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Chapter 10 - Energy and Work (Cont’d) w./ QuickCheck Questions © 2015 Pearson Education, Inc. Anastasia Ierides Department of Physics and Astronomy University of New Mexico October 22, 2015 Review of Last Time Impulse & Momentum; conservation of momentum Isolated system; internal & external forces; collisions (elastic/inelastic); explosions Energy - different forms (kinetic, work, etc.) Energy transformations - from one kind of energy to another QuickCheck Question 10.1 A child is on a playground swing, motionless at the highest point of her arc. What energy transformation takes place as she swings back down to the lowest point of her motion? A. B. C. D. E. K → Ug Ug → K Eth → K Ug → Eth K → Eth QuickCheck Question 10.1 A child is on a playground swing, motionless at the highest point of her arc. What energy transformation takes place as she swings back down to the lowest point of her motion? A. B. C. D. E. K → Ug Ug → K Eth → K Ug → Eth K → Eth Hint: Going from a high point back down to ground where she is moving QuickCheck Question 10.2 A skier is gliding down a gentle slope at a constant speed. What energy transformation is taking place? A. B. C. D. E. K → Ug Ug → K Eth → K Ug → Eth K → Eth QuickCheck Question 10.2 A skier is gliding down a gentle slope at a constant speed. What energy transformation is taking place? A. B. C. D. E. K → Ug Ug → K Eth → K Ug → Eth K → Eth Since the skier glides at a constant speed, the kinetic energy doesn’t change; but he is going downward meaning that his gravitational potential energy is decreasing. This energy must then be transformed into thermal energy created by friction between the skis and the slope. QuickCheck 10.3 A tow rope pulls a skier up the slope at constant speed. What energy transfer (or transfers) is taking place? A. B. C. D. E. W → Ug W→K W → Eth Both A and B. Both A and C. QuickCheck 10.3 A tow rope pulls a skier up the slope at constant speed. What energy transfer (or transfers) is taking place? A. B. C. D. E. W → Ug W→K W → Eth Both A and B. Both A and C. Since the skier is traveling at a constant speed, the kinetic energy doesn’t change, which means that this energy must then be transformed into thermal energy created by friction between the skis and the slope. He is also going upward meaning that he is gaining gravitational potential energy. The Law of Conservation of Energy The total energy of an isolated system remains constant! QuickCheck Question 10.4 A crane lowers a girder into place at constant speed. Consider the work Wg done by gravity and the work WT done by the tension in the cable. Which is true? A. B. C. D. E. Wg > 0 and WT > 0 Wg > 0 and WT < 0 Wg < 0 and WT > 0 Wg < 0 and WT < 0 Wg = 0 and WT = 0 QuickCheck Question 10.4 A crane lowers a girder into place at constant speed. Consider the work Wg done by gravity and the work WT done by the tension in the cable. Which is true? A. B. C. D. E. Wg > 0 and WT > 0 Wg > 0 and WT < 0 Wg < 0 and WT > 0 Wg < 0 and WT < 0 Wg = 0 and WT = 0 The downward force of gravity is in the direction of motion ⇒ positive work. The upward tension is in the direction opposite the motion ⇒ negative work. QuickCheck Question 10.5 Robert pushes the box to the left at constant speed. In doing so, Robert does ______ work on the box. A. positive B. negative C. zero QuickCheck Question 10.5 Robert pushes the box to the left at constant speed. In doing so, Robert does ______ work on the box. A. positive B. negative C. zero Force is in the direction of displacement ⇒ positive work QuickCheck Question 10.9 Ball A has half the mass and eight times the kinetic energy of ball B. What is the speed ratio vA/vB? A. B. C. D. E. 16 4 2 1/4 1/16 QuickCheck Question 10.9 Ball A has half the mass and eight times the kinetic energy of ball B. What is the speed ratio vA/vB? A. B. C. D. E. 16 4 2 1/4 1/16 Remember: Rearranging to solve vA = (2 KA/mA)1/2 for the velocity, v, vB (2 KB/mB)1/2 v = (2 K/m)1/2 vA/vB = [(KA/KB)(mB/mA)]1/2 = [(8)(2)]1/2 = [16]1/2 =4 QuickCheck Question 10.10 A light plastic cart and a heavy steel cart are both pushed with the same force for a distance of 1.0 m, starting from rest. After the force is removed, the kinetic energy of the light plastic cart is ________ that of the heavy steel