WORK & ENERGY

1. WORK & ENERGY

2. WORK & ENERGY • Work, in a physics sense, has a precise definition, unlike the common use of the word. When you do your home “work” you probably, from a physics stand point, did no work at all ! • Work is defined as force applied in the direction of the motion multiplied times the distance moved. • When work is done by moving an object in a horizontal direction, work equals the applied force times the cosine of the angle of the applied force times the distance the object is moved. • W = F (cos ) x s, (s stands for distance) • Work is a scalar quantity (it has no direction). The sign of a work quantity (positive or negative) indicates the direction of energy flow as into or out of a system but does not give it a direction as in a vector quantity.

3. Energy Units work = force x distance - MKS 1 joule = 1 newton x 1 meter CGS 1 erg = 1 dyne x 1 centimeter Eng 1 foot - pound = 1 pound x 1 foot

4. WORK & ENERGY • The terms work and energy are interchangeable. Energy is defined as the ability to do work. • Kinds of work and energy • (1) mechanical work – work done by applying a force over a distance • (2) work of friction – work required to overcome friction • (3) gravitational potential energy – energy needed to lift an object against the force of gravity • (4) elastic potential energy – the energy stored in a compressed or stretched spring • (5) kinetic energy – energy an object has because of its motion (velocity)

5.  Vertical component = Applied force x sin Horizontal component = Applied force x cos   W = F x s cos Mechanical Work Applied force Vertical component  Horizontal component Distance moved The horizontal force component is in the direction of the motion Work = force in direction of motion x distance moved

6. W = F x s f f Work of Friction FORCE OF FRICTION = 0 WORK DONE = 0 FORCE OF FRICTION > 0 WORK DONE = FFRICTION x DISTANCE Recall: Ffriction = coefficient of friction x Fnormal and on a horizontal surface: Fnormal = weight of object = mass x gravity

7. GRAVITATIONAL POTENTIAL ENERGY • When an object is lifted, work is done against the force of gravity (the weight of the object). • Since weight is a force and the height to which an object is lifted is a distance, then force times distance equals work done. • Weight of an object can be calculated using mass time gravity. When objects are lifted near the surface of the earth, gravity is assumed to be constant at 9.8 m/s2 (32 ft/s2). • If object are lifted well beyond the earth’s surface gravity diminishes to progressively smaller values and the work done in the lifting becomes less and less.

8.   P.E. = mg h GRAVITATIONAL POTENTIAL ENERGY • Potential energy change equals weight times change in height. • Weight equals mass times gravity • Potential energy change equals mass times gravity times height (distance lifted) Gravitational Potential Energy

9. POTENTIAL ENERGY MEASUREMENT OF POTENTIALENERGY IS RELATIVE Who are they kidding ?? If this point was used as reference, One, Two and Three would all have negative potential energies. Three I sure am ! Two is at a higher potential energy than One but lower than Three. Two’s potential energy is negative relative to Three’s and positive relative to One’s. Boy Two! You’re at a High potential Two One

10. Gravity An Inverse Square Law 16.7 lbs scale 1/9 wt Three Radius of Earth = 12000 miles 37.5 lbs ¼ wt scale Two Radius of Earth = 8000 miles 150 lbs Normal wt scale Radius of Earth = 4000 miles

11. Calculating Work in Different Gravitational Fields • Potential energy changes are different in different gravitational field because the value of g changes. • As seen in the previous slide, at an altitude of one earth radii above the earth (4000 miles) gravity is ¼ of normal gravity (1/4 x 9.8 m/s2 = 2.45 m/s2). At two earth radii altitude, gravity is 1.09 m/s2. • An object of mass 10 kg is lifted 5 meters on earth. The work done (potential energy increase) is (P.E. = mgh) 10 kg x 9.8 m/s2 x 5 m = 490 joules. • At one earth radii, work done is 10 kg x 2.45 m/s2 x 5 m = 122.5 joules (1/4 of the work done in lifting the same object on earth) • At two earth radii above the earth (8000 miles altitude) the work done on the same object is 10 kg x 1.09 m/s2 x 5 m = 54.4 joules (1/9 of the work required to lift the object on earth)

