Chapter 6 Work and Energy © 2014 Pearson Education, Inc.

Contents of Chapter 6 Kinetic Energy, and the Work-Energy Principle Potential Energy Conservative and Nonconservative Forces Mechanical Energy and Its Conservation Problem Solving Using Conservation of Mechanical Energy Other Forms of Energy and Energy Transformations; the Law of Conservation of Energy Energy Conservation with Dissipative Forces: Solving Problems © 2014 Pearson Education, Inc.

Work is force x distance but the force must move the object in same direction that the force is being applied.

6-3 Kinetic Energy and the Work-Energy Principle Energy was traditionally defined as the ability to do work. We now know that not all forces are able to do work; however, we are dealing in these chapters with mechanical energy, which does follow this definition. © 2014 Pearson Education, Inc.

6-3 Kinetic Energy and the Work-Energy Principle If we write the acceleration in terms of the velocity and the distance, we find that the work done here is We define the kinetic energy: (6-2) (6-3) © 2014 Pearson Education, Inc.

6-3 Kinetic Energy and the Work-Energy Principle This means that the work done is equal to the change in the kinetic energy: If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases. (6-4) © 2014 Pearson Education, Inc.

6-3 Kinetic Energy and the Work-Energy Principle Because work and kinetic energy can be equated, they must have the same units: kinetic energy is measured in joules. © 2014 Pearson Education, Inc.

Example 6-4 A 145g baseball is thrown with a speed of 25m/s. What is its KE? Given:mass=145g=0.145kg; v=25m/s Formula: KE=½mv2 Substitution: KE=½(0.145kg)(25m/s)2 Answer w/unit: 45.3J B) How much work was done on the ball to make it reach this speed , if it started from rest? Given:mass=145g=0.145kg; final v=25m/s;initial vo=0m/s Formula: Wnet= ½mv22 - ½mv12 Substitution: Wnet=½(0.145kg)(25m/s)2 - ½(0.145kg)(0m/s)2 Remember: Use W-E Theorem only when solving for net work!

Example 6-5 How much work is required to accelerate a 1000kg car from 20m/s to 30m/s? Given:mass=1000kg; final v=30m/s; initial vo=20m/s Formula: Wnet= ½mv22 - ½mv12 Substitution: Wnet=½(1000kg)(30m/s)2 - ½(1000kg)(20m/s)2 Answer w/unit: 2.50 x 105J

Example 6-6 An automobile traveling 60km/h can brake to a stop within a distance of 20m. If the car is going twice as fast, what is its stopping distance? The maximum braking force is approximately independent of speed. Since speed is squared in the equation and the speed is doubled, the stopping distance will be 4x as great. 20m x 4 = 80m

6-4 Potential Energy An object can have potential energy by virtue of its surroundings. Familiar examples of potential energy: A wound-up spring A stretched elastic band An object at some height above the ground © 2014 Pearson Education, Inc.

6-4 Potential Energy In raising a mass m to a height h, the work done by the external force is We therefore define the gravitational potential energy: (6-5a) (6-6) © 2014 Pearson Education, Inc.

6-4 Potential Energy This potential energy can become kinetic energy if the object is dropped. Potential energy is a property of a system as a whole, not just of the object (because it depends on external forces). If PEG = mgy, where do we measure y from? It turns out not to matter, as long as we are consistent about where we choose y = 0. Only changes in potential energy can be measured. © 2014 Pearson Education, Inc.

Example 6-7 A 1000kg roller-coaster car moves from point A to point B and then to point C. What is its gravitational PE at B and C relative to point A? Given:mass=1000kg; yB=10m; yC=-15m; g=9.81m/s2 Formula: PE=mgy Substitution: PEB=1000kg (9.81m/s2) (10m) PEC=1000kg (9.81m/s2) (-15m) Answer w/unit: PEB= 9.80 x 104J PEC= -1.47 x 105J B A C

Example 6-7 A 1000kg roller-coaster car moves from point A to point B and then to point C. b) What is the change in PE when it goes from B to C? Given: PEB= 9.80 x 104J; PEC=-1.47 x 105J Formula: ΔPE=PEf – Pei=PEC - PEB Substitution: ΔPE=-1.47 x 105J – 9.80 x 104J Answer w/unit: -2.45 x 105J B A C

Example 6-7 A 1000kg roller-coaster car moves from point A to point B and then to point C. c) Repeat parts a) and b), but take the reference point to be at point 3. a) Given:mass=1000kg; yB=25m; yC=0m; g=9.81m/s2 Formula: PE=mgy Substitution: PEB=1000kg (9.81m/s2) (25m) PEC=1000kg (9.81m/s2) (0m) Answer w/unit: PEB= 2.45 x 105J; PEC= 0J b) Given: PEB= 2.45 x 105J; PEC=0J Formula: ΔPE=PEf – Pei=PEC - PEB ΔPE=0J – 2.45 x 105J -2.45 x 105J B A C

6-4 Potential Energy Potential energy can also be stored in a spring when it is compressed; the figure below shows potential energy yielding kinetic energy. © 2014 Pearson Education, Inc.