cart. A. B. C. D. greater than equal to less than Can’t say. It depends on how big the force is. QuickCheck Question 10.10 A light plastic cart and a heavy steel cart are both pushed with the same force for a distance of 1.0 m, starting from rest. After the force is removed, the kinetic energy of the light plastic cart is ________ that of the heavy steel cart. Same force, same distance ⇒ same work done Same work ⇒ change of kinetic energy A. B. C. D. greater than equal to less than Can’t say. It depends on how big the force is. QuickCheck Question 10.11 Each of the boxes shown is pulled for 10 m across a level, frictionless floor by the force given. Which box experiences the greatest change in its kinetic energy? QuickCheck Question 10.11 Each of the boxes shown is pulled for 10 m across a level, frictionless floor by the force given. Which box experiences the greatest change in its kinetic energy? D Work-energy equation: All have same d, so largest work (and hence ∆K = W = Fd largest ∆K) corresponds to largest force. QuickCheck Question 10.12 Each of the 1.0 kg boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force at the noted angle. Which box has the highest final speed? QuickCheck Question 10.12 Each of the 1.0 kg boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force at the noted angle. Which box has the highest final F|| = 5√3 N F|| = 10√3 N F|| = 5√2 N speed? B Same d; only need to focus on F|| = F cos θ F|| = 10√2 N Work-energy equation: ∆K = W = Fd cos θ cos (30°) = √3 /2 cos (45°) = √2 /2 cos (60°) = 1/2 F|| = 5 N QuickCheck Question 10.13 Rank in order, from largest to smallest, the gravitational potential energies of the balls. A. B. C. D. 1>2=4>3 1>2>3>4 3>2>4>1 3>2=4>1 QuickCheck Question 10.13 Rank in order, from largest to smallest, the gravitational potential energies of the balls. A. B. C. D. 1>2=4>3 1>2>3>4 3>2>4>1 3>2=4>1 QuickCheck Question 10.14 Starting from rest, a marble first rolls down a steeper hill, then down a less steep hill of the same height. For which is it going faster at the bottom? A. B. C. D. Faster at the bottom of the steeper hill. Faster at the bottom of the less steep hill. Same speed at the bottom of both hills. Can’t say without knowing the mass of the marble. QuickCheck Question 10.14 Starting from rest, a marble first rolls down a steeper hill, then down a less steep hill of the same height. For which is it going faster at the bottom? Think energy transformation! The work required to bring both balls down from the same height is the same, that is, Ug = mgy. A. B. C. D. Work-energy equation: ∆K = W = Fd, where F = w = mg and d = y. Faster at the bottom of the steeper hill. Faster at the bottom of the less steep hill. Same speed at the bottom of both hills. Can’t say without knowing the mass of the marble. QuickCheck Question 10.15 A small child slides down the four frictionless slides A–D. Rank in order, from largest to smallest, her speeds at the bottom. A. B. C. D. vD > vA > vB > vC vD > vA = vB > vC vC > vA > vB > vD vA = vB = vC = vD QuickCheck Question 10.15 A small child slides down the four frictionless slides A–D. Rank in order, from largest to smallest, her speeds at the bottom. A. B. C. D. vD > vA > vB > vC vD > vA = vB > vC vC > vA > vB > vD vA = vB = vC = vD The work required to slide down all the slides from the same height is the same, that is, Ug = mgh. Work-energy equation: ∆K = W = Fd, where F = w = mg and d = h. Same height ⇒ same work, W! ⇒ same change in kinetic energy QuickCheck Question 10.16 Three balls are thrown from a cliff with the same speed but at different angles. Which ball has the greatest speed just before it hits the ground? A. B. C. D. Ball A. Ball B. Ball C. All balls have the same speed. QuickCheck Question 10.16 Three balls are thrown from a cliff with the same speed but at different angles. Which ball has the greatest speed just before it hits the ground? Work-energy equation: ∆K = W = Fd, where F = w = mg and d = h. h Same height ⇒ same work, W! A. Ball A. ⇒ same change in kinetic energy B. Ball B. C. Ball C. D. All balls have the same speed. Thermal Energy The sum of the kinetic energy of a substance’s atoms and molecules and the elastic potential energy stored in their molecular bonds is the substance’s thermal energy. Thermal Energy Friction on a moving