12. KINETIC ENERGY • Kinetic energy is the energy of motion. In order to possess kinetic energy an object must be moving. • As the speed (velocity) of an object increases its kinetic energy increases. The kinetic energy content of a body is also related to its mass. The most massive objects at the same speed contain the most kinetic energy. • Work = force x distance (W = F x s ) • Recall that F = mass x acceleration (F = m x a) • Therefore: Work = m x a x s • Also, for an object initially at rest, recall that acceleration equals the final velocity squared divided by twice the distance traveled: a = v2 / (2 s) • Work = m (v2 / (2 s)) s, canceling out the distance term (s) gives, Work = (m v2 ) / 2 or 1/2 m v2 • Since the object is in motion, the work content is called kinetic energy and therefore: K.E. = 1/2 m v2

13. Kinetic Energy & Motion High kinetic energy. High velocity ! Kinetic energy = 0 No motion !

14. ELASTIC POTENTIAL ENERGY • Elastic potential energy refers to the energy which is stored in stretched of compressed items such as springs or rubber bands. • The elongation or compression of elastic bodies is described by Hooke’s Law. This law relates the force applied to the elongation or compression experienced by the body. • In plain words, Hooke’s Law says, “the harder you pull on a spring, the more it stretches”. This relationship is given by the equation: F = k x. • F is the applied force, k is a constant called the spring constant or Hooke’s constant and x is the elongation of the spring. • Springs with large k values are hard to stretch or compress such as a car spring. Those with small constants are easy of stretch or compress such as a slinky spring.

15. 200 grams 400 grams 600 grams Hooke's Law F = - k x \ Elongation of spring FORCE(N) Slope = spring constant ELONGATION (M)

16. Constant force F o r c e (N) Constant force Distance (M) x Distance moved Work & Force vs Distance Graphs Work = force x distance Force x distance equals area under the graph Work = area under a force versus distance graph

17. Work to Stretch a Spring Area under the graph gives the work to stretch the spring. Work needed to stretch the spring to x2 is ½ F2 times x2 Work needed to stretch the spring to x1 is ½ F1 times x1 FORCE(N) F1 Work needed to stretch the spring from x1 to x2 is (½ F2 times x2) – ( ½ F1 times x1) F1 Since F = kx and W = ½ Fx, W = ½ (kx) x or W = ½ kx2 and work from x1 to x2 is given by: W = ½ k (x22 – x12) X1 X2 ELONGATION (M)

18. 2 K.E. = 1/2 mv  W = F x d cos 2 W = 1/2 k x sp Work & Energy Equations Work Kinetic Energy Gravitational Potential Energy P.E. = mgh Work Stored in a Spring

19. WORK - Finding Area Under a Non linear Force vs Displacement Curve BOX METHOD F O R C E (N) WORK = AREA UNDER THE CURVE W =  F  X (SUM OF THE BOXES) AS THE NUMBER OF BOXES INCREASES, THE ERROR DECREASES! AREA MISSED - INCREASING THE NUMBER BOXES WILL REDUCE THIS ERROR! WIDTH OF EACH BOX =  X X1 X2 DISPLACEMENT (M)

20. Finding Area Under Curves Mathematically • Areas under force versus distance graphs (work) can be found mathematically. The process requires that the equation for the graph be known and integral calculus be used. • Recall that integration is also referred to as finding the antiderivative of a function. • The next slide reviews the steps in finding the integral of the basic function, y = kxn.