6-4 Potential Energy The force required to compress or stretch a spring is: where k is called the spring constant, and needs to be measured for each spring. (6-8) © 2014 Pearson Education, Inc.

6-4 Potential Energy The force increases as the spring is stretched or compressed further. We find that the potential energy of the compressed or stretched spring, measured from its equilibrium position, can be written: (6-9) © 2014 Pearson Education, Inc.

Elastic Potential Energy Problew (complete in notebook) A spring has a a spring constant k of 88.0 N/m. How much must this spring be compressed to store 45.0J of potential energy? Given:spring constant (k)=88.0 N/m; PEel=45.0J Formula: PEel=½kx2 Substitution: 45.0J=½(88.0 N/m)(x)2 Answer w/unit: 1.01m

6-5 Conservative and Nonconservative Forces If friction is present, the work done depends not only on the starting and ending points, but also on the path taken. Friction is called a nonconservative force. © 2014 Pearson Education, Inc.

6-5 Conservative and Nonconservative Forces Potential energy can only be defined for conservative forces. © 2014 Pearson Education, Inc.

6-5 Conservative and Nonconservative Forces Therefore, we distinguish between the work done by conservative forces and the work done by nonconservative forces. We find that the work done by nonconservative forces is equal to the total change in kinetic and potential energies: (6-10) © 2014 Pearson Education, Inc.

6-6 Mechanical Energy and Its Conservation If there are no nonconservative forces, the sum of the changes in the kinetic energy and in the potential energy is zero—the kinetic and potential energy changes are equal but opposite in sign. This allows us to define the total mechanical energy: E = KE + PE And its conservation: (6-12b) © 2014 Pearson Education, Inc.

6-7 Problem Solving Using Conservation of Mechanical Energy In the image on the left, the total mechanical energy is: E = KE + PE = ½ mv2 + mgy The energy buckets on the right of the figure show how the energy moves from all potential to all kinetic. © 2014 Pearson Education, Inc.

Example 6-8 If the original height of the stone in the figure is h=3.0m, calculate the stone’s speed when it has fallen to 1.0m above the ground. Given: y1=3.0m; y2=1.0m; v1=0m/s Strategy: Apply conservation of ME and with only gravity acting on the stone, we choose the ground as a reference level where h=0. Formula: ½mv21 + mgy1= ½mv22 + mgy2 (m’s cancel out (same mass) and since v1=0, the first part of the equation cancels) Substitution: 0 + gy1= ½v22 + gy2 (9.81m/s2)3.0m = ½v22 + 9.81m/s2(1.0m) Answer w/unit: 6.26m/s

6-7 Problem Solving Using Conservation of Mechanical Energy If there is no friction, the speed of a roller coaster will depend only on its height compared to its starting height. © 2014 Pearson Education, Inc.

Example 6-9 Assuming the height of the hill in the figure is 40m, nad the rollercoaster car starts from rest at the top, calculate The speed of the rollercoaster car at the bottom of the hill at what height it will have half this speed. Given: y1=40m; y2=0m; v1=0m/s Formula: ½mv21 + mgy1= ½mv22 + mgy2 (m’s cancel out (same mass) and since v1 and y2=0, those parts of the equation cancels) Substitution: 0 + gy1= ½v22 + 0 (9.81m/s2)40m = ½v22 Answer w/unit: 28.0m/s Given: y1=40m; v1=0m/s; v2=14.0m/s Formula: ½mv21 + mgy1= ½mv22 + mgy2 (m’s cancel out (same mass) and since v1, that part of the equation cancels) Substitution: 0 + gy1= ½v22 + gy2 ; (9.81m/s2)40m = ½(14.0)2 + (9.81m/s2)y2 Answer w/unit: 30.0m

Example 6-10 Two water slides at a pool are shaped differently but start at the same height. Two riders, Paul and Kathleen, start from rest at the same time on different slides. Ignore friction and assume both slides have the same path length. a) Which rider is traveling faster at the bottom? Since both riders are traveling the same length, they both will be traveling at the same speed at the bottom. b) Which rider makes it to the bottom first? Kathleen will make it to the bottom first. She is at a lower elevation than Paul throughout the whole slide. Kathleen

Example 6-11 Estimate the KE and the speed required for a 70kg pole vaulter to just pass over a bar 5.0m high. Assume the vaulter’s center of mass is initially 0.90m off the ground and reaches its maximum height at the level of the bar itself. Strategy: Find speed then KE. Find Speed Given: y1=0.90m; y2=5m; v2=0m/s(max height) Formula: (9.81m/s2)0.90m + mgy1= ½mv22 + mgy2 (m’s cancel out (same mass) and since v2, that part of the equation cancels) Substitution: ½v21 + gy1= 0 + gy2 ½v21 + (9.81m/s2)0.90m = (9.81m/s2)5m Answer w/unit: 8.96m/s Find KE Given: mass=70kg; speed=8.96m/s Formula: KE= ½mv2 Substitution: KE = ½(70kg)(8.96m/s)2 Answer w/unit: 2810J

6-7 Problem Solving Using Conservation of Mechanical Energy For an elastic force, conservation of energy tells us: (6-14) © 2014 Pearson Education, Inc.