object does work Thermal Energy Friction on a moving object does work Work due to friction creates thermal energy Example 10.9: Creating thermal energy by rubbing A 0.30 kg block of wood is rubbed back and forth against a wood table 30 times in each direction. The block is moved 8.0 cm during each stroke and pressed against the table with a force of 22 N. How much thermal energy is created in this process? Example 10.9: Creating thermal energy by rubbing A 0.30 kg block of wood is rubbed back and forth against a wood table 30 times in each direction. The block is moved 8.0 cm during each stroke and pressed against the table with a force of 22 N. How much thermal energy is created in this process? PREPARE The hand holding the block does work to push the block back and forth. Work transfers energy into the block + table system, where it appears as thermal energy according to Equation 10.16. The force of friction can be found from the model of kinetic friction introduced in Chapter 5, fk = µkn; from Table 5.2 the coefficient of kinetic friction for wood sliding on wood is µk = 0.20. Example 10.9: Creating thermal energy by rubbing PREPARE To find the normal force n acting on the block, we draw the free-body diagram of the figure, which shows only the vertical forces acting on the block. Example 10.9: Creating thermal energy by rubbing PREPARE To find the normal force n acting on the block, we draw the free-body diagram of the figure, which shows only the vertical forces acting on the block. Example 10.9: Creating thermal energy by rubbing PREPARE To find the normal force n acting on the block, we draw the free-body diagram of the figure, which shows only the vertical forces acting on the block. SOLVE We want to use ΔEth = fk Δx, where the kinetic friction is fk = µkn. Example 10.9: Creating thermal energy by rubbing PREPARE To find the normal force n acting on the block, we draw the free-body diagram of the figure, which shows only the vertical forces acting on the block. SOLVE We want to use ΔEth = fk Δx, where the kinetic friction is fk = µkn. The block is not accelerating in the y-direction, so from the free-body diagram Newton’s second law gives Example 10.9: Creating thermal energy by rubbing PREPARE To find the normal force n acting on the block, we draw the free-body diagram of the figure, which shows only the vertical forces acting on the block. SOLVE We want to use ΔEth = fk Δx, where the kinetic friction is fk = µkn. The block is not accelerating in the y-direction, so from the free-body diagram Newton’s second law gives Σ Fy = n − w − F = may = 0 or n = w + F = mg + F = (0.30 kg)(9.8 m/s2) + 22 N = 24.9 N Example 10.9: Creating thermal energy by rubbing SOLVE The friction force is then fk = µkn = (0.20)(24.9 N) = 4.98 N. Example 10.9: Creating thermal energy by rubbing SOLVE The friction force is then fk = µkn = (0.20)(24.9 N) = 4.98 N. The total displacement of the block is 2 × 30 × 8.0 cm = 4.8 m. Thus the thermal energy created is ΔEth = fk Δx = (4.98 N)(4.8 m) = 24 J Example 10.9: Creating thermal energy by rubbing SOLVE The friction force is then fk = µkn = (0.20)(24.9 N) = 4.98 N. The total displacement of the block is 2 × 30 × 8.0 cm = 4.8 m. Thus the thermal energy created is ΔEth = fk Δx = (4.98 N)(4.8 m) = 24 J ASSESS This modest amount of thermal energy seems reasonable for a person to create by rubbing. In Class Exercise Vigorously rub the top of your leg with your dominant hand’s palm for about 10 seconds. If you then pass the fingers of your other hand over the spot where you rubbed, you’ll feel a distinct warm area. In Class Exercise Vigorously rub the top of your leg with your dominant hand’s palm for about 10 seconds. If you then pass the fingers of your other hand over the spot where you rubbed, you’ll feel a distinct warm area. Congratulations! You’ve just set some 100,000,000,000,000,000,000,000 atoms into motion! Using the Law of Conservation of Energy Using the Law of Conservation of Energy, W = ∆E We use the law of conservation of energy to develop a before-and-after perspective: ∆K + ∆Ug + ∆Us + ∆Eth = W Using the Law of Conservation of Energy, W = ∆E We use the law of conservation of energy to develop a before-and-after perspective: ∆K + ∆Ug + ∆Us + ∆Eth = W Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i + W Using the Law of Conservation of Energy, W = ∆E We use the law of conservation of energy to develop a before-and-after perspective: ∆K + ∆Ug + ∆Us + ∆Eth = W Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i + W Analogous to before-and-after of the law of conservation of momentum Using the Law of Conservation of Energy, W = ∆E We use the law of conservation of energy to develop a before-and-after perspective: ∆K + ∆Ug + ∆Us + ∆Eth = W Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i + W Analogous to before-and-after of the law of conservation of momentum In an isolated system, W = 0, because ∆E = 0: Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i Using the Law of Conservation of Energy, W = ∆E Using the Law of Conservation of Energy, W = ∆E How to Choose an Isolated System QuickCheck Question 10.18 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be A. B. C. D. E. 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s QuickCheck Question 10.18 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be We initially have spring potential energy, Us = 1/2 k x2, and produce kinetic energy, K = 1/2 m v2. A. B. C. D. E. 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s QuickCheck Question 10.18 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be We initially have spring potential energy, Us = 1/2 k x2, and produce kinetic energy, K = 1/2 m v2. A. B. C. D. E. 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Equating the two we have, x2 ∝ v2 QuickCheck Question 10.18 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be We initially have spring potential energy, Us = 1/2 k x2, and produce kinetic energy, K = 1/2 m v2. A. B. C. D. E. 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Equating the two we have, x2 ∝ v2 Double the displacement, x ⇒ double the speed, v QuickCheck Question 10.18 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be We initially have spring potential energy, Us = 1/2 k x2, and produce kinetic energy, K = 1/2 m v2. A. B. C. D. E. 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Equating the two we have, x2 ∝ v2 Double the displacement, x ⇒ double the speed, v QuickCheck Question 10.19 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be A. B. C. D. E. 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s QuickCheck Question 10.19 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be Again, we begin with Us = 1/2 k x2, and produce K = 1/2 m v2. A. B. C. D. E. 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Equating the two we have, v ∝ k1/2 QuickCheck Question 10.19 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be Again, we begin with Us = 1/2 k x2, and produce K = 1/2 m v2. A. B. C. D. E. 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Equating the two we have, v ∝ k1/2 Double the spring constant, k ⇒ multiply the speed by 21/2 ≈ 1.4 QuickCheck Question 10.19 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be Again, we begin with Us = 1/2 k x2, and produce K = 1/2 m v2. A. B. C. D. E. 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Equating the two we have, v ∝ k1/2 Double the spring constant, k ⇒ multiply the speed by 21/2 ≈ 1.4 Example 10.11: Speed at the bottom of a water slide While at the county fair, Katie tries the water slide, whose shape is shown in the figure. The starting point is 9.0 m above the ground. She pushes off with an initial speed of 2.0 m/s. If the slide is frictionless, how fast will Katie be traveling at the bottom? Example 10.11: Speed at the bottom of a water slide PREPARE Table 10.2 showed that the system consisting of Katie and the earth is isolated because the normal force of the slide is perpendicular to Katie’s motion and does no work. If we assume the slide is frictionless, we can use the conservation of mechanical energy equation. Example 10.11: Speed at the bottom of a water slide PREPARE Table 10.2 showed that the system consisting of Katie and the earth is isolated because the normal force of the slide is perpendicular to Katie’s motion and does no work. If we assume the slide is frictionless, we can use the conservation of mechanical energy equation. Conservation of mechanical energy gives Kf + (Ug)f = Ki + (Ug)i or SOLVE Example 10.11: Speed at the bottom of a water slide PREPARE Table 10.2 showed that the system