21. n Y = k X n n + 1 n   d y = k X dx = k X + C --------------- n + 1 INTEGRATION – THE ANTIDERIVATIVE INTEGRATION IS THE REVERSE PROCESS OF FINDING THE DERIVATIVE. IT CAN ALSO BE USED TO FIND THE AREA UNDER A CURVE. THE GENERAL FORMAT FOR FINDING THE INTEGRAL OF A SIMPLE POWER RELATIONSHIP, Y = KXn ADD ONE TO THE POWER n + 1 DIVIDE THE EQUATION BY THE N + 1 + C ---------------  is the symbol for integration n + 1 ADD A CONSTANT

22. 3 3 , dy = 5 X dX Y = 5 X 4  d y = = 5 X --------------- 4 n n + 1 n   d y = k X dx = k X + C --------------- n + 1 APPLYING THE INTEGRAL FORMULA GIVEN THE EQUATION FORMAT TO FIND THE INTEGRAL 3 + 1 3 5 X + C --------------- 3 + 1 Integration can be used to find area under a curve between two points. Also, if the original equation is a derivate, then the equation from which the derivate came can be determined.

23. 4 = 5 X + C --------------- 4 4 4 = 5 (5) 5 (2) + C + C --------------- --------------- 4 4 APPLYING THE INTEGRAL FORMULA Find the area between x = 2 and x = 5 for the equation y = 5X3. First find the integral of the equation as shown on the previous frame. The integral was found to be 5/4 X4 + C. The values 5 and 2 are called the limits. each of the limits is placed in the integrated equation and the results of each calculation are subtracted (lower limit from upper limit) 5 Area 2 Area = 761.25 -

24. x2 | x1 Finding the Equation for Work Stored in Spring using Integration Hooke’s Law F = kx Work =  Fdx =  kx dx Work = k x1+1/ (1+1) = k x2 / 2 + C W = ½ k x2 Note that the work equation is the same as that found by area under the curve methods used in previous slides

25. Conservation of Energy • A most fundamental law of physics is the “Law of Conservation of Energy”. It is the basis upon which much of physics is built. • “Energy (ability to do work) cannot be created or destroyed, only changed in form”. This means that heat can be converted of electricity, electricity can be converted to motion, motion can be converted to heat, etc. In each and every case, all energy is conserved and can be accounted for as equal before and after the process. • More fundamentally, potential energy can be converted to kinetic energy, kinetic energy to work of friction, elastic energy to kinetic energy, etc. all without net energy loss.

26. Conservation of Energy “The energy content of the universe is constant” Energy change for a falling stone Stone initially at rest (height is greatest velocity is zero). Potential energy is maximum. Kinetic energy is zero. Stone is half way to the ground. Potential energy is 1/2 maximum. Kinetic energy is 1/2 maximum . Stone is just about to hit the ground. Potential energy is zero. Kinetic energy is maximum . All potential energy has been converted to kinetic energy. ground

27. Potential Energy Kinetic Energy PE = mg h High potential energy Low kinetic energy h KE = ½ mv2 Low potential energy High kinetic energy

28. Power • Power is work divided by the time required to perform the work. If two different energy sources do the same quantity of work, the one requiring the least time is the more powerful.Power can be measured in watts (joules / second) or horsepower (550 ft lbs / second). • Power = work / time • Work = force x distance • Power = (force x distance) / time • Since distance divided by time equals velocity • Power = force x velocity P = W / t P = F x v

29. Power Power = work / time power units MKS 1 watt = 1 joule / 1 second Eng 1 horsepower = 550 foot pounds / 1 second 1 horsepower = 746 watts

30. 500 N 420 N 5 m Work & Energy ProblemsA horizontal force pulls a box 5 meters across a floor with a force of 420 N. The box weighs 500 N. How much work is done ? Work = F cos  x s Since the weight of the box is not the applied force it is not related to the work done. Horizontal means that = 00, cos 00 = 1.0 W = 420 N x 1.0 x 5 m = 2100 joules (Recall : 1 j = 1 N x 1 m) or 2.1 kilojoules

31. 60 kg Ff ? N 12 m Work & Energy ProblemsA 60 kg box is pushed across a floor with a coefficient of sliding friction equaling 0.30. If the box moves 12 meters at constant speed, how much work is done ? Constant speed means no acceleration therefore the net force must be zero. The applied force must equal the force of friction w = mg w = 60 x 9.8 = 588 N Recall: Ffriction = coefficient of friction x Fnormal and on a horizontal surface: Fnormal = weight of object = mass x gravity Ff = 0.30 x 588 N = 176 N Work = F cos  x s,  = 00 W = 176 N x (1.0) x 12 = 2112 j

