Work, Energy, and Power

1. Work, Energy, and Power

2. Work • Work tells us how much a force or combination of forces changes the energy of a system. • Work is the bridge between force (a vector) and energy (a scalar). • W = F ∆r cos θ • F: force (N) • ∆r : displacement (m) • θ : angle between force and displacement

3. 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)

4. 20 N Moves 20m For a force in line with the motion: 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? The work lowering the weight is -461.5J

5. F θ If force is at an angle, only the portion of the force in the direction of the motion creates work: 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?

6. Force and direction of motion both matter in defining work! • There is no workdone by a force if it causes no displacement • Forces can do positive, negative, or zerowork. When a box is pushed on a flat floor for example • The normal forceand gravity do not work, since they are perpendicular to the direction of motion • The person pushing the box does positivework, since she is pushing in the direction of motion. • Friction does negative work, since it points oppositethe direction of motion Cos (90º) = 0 Cos (0º) =+1 Cos (180º) =-1

7. FBD W =Fcos(θ)d Fapp • 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? • 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? W =Fcos(θ)·0m = 0J mg FBD W =Fcos(θ)d Fapp cos(90º) = 0 W =Fcos(90º)·1000m = 0J mg

8. 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

9. 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 workon a system, it tries to increasemechanical energy. If an applied force does negative work, it tries to decreasemechanical energy. The two forms of mechanical energy are called potentialand kineticenergy. v = instantaneous velocity (initial or final) – work is done to accelerate objects and change their velocity

10. FBD Wj =Fcos(θ)d • 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? • How much work does gravity do when Jane lifts Tarzan? d = 9.0m Fapp Wj =mgcos(θ)d mg Wj =70kg·9.8m/s2·cos(0º)·9m = +6174J Wnet = Wj+Wg = 0 Wg = -6174J

11. 15m • 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? • How much work does friction do as Joe pushes the box? ΣF = 0 FBD W =Fapp·cos(θ)d fk Fapp W =fk·cos(θ)d ΣF = fk – Fapp = 0 W =μFN·cos(θ)d fk = Fapp W =μmg·cos(θ)d W =0.5·10kg·9.8m/s2·cos(0º) ·15m=735J Wg = -735J Wnet = Wj + Wg = 0

12. 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

13. Kinetic energy • Energy due to motion • K = ½ m v2 • K: Kinetic Energy • m: mass in kg • v: speed in m/s • Unit: Joules

14. How do we get to the kinetic from the equations we already know?

15. m=10.0g =0.010kg K=1/2mv2 v = 1.2km = 1200m • A 10.0 g bullet has a speed of 1.2 km/s. • What is the kinetic energy of the bullet? • What is the bullet’s kinetic energy if the speed is halved? • What is the bullet’s kinetic energy if the speed is doubled? K=1/2·0.01Kg·(1200m/s)2= 7200J K=1/2m(1/2v)2 K=1/4(1/2mv2) K=1/2m1/4(v)2 K=1/4·7200J = 1800J K=1/2m(2v)2 K=4(1/2mv2) K=1/2m4(v)2 K=4·7200J = 28000J

16. 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? Why did the angle not impact the work done? Because does not matter what direction we take for work. We look at the beginning energy and the final energy

17. 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 Units: Joules (J) For springs: Us = ½ k x2 k: spring force constant x: displacement from equilibrium position

18. Law of Conservation of Energy • In any isolated system, the total energy remains constant. • Energy can neither be created nor destroyed, but can only be transformed from one type of energyto another.

19. Conservation of Mechanical Energy • Total mechanical energy (kinetic + potential) of an object remains constant provided the net work done is by gravity • Conservative forces and Potential energy • Wc= -∆U • If a conservative force does positive work on a system, potential energy is lost. • If a conservative force does negative work, potential energy is gained.

20. For gravity • Wg = -∆Ug = -(mghf– mghi) • For springs • Ws = - ∆ Us = -(½ k xf2 – ½ k xi2)

21. h = 40 m Ug= 25,000J • As an Acapulco cliff diver drops to the water from a height of 40.0 m, his gravitational potential energy decreases by 25,000 J. How much does the diver weigh? W = ? m U = mgh mg= U / h W = mg W= U / h = 25000 J / 40.0 m = 625 N

22. Simulation: http://phet.colorado.edu/en/simulation/energy-skate-park • ..\..\simulations\energy-skate-park_en.jar

23. 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? Equation from previous section:

24. 38 m/s 70m 35m 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.