consisting of Katie and the earth is isolated because the normal force of the slide is perpendicular to Katie’s motion and does no work. If we assume the slide is frictionless, we can use the conservation of mechanical energy equation. Conservation of mechanical energy gives Kf + (Ug)f = Ki + (Ug)i or SOLVE Example 10.11: Speed at the bottom of a water slide Taking yf = 0 m, we have Example 10.11: Speed at the bottom of a water slide Taking yf = 0 m, we have which we can solve to get Example 10.11: Speed at the bottom of a water slide Taking yf = 0 m, we have which we can solve to get Notice that the shape of the slide does not matter because gravitational potential energy depends only on the height above a reference level. Example 10.13: Pulling a bike trailer Monica pulls her daughter Jessie in a bike trailer. The trailer and Jessie together have a mass of 25 kg. Monica starts up a 100-mlong slope that’s 4.0 m high. On the slope, Monica’s bike pulls on the trailer with a constant force of 8.0 N. They start out at the bottom of the slope with a speed of 5.3 m/s. What is their speed at the top of the slope? Example 10.13: Pulling a bike trailer PREPARE Taking Jessie and the trailer as the system, we see that Monica’s bike is applying a force to the system as it moves through a displacement; that is, Monica’s bike is doing work on the system. Thus we’ll need to use the full version of the law of energy conservation. Example 10.13: Pulling a bike trailer SOLVE If we assume there’s no friction, so that ΔEth = 0, then Kf + (Ug)f = Ki + (Ug)i + W or Example 10.13: Pulling a bike trailer Taking yi = 0 m and writing W = Fd, we can solve for the final speed: Example 10.13: Pulling a bike trailer from which we find that vf = 3.7 m/s. Note that we took the work to be a positive quantity because the force is in the same direction as the displacement. Example 10.13: Pulling a bike trailer ASSESS A speed of 3.7 m/s—about 8 mph—seems reasonable for a bicycle’s speed. Jessie’s final speed is less than her initial speed, indicating that the uphill force of Monica’s bike on the trailer is less than the downhill component of gravity. Energy in Collisions Two types of collisions: elastic and perfectly inelastic Energy in Collisions Two types of collisions: elastic and perfectly inelastic Momentum is conserved in all collisions, but mechanical energy is only conserved in a perfectly elastic collision Energy in Collisions Two types of collisions: elastic and perfectly inelastic Momentum is conserved in all collisions, but mechanical energy is only conserved in a perfectly elastic collision In an inelastic collision, some mechanical energy is converted to thermal energy Example 10.15: Energy transformations in a perfectly inelastic collision The figure shows two train cars that move toward each other, collide, and couple together. In Example 9.8, we used conservation of momentum to find the final velocity shown in the figure from the given initial velocities. How much thermal energy is created in this collision? Example 10.15: Energy transformations in a perfectly inelastic collision PREPARE We’ll choose our system to be the two cars. Because the track is horizontal, there is no change in potential energy. Thus the law of conservation of energy is Kf + ΔEth = Ki. Example 10.15: Energy transformations in a perfectly inelastic collision PREPARE We’ll choose our system to be the two cars. Because the track is horizontal, there is no change in potential energy. Thus the law of conservation of energy is Kf + ΔEth = Ki. The total energy before the collision must equal the total energy afterward, but the mechanical energies need not be equal. Example 10.15: Energy transformations in a perfectly inelastic collision SOLVE The initial kinetic energy is Example 10.15: Energy transformations in a perfectly inelastic collision SOLVE The initial kinetic energy is Because the cars stick together and move as a single object with mass m1 + m2, the final kinetic energy is Example 10.15: Energy transformations in a perfectly inelastic collision From the conservation of energy equation on the previous slide, we find that the thermal energy increases by ΔEth = Ki − Kf = 4.7 × 104 J − 1900 J = 4.5 × 104 J Example 10.15: Energy transformations in a perfectly inelastic collision From the