32. 1.5’ 7.0 ft 5’6’’ Work & Energy ProblemsHow much work is needed to lift a 100 lb barbell from the floor 1.5 feet over the head of a 5ft 6 inch man? Lifting always involves changing gravitational potential energy. P.E. = mgh weight is m x g and is expressed in pounds. P.E. = mgh P.E. = 100 lbs x 7 ft = 700 ft-lbs

33. 8 m h 200 Work & Energy ProblemsA 20.0 kg crate is pulled 8.00 meters up a frictionless incline with a 200 angle. How much work is done ? The crate is actually being lifted against gravity. Although it is pulled 8 m it is lifted only the vertical distance h. h = 8 x sin 200 = 2.74 m P.E. = mgh P.E. = 20 kg x 9.8 m/s2 x 2.74 m P.E. = 536 j How much work is used If the force of friction Against the crate is 10.0 N ? The crate is moved 8.00 meters against friction forces. Work of friction = Ff x s, Wf = 10.0 N x 8.00 m = 80j Work total = P.E. + Wf Work total = 536 j + 80 j = 616 j

34. 12 ft h 300 Work & Energy ProblemsA box slides down a frictionless incline 12 feet long with an angle of 300. What is its speed at the bottom? The energy in the box is being converted from potential (it is elevated) to kinetic as it slides. Conservation of energy tells us that all potential becomes kinetic. P.E.box = K.E.box mgh = ½ mv2 , since mass appears on both sides it can be divided out leaving: gh = ½ v2, rearranging the equation gives: v = (2 x g x h)1/2 h is the vertical height = 12 sin 300 = 12 x 0.5 = 6 ft v = (2 x 32 ft/s2 x 6 ft)1/2 = 19.6 ft/s

35. 10 ft h 300 Work & Energy ProblemsA 2.0 slug crate is pushed with a force of 100 lbs, 10 ft up an incline of 300 which has a coefficient of friction of 0.10. Find the speed of the crate. Conservation of energy tells us that all work in must equal all work out 100 lbs Work in = work to push the crate Work out = P.E.crate + Wf + K.E.crate • Wpush = F x s = 100 lbs x 10 ft = 1000 ft lbs • P.E. = mgh = 2.0 slugs x 32 ft/s2 x (10 sin 300) ft = 320 ft lbs • Wf = Ff x s= FN s =  (mg x cos 300) s • Wf = 0.10 x 2.0 slug x 32 ft/s2 x 0.866 x 10 ft = 55.4 ft lbs • K.E. = ½ mv2 = ½ (2) v2 = v2 • Work out = P.E.crate + Wf + K.E.crate • = 320 + 55.4 + v2 • v = (1000 – 320 – 55.4)1/2 = 25 ft/s

36. Tension (T) Vertical component  Horizontal component Work & Energy ProblemsA boy pulls a 30 lb cart 20 feet with a rope making a 600 angle with the horizontal. The tension is the rope is 25 lbs. Disregard friction. How much work is done ? Recall that only the component of the applied force in the direction of the motion does work. Work = F cos  x s W = 25 lbs x cos 600 x 20 = 43.3 ft lbs How much work is done if  = 0.2 and the cart moves at constant speed? Upward pull of rope = T sin 600 Pup = 25 lbs x 0.5 = 12.5 lbs FN = w – Pup = 30 – 12.5 = 17.5 lbs Wf = Ff x s= FN s Wf = 0.2 x 17.5 lbs x 20 ft = 70 ft lbs Recall: Wf = Ff x s= FN s The normal force of the cart is reduced by the upward pull of the rope.

