (59.4m 0m) 34.121m / s Equation from previous section: v v 2ax 2 2 0 Ex: A motorcycle is trying to leap across a canyon by driving off the higher cliff at 38 m/s. Find the impact velocity at which the cycle strikes the ground at the top of the far side of the cliff. 38 m/s 70m 35m 1/ 2mv2f mghf 1/ 2mv02 mgh0 v f v02 2 g (h0 h f ) (0 m s) 2 2 9.8 m s 2 (70m 35m) 26.160m / s 38m/s θ 26.16m/s a2 + b2 = c2 c2=√( a2 + b2) c2=√( (38m/s)2 + (26.16m/s)2) = 46.13m/s A=O tan θ θ = tan-1(O/A) θ = tan-1((26.16m/s)/(38m/s)) = 34.54º Work Energy Theorem Result of doing work is a change in kinetic energy. Theorem relates work to this change. Deals only with the work done by the NET FORCE, not individual forces. The net work due to all forces equals the change in the kinetic energy of a system. W1+W2+W3+ …..Wn = ΔK Wnet = ΔK Wnet: work due to all forces acting on an object ΔK: change in kinetic energy (Kf – Ki) A 15-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. What would the speed of the acorn have been if there had been no air resistance and determine if 11.0m/s is real world velocity or not? W =Fapp·cos(θ)d v2f =v2i +2ad vf =√((0m/s)+2 · -9.8m/s2 · 10m)= 14 m/s W =0.015kg·9.8m/s2cos(0º) · 10m = 1.47J KE=0.5mv2 v =√ (KE/0.5m)= √ (1.47J/(0.5·0.015kg))= 14m/s mg d, define down direction as 0º Did air resistance do positive, negative or zero work on the acorn? Why? F W =Fapp·cos(θ)d Fapp = mg r W =Fr·cos(180º)d = -J d, define down direction as 0º A 15-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. How much work was done by air resistance? Wnet = ΔK Wg + Wd = Kf Wnet = Wg + Wd ΔK = Kf - Ki Wg + Wd = Kf-Ki Wd = 1/2mv2- Wg Wd = 1/2·0.015kg·(11 m/s)2- 1.47J Wd = -0.5625J Ki = 0 What was the average force of air resistance? Wd =Fd·cos(θ)d Fd=Wd/(cos(θ)d) Fd=-0.5625J / cos(180º)·10m Fd= 0.05625 N 6.7 Power Rate at which work is done by a force. Rate at which energy is changing. W F d Power t t Net force can sometimes be found using Newton’s 2nd law, F=ma. If the object is moving up or down with constant speed, the force is the weight. SI unit for Power is the Watt. 1 Watt = 1 Joule/s Named after the Scottish engineer James Watt (17761819) who perfected the steam engine. British system horsepower 1 hp = 746 W The kilowatt-hour is a commonly used unit by the electrical power company. Power companies charge you by the kilowatt-hour (kWh), but this not power, it is really energy consumed. 1 kW = 1000 W 1 h = 3600 s 6 1 kWh = 1000J/s • 3600s = 3.6 x 10 J Ex: A 1100kg car starts from rest and accelerates for 5 seconds at 4.6m/s/s. How much power does the car generate? F ma 1100kg 4.6m / s 2 5060 N d 1 / 2at 2 .5 4.6m / s 2 (5s) 2 57.5m W F d 5060 N 57.5m Power 58190 J / s t t 5s A record was set for stair climbing when a man ran up the 1600 steps of the Empire State Building in 10 minutes and 59 seconds. If the height gain of each step was 0.20 m, and the man’s mass was 70.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower. H of 1 step is 0.2 m h = 0.2m · 1600 steps = 320 m W =Fapp·cos(θ)d Fapp =mg W =mg·cos(θ)d W =70kg·9.8m/s2·cos(0º) · 320m = 219520 J t = 10min · 60s = P= 219520J / 659s P= W/t 600s + 59s = 659s = 333.111 W Calculate the power output of a 1.0 g fly as it walks straight up a window pane at 2.5 cm/s. The window pane is 0.4 meters tall. K=1/2mv2 m = 1.0g = 0.001kg v = 2.5cm/s = 1 m / 100cm = 0.025m/s K=1/2 · 0.001 kg (0.025m/s)2 = 3.13 x 10 -7 J KE= W = 3.13 x 10-7 J v = d/t t=v/d t=0.4 m / 0.025m/s = 16s P= W/t P= 3.13 x 10-7 J / 16s = 1.956x10-8 W Constant force and work The force shown is a constant force. W = F·Δd can be used to calculate the work done by this force when it moves an object from xa to xb. The area under the curve from xa to xb can also be used to calculate the work done by the force when it moves an object from xa to xb Area of a square = l·w W = l·w =F·Δd = area under the curve F(x) Δd F x xa xb The force shown is a variable force. W = F·Δd CANNOT be used to calculate the work done by this force! The area under the curve from xa to xb can STILL be used to calculate the work done by the force when it moves an object from xa to xb Area of a square = l·w Area of a triangle = 0.5bh W = l·w + 0.5bh =F·Δd = area under the curve F(x) x xa xb When a spring is stretched or compressed from its equilibrium position, it does negative work, since the spring pulls opposite the direction of motion. Ws = - ½ k x2 Ws: work done by spring (J) k: force constant of spring (N/m) x: displacement from equilibrium (m) The force doing the stretching does positive work equal to the magnitude of the work done by the spring. Wapp = - Ws = ½ k x2 0 F(N) M 200 100 M 0 -100 x -200 0 1 Fs Ws = negative area = - ½ kx2 Fs = -kx (Hooke’s Law) 2 x(m) 3 4 5 Cont with slide 25