conservation of energy equation on the previous slide, we find that the thermal energy increases by ΔEth = Ki − Kf = 4.7 × 104 J − 1900 J = 4.5 × 104 J This amount of the initial kinetic energy is transformed into thermal energy during the impact of the collision. Example 10.15: Energy transformations in a perfectly inelastic collision From the conservation of energy equation on the previous slide, we find that the thermal energy increases by ΔEth = Ki − Kf = 4.7 × 104 J − 1900 J = 4.5 × 104 J This amount of the initial kinetic energy is transformed into thermal energy during the impact of the collision. ASSESS About 96% of the initial kinetic energy is transformed into thermal energy. This is typical of many real-world collisions. Elastic Collisions Elastic collisions obey conservation of momentum and conservation of mechanical energy Elastic Collisions Elastic collisions obey conservation of momentum and conservation of mechanical energy Momentum: m1 (v1x )i = m1 (v1x )f + m2 (v2 x )f Energy: 1 1 1 2 2 2 m1 (v1x )i = m1 (v1x )f + m2 (v2 x )f 2 2 2 Example 10.16: Velocities in an air hockey collision On an air hockey table, a moving puck, traveling to the right at 2.3 m/s, makes a head-on collision with an identical puck at rest. What is the final velocity of each puck? Example 10.16: Velocities in an air hockey collision PREPARE We’ve shown the final velocities in the picture, but we don’t really know yet which way the pucks will move. Because one puck was initially at rest, we can use Equations 10.20 to find the final velocities of the pucks. The pucks are identical, so we have m1 = m2 = m. Example 10.16: Velocities in an air hockey collision SOLVE We use Equations 10.20 with m1 = m2 = m to get The incoming puck stops dead, and the initially stationary puck goes off with the same velocity that the incoming one had. Example 10.17: Protecting your head A bike helmet—basically a shell of hard, crushable foam—is tested by being strapped onto a 5.0 kg headform and dropped from a height of 2.0 m onto a hard anvil. What force is encountered by the headform if the impact crushes the foam by 3.0 cm? Example 10.17: Protecting your head A bike helmet—basically a shell of hard, crushable foam—is tested by being strapped onto a 5.0 kg headform and dropped from a height of 2.0 m onto a hard anvil. What force is encountered by the headform if the impact crushes the foam by 3.0 cm? PREPARE The work-energy equation can be used to calculate the force on the headform. We’ll choose the headform and the earth to be the system; the foam in the helmet is part of the environment. We make this choice so that the force on the headform due to the foam is an external force that does work W on the headform. Example 10.17: Protecting your head SOLVE The headform starts at rest, speeds up as it falls, then returns to rest during the impact. Overall, Kf = Ki. Furthermore, ΔEth = 0 because there’s no friction to increase the thermal energy. Only the gravitational potential energy changes, so the work-energy equation is (Ug)f − (Ug)i = W Example 10.17: Protecting your head The upward force of the foam on the headform is opposite the downward displacement of the headform. We see that the work done is negative: W = −Fd, assuming that the force is constant. Using this result in the work-energy equation and solving for F, we find Example 10.17: Protecting your head Taking our reference height to be y = 0 m at the anvil, we have (Ug)f = 0. We’re left with (Ug)i = mgyi, so Example 10.17: Protecting your head Taking our reference height to be y = 0 m at the anvil, we have (Ug)f = 0. We’re left with (Ug)i = mgyi, so This is the force that acts on the head to bring it to a halt in 3.0 cm. More important from the perspective of possible brain injury is the head’s acceleration: Example 10.17: Protecting your head ASSESS The accepted threshold for serious brain injury is around 300g, so this helmet would protect the rider in all but the most serious accidents. Without the helmet, the rider’s head would come to a stop in a much shorter distance and thus be subjected to a much larger acceleration. Example 10.17: Protecting your head ASSESS The accepted threshold for serious brain injury is around 300g, so this helmet would protect the rider in all but the most serious accidents. Without the