37. Work & Energy ProblemsElectrons in a TV tube have a mass of 9.11 x 10-28 grams and a velocity of 3 x 107 m/s. What is there kinetic energy? 3 x 107m/s e K.E. = ½ mv2 Using MKS units (meters, kilograms, seconds), 9.11 x 10-28 grams = 9.11 x 10-31 kg. K.E. = ½ (9.11 x 10-31 kg)(3 x 107 m/s)2 K.E. = 4.1 x 10-16 joules / electron

38. 2.0 m Work & Energy ProblemsA bug crawls up a flight of stair 2.0 meters high in 5.0 minutes. His mass is 7.0 grams. What is his power output? P = W / t The bug is lifting himself P.E. = mgh Using the CGS system 2.0 meters = 200 cm 5.0 minutes = 300 seconds W = P.E. = 7.0 g x 980 cm/s2 x 200 cm W = 1.37 x 106 ergs 1 joule = 10,000,000 ergs (107) W = 0.137 joules P = 0.137 j / 300 sec = 4.6 x 10-4 watts

39. 5.0 cm 100 grams 2.0 cm 7.0 cm Work & Energy ProblemsA spring is stretched 5.0 cm when a mass of 100 grams is hung on it. How much work is needed to spring the same spring from 2.0 cm to 7.0 cm ? Hooke’s Law F = - kx k = F/x, F = mg k = (0.1 kg x 9.8 m/s2) / 0.05 m k = 19.6 N/m Work =  Fdx W = ½ k (x22 – x12) W = ½ (19.6 N/m)(0.0702 – 0.0202 ) 0.0441 joules

40. Work & Energy ProblemsThe spring in the previous problem is compressed 3.0 cm lying on the horizontal. It is released against a 50.0 gram toy cart. What is the speed of the cart leaving the spring? v Work stored in spring is released as kinetic energy to the cart Wspring =  K.E. of cart 0.0441 joules = ½ mv2 50.0 grams = 0.050 kg (MKS) 0.0441 joules = ½ (0.050 kg) v2 v = (0.0441 / 0.025)1/2 = 1.33 m/s or 133 cm/sec

41. 3000 lbs 30 mph h d 150 Work & Energy ProblemsA 3000 lb car ascends a 15 0 hill at a constant speed of 30 mph. What is the power output of the car ? (Neglect friction) 30 mph = (30 x 5280) / 3600 = 44 ft /s In 1.0 second, distance = 44 ft Work in = Work out Work in = lifting car +  K.E. of car Wf = 0 P.E. = mgh ,w = mg h = d x sin 15 0 P.E. = 3000 lbs x 44 ft x sin 15 0 P.E. =34,200 ft lbs K.E. = ½ mv2 At constant speed, K.E.=0 Work in = lifting car + K.E. of car W = 34,200 + 0 = 34,200 ft lbs P = W / t = 34,200 ft lbs / 1 sec 1 hp = 550 ft lbs / sec 34,200 / 550 = 62.2 hp

42. Now it's time for you to try some problems on your own ! The problems are similar to the ones which have been solved so look back and review the appropriate problem if you get stuck !

43. Find the power output of an 80 lb girl who climbs a 12 ft rope in 7.5 seconds ? (A) 0.13 hp (B) 0.23 hp (C) 2.0 hp (D) 960 hp Click here for answers What is the velocity of a car with kinetic energy of 360 kj ? The mass of The car is 5.0 metric tons. (A) 11.9 m/s (B) 36 m/s (C) 1.2 m/s (D) 24 m/s A mass of 16 slug is elevated 20 feet. What is its change in potential energy? (A) 510 watts (B) 320 joules (C) 320 ft lbs (D) 5120 ft lbs A spring is stretched from its normal length by 30 mm using a force of 0.40 N. How energy is stored in the spring? (A) 6.0 x 10-3 j (B) 3.0 x 10-5 j (C) 7.2 x 10-2 j (D) none of these A man pushes with a force of 200 N and moves a box 8 meters up a 200 incline . How much work does he do ? (A) 1600 j (B) 1500 j (C) 550 j (D) 200 ft lbs

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