Example 6-12 A dart of mass 0.100kg is pressed against the spring of a toy dart gun. The spring (k=250 N/m) is compressed 6.0cm and released. If the dart detaches from the spring when the latter reaches its normal length (x=0), what speed does the dart acquire? Given: m=0.100kg; k=250 N/m; x=6.0cm=0.06m; v1=0m/s; x2=0m Formula: ½mv21 + ½kx12 = ½mv22 + ½kx22 (Since v1 and x2=0, those parts of the equation cancels) Substitution: 0 + ½kx12 = ½mv22 + 0 ½(250 N/m)(0.06m)2 = ½(0.100kg)v22 Answer w/unit: 3.00m/s

Example 6-13 Given: m=2.60kg; y1=55.0cm=0.55m; For this problem, y is used instead of x in the elastic equation for consistency and image. A ball of mass m=2.60kg, starting from rest, falls a vertical distance h=55.0cm before striking a vertical coiled spring, which it compresses an amount Y=15.0cm. Determine the spring constant of the spring. Assume the spring has negligible mass. Measure all distances from the point where the ball first touches the uncompressed spring. (y=o at this point) Given: m=2.60kg; y1=55.0cm=0.55m; y3=-15cm=-0.15m; v1=0m/s; y2=0m; v3=0m/s Formula: ½mv21+mgy1+½ky22 = ½mv23+mgy3+½ky32 (Since v1, y2, and v3=0, those parts of the equation cancels) Substitution: 0+mgy1+0=0+mgy3+½ky32 2.60kg(9.81m/s2)0.55m=(2.60kg)(9.81m/s2)(-0.15m)+½k-0.15m Answer w/unit: 1590 N/m

Example 6-14 Your looking for total height fell (d). Y1=15m+x and x2=the length stretched. Dave jumps off a bridge with a bungee cord tied around his ankle. He falls for 15m before the bungee cord begins to stretch. Dave’s mass is 75kg and we assume the cord obeys Hooke’s law, F=-kx, with k=50N/m. If we neglect air resistance, estimate how far below the bridge Dave will fall before coming to a stop. Ignore the mass of the cord. Given: m=75kg; y1=15m+x; k=50N/m; v1=0m/s; y2=0m; v2=0m/s Formula: ½mv21+mgy1+½ky22 = ½mv23+mgy3+½ky32 (Since v1, y2, and v2=0, those parts of the equation cancels) Substitution: 0+mgy1+0=0+0+½kx22 75kg(9.8m/s2)0.15m+x=½50x22=11025m+735x=25x2 0=25x2-735x-11025; Answers from Quad. Eq: 40.3 and -10.9 Total distance=15m + x=15m + 40.3m= Answer w/unit: 55.3m You should notice the need for a Quadratic equation. Out of your two answers, only 40.3 makes sense.

6-8 Other Forms of Energy and Energy Transformations; the Law of Conservation of Energy Some other forms of energy: Electric energy, nuclear energy, thermal energy, chemical energy. Work is done when energy is transferred from one object to another. Accounting for all forms of energy, we find that the total energy neither increases nor decreases. Energy as a whole is conserved. © 2014 Pearson Education, Inc.

6-9 Energy Conservation with Dissipative Processes; Solving Problems If there is a nonconservative force such as friction, where do the kinetic and potential energies go? They become heat; the actual temperature rise of the materials involved can be calculated. © 2014 Pearson Education, Inc.

6-9 Energy Conservation with Dissipative Processes; Solving Problems Problem solving: Draw a picture. Determine the system for which energy will be conserved. Figure out what you are looking for, and decide on the initial and final positions. Choose a logical reference frame. Apply conservation of energy. Solve. © 2014 Pearson Education, Inc.

Example 6-15 The rollercoaster car in example 6-9 is found to reach a vertical height of only 25m on the second hill before coming to a stop. It traveled a total distance of 400m. Estimate the average friction force on the car, whose mass is 1000kg. Determine the system for which energy will be conserved. The system is the rollercoaster and the Earth. The forces acting are gravity and friction. PE is from gravity and friction will be determined by work formula (Ffrd). Figure out what you are looking for, and decide on the initial and final positions. Point 1 is when the car begins and point 2 is when it stops at 25m. Choose a logical reference frame. At lowest point, y=0. Apply conservation of energy. Since ME isn’t conserved because of friction, you will use the conservation equation that includes it. Solve. given: h1=40m; h2=25m; d=400m; m=1000kg;v1=0m/s;v2=0m/s formula: ½mv12+mgh1=½mv22+mgh2+Ffrd Substitution: 0+1000kg(9.81m/s2)40m=0+1000kg(9.81m/s2)25m+Ffr400m Answer w/unit: 368N

Summary of Chapter 6 Kinetic energy is energy of motion: KE = ½ mv2 Potential energy is energy associated with forces that depend on the position or configuration of objects. The net work done on an object equals the change in its kinetic energy. If only conservative forces are acting, mechanical energy is conserved. © 2014 Pearson Education, Inc.