25. 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

26. 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. Wnet = ΔK Wnet: work due to all forces acting on an object ΔK: change in kinetic energy (Kf – Ki) W1+W2+W3+ …..Wn = ΔK

27. 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? • Did air resistance do positive, negative or zero work on the acorn? Why? W =Fapp·cos(θ)d v2f =v2i +2ad vf =√((0m/s)+2 · -9.8m/s2·-10m)= 14 m/s Fapp =mg 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 d, define down direction as 0º mg W =Fapp·cos(θ)d Fr W =Fr·cos(180º)d = -J d, define down direction as 0º

28. Wnet =ΔK Wg + Wd = Kf • 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? • What was the average force of air resistance? Wnet =Wg + Wd Wd = 1/2mv2-Wg ΔK = Kf - Ki Wd = 1/2·0.015kg·(11 m/s)2- 1.47J Wg + Wd = Kf-Ki Wd = -0.5625J Ki = 0 Wd =Fd·cos(θ)d Fd=-0.5625J / cos(180º)·10m Fd= 0.05625 N Fd=Wd/(cos(θ)d)

29. Power Rate at which work is done by a force. Rate at which energy is changing. 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.

30. SI unit for Power is the Watt. • 1 Watt = 1 Joule/s • Named after the Scottish engineer James Watt (1776-1819) 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 • 1 kWh = 1000J/s • 3600s = 3.6 x 106J

31. A 1100kg car starts from rest and accelerates for 5 seconds at 4.6m/s/s. How much power does the car generate?

32. 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 = 600s + 59s = 659s P= 219520J / 659s = 333.111 W P= W/t

33. K=1/2mv2 m = 1.0g = 0.001kg • 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. 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

34. 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

35. 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

36. 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 workequal to the magnitude of the work done by the spring. • Wapp = - Ws = ½ k x2

37. 0 Fs(N) M 200 100 M 0 Fapp -100 Ws = negative area = - ½ kx2 x -200 0 1 2 3 4 5 Fs = -kx (Hooke’s Law) Fs x(m) Wapp = F∙ ∆r cos (0) = +W Ws = -F ∙-∆r cos (180) = -W

38. Fs = -kx Ws = -1/2kx2 Fapp Fs = -k(-x) k= -2Ws / x2 It takes 180 J of work to compress a certain spring 0.10 m. a)What is the force constant of the spring? b) To compress the spring an additional 0.10 m, does it take 180 J, more than 180 J, or less than 180 J? Verify your answer with a calculation. Fs = k(x) Fs k= -2(-180 J) / (0.1m)2 = 36000 N/m Ws = -1/2k(2x)2 Ws = -1/2kx2 Ws = -1/2k4 (x)2 Ws = -1/2∙36000∙(0.2)2 = -720 J Ws = 4(-1/2k(x)2) 720 / 180 = 4

39. Ws = 1/2kx2 A vertical spring (ignore its mass) whose spring constant is 900 N/m, is attached to a table and is compressed 0.150 m. a) What speed can it give to a 0.300 kg ball when released? Fs Work is going to change the kinetic energy of the ball (Work-Energy theorm ∆r = 0.15 m W = ∆K v0 = 0 m/s W = Kf – Ki W = 1/2 mvf2 – 1/2 mvi2 1/2kx2 = 1/2 mvf2 v = √(kx2 /m) = √(900 n/m∙ (0.15m)2 /0.3kg) = 9.22 m/s

40. v = 0 m/s W = ∆K W = Kf – Ki b) How high above its original position (spring compressed) will the ball fly? F ∆r cos (180) = - 1/2 mvi2 -F ∆r = - 1/2 mvi2 mg ∆r = 1/2 mvi2 g ∆r = ½ vi2 ∆r = ½ vi2/ g = ½ (9.22 m/s)2/ 9.8 m/s2 = 4.34 m We define the direction in the direction of the applied force, so the Fspring is in opposite direction

41. Force types • Forces acting on a system can be divided into two types according to how they affect potential energy. • Conservative forces can be related to potential energy changes. • Non-conservative forces cannot be related to potential energy changes. • So, how exactly do we distinguish between these two types of forces?

42. Conservative forces • Work is path independent. • Work can be calculated from the starting and ending points only. • The actual path is ignored in calculations. • Work along a closed path is zero. • If the starting and ending points are the same, no work is done by the force. • Work changes potential energy. • Examples: • Gravity • Spring force • Conservation of mechanical energy holds!

43. Non-conservative forces • Work is path dependent. • Knowing the starting and ending points is not sufficient to calculate the work. • Work along a closed path is NOT zero. • Work changes mechanical energy. • Examples: • Friction • Drag (air resistance) • Conservation of mechanical energy does not hold!