helmet, the rider’s head would come to a stop in a much shorter distance and thus be subjected to a much larger acceleration. Power Power is the rate at which energy is transformed or transferred Power Power is the rate at which energy is transformed or transferred Power Power is the rate at which energy is transformed or transferred The unit of power is the watt: 1 watt = 1 W = 1 J/s Output Power of a Force A force doing work transfers energy Output Power of a Force A force doing work transfers energy The rate that this force transfers energy is the output power of that force: Output Power of a Force A force doing work transfers energy The rate that this force transfers energy is the output power of that force: QuickCheck Question 10.20 Four students run up the stairs in the time shown. Which student has the largest power output? QuickCheck Question 10.20 Four students run up the stairs in the time shown. Which student has the largest power output? QuickCheck Question 10.20 Four students run up the stairs in the time shown. Which student has the largest power output? The work, W, done by the student is against gravity, where each student gains gravitational potential energy, Ug = mgh. QuickCheck Question 10.20 Four students run up the stairs in the time shown. Which student has the largest power output? Ug = 7840 J Ug = 7840 J Ug = 6272 J Ug = 15680 J The work, W, done by the student is against gravity, where each student gains gravitational potential energy, Ug = mgh. QuickCheck Question 10.20 Four students run up the stairs in the time shown. Which student has the largest power output? Ug = 7840 J P = 784 W Ug = 7840 J P = 980 W Ug = 6272 J P = 784 W Ug = 15680 J P = 627.2 W The work, W, done by the student is against gravity, where each student gains gravitational potential energy, Ug = mgh. QuickCheck Question 10.20 Four students run up the stairs in the time shown. Which student has the largest power output? B. Ug = 7840 J P = 784 W Ug = 7840 J P = 980 W Ug = 6272 J P = 784 W Ug = 15680 J P = 627.2 W The work, W, done by the student is against gravity, where each student gains gravitational potential energy, Ug = mgh. QuickCheck Question 10.21 Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? QuickCheck Question 10.21 Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? QuickCheck Question 10.21 Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? Remember: F = ma = m Δv/Δt QuickCheck Question 10.21 Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? Remember: F = ma = m Δv/Δt QuickCheck Question 10.21 Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? Remember: F = ma = m Δv/Δt = m v/Δt QuickCheck Question 10.21 Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? Remember: F = ma = m Δv/Δt = m v/Δt = mv2/Δt QuickCheck Question 10.21 Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? Remember: F = ma = m Δv/Δt = m v/Δt v2 = 9 v2 = 4 v2 = 4 v2 = 1 v2 = 4 = mv2/Δt m/Δt = .05 m/Δt = .1 m/Δt = .1 m/Δt = .2 m/Δt = .1 QuickCheck Question 10.21 Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power? Remember: F = ma = m Δv/Δt = m v/Δt v2 = 9 v2 = 4 v2 = 4 v2 = 1 v2 = 4 = mv2/Δt m/Δt = .05 m/Δt = .1 m/Δt = .1 m/Δt = .2 m/Δt = .1 Example 10.18: Power to pass a truck Your 1500 kg car is behind a truck traveling at 60 mph (27 m/ s). To pass the truck, you speed up to 75 mph (34 m/s) in 6.0 s. What engine power is required to do this? Example 10.18: Power to pass a truck Your 1500 kg car is behind a truck traveling at 60 mph (27 m/ s). To pass the truck, you speed up to 75 mph (34 m/s) in 6.0 s. What engine power is required to do this? PREPARE Your engine is transforming the chemical energy of its fuel into the kinetic energy of the car. We can calculate the rate of transformation by finding the change ΔK in the kinetic energy and using the known time interval. Example 10.18: Power to pass a truck Your 1500 kg car is behind a truck traveling at 60 mph (27 m/ s). To pass the truck, you speed up to 75 mph (34 m/s) in 6.0 s. What engine power is required to do this? PREPARE Your engine is transforming the chemical energy of its fuel into the kinetic energy of the car. We can calculate the rate of transformation by finding the change ΔK in the kinetic energy and using the known time interval. SOLVE We have Example 10.18: Power to pass a truck SOLVE so that ΔK = Kf − Ki = (8.67 × 105 J) − (5.47 × 105 J) = 3.20 × 105 J Example 10.18: Power to pass a truck SOLVE so that ΔK = Kf − Ki = (8.67 × 105 J) − (5.47 × 105 J) = 3.20 × 105 J To transform this amount of energy in 6 s, the power required is Example 10.18: Power to pass a truck SOLVE so that ΔK = Kf − Ki = (8.67 × 105 J) − (5.47 × 105 J) = 3.20 × 105 J To transform this amount of energy in 6 s, the power required is This is about 71 hp. This power is in addition to the power needed to overcome drag and friction and cruise at 60 mph, so the total power required from the engine will be even greater than this. Chapter 11 - Using Energy w./ QuickCheck Questions © 2015 Pearson Education, Inc. Basic Energy Model Work and heat are energy transfers that change the system’s total energy If the system is isolated, the total energy is conserved Stop To Think: Review Christina throws a javelin into the air. As she propels it forward from rest, she does 270 J of work on it. At its highest point, its gravitational potential energy has increased by 70 J. What is the javelin’s kinetic energy at this point? A. B. C. D. E. 270 J 340 J 200 J –200 J –340 J Stop To Think: Review Christina throws a javelin into the air. As she propels it forward from rest, she does 270 J of work on it. At its highest point, its gravitational potential energy has increased by 70 J. What is the javelin’s kinetic energy at this point? W = 270 J ΔUg = 70 J A. 270 J B. C. D. E. 340 J 200 J –200 J –340 J Stop To Think: Review Christina throws a javelin into the air. As she propels it forward from rest, she does 270 J of work on it. At its highest point, its gravitational potential energy has increased by 70 J. What is the javelin’s kinetic energy at this point? W = 270 J ΔUg = 70 J A. 270 J B. C. D. E. 340 J 200 J –200 J –340 J Using the work energy theorem we have W = ∆K + ΔUg Stop To Think: Review Christina throws a javelin into the air. As she propels it forward from rest, she does 270 J of work on it. At its highest point, its gravitational potential energy has increased by 70 J. What is the javelin’s kinetic energy at this point? W = 270 J ΔUg = 70 J A. 270 J B. C. D. E. 340 J 200 J –200 J –340 J Using the work energy theorem we have W = ∆K + ΔUg or ∆K = W - ΔUg Stop To Think: Review Christina throws a javelin into the air. As she propels it forward from rest, she does 270 J of work on it. At its highest point, its gravitational potential energy has increased by 70 J. What is the javelin’s kinetic energy at this point? W = 270 J ΔUg = 70 J A. 270 J B. C. D. E. 340 J 200 J –200 J –340 J Using the work energy theorem we have W = ∆K + ΔUg or ∆K = W - ΔUg Transforming Energy Energy cannot be created or destroyed, but it can be transformed Transforming Energy Energy cannot be created or destroyed, but it can be transformed The energy in these transformations is not lost, but converted to other forms that are typically less useful to us Transforming Energy Energy cannot be created or destroyed, but it can be transformed The energy in these transformations is not lost, but converted to other forms that are typically less useful to us Transforming Energy Energy cannot be created or destroyed, but it can be transformed The energy in these transformations is not lost, but converted to other forms that are typically less useful to us Transforming Energy Energy cannot be created or destroyed, but it can be transformed The energy in these transformations is not lost, but converted to other forms that are typically less useful to us Transforming Energy The work-energy equation includes work, an energy transfer. We now include electric, radiant, and chemical energy in our definition of work Transforming Energy The work-energy equation includes work, an energy transfer. We now include electric, radiant, and chemical energy in our definition of work Work is positive when energy is transferred into the system and negative when energy is transferred out of the system Transforming Energy The work-energy equation includes work, an energy transfer. We now include electric, radiant, and chemical energy in our definition of work Work is positive when energy is transferred into the system and negative when energy is transferred out of the system When other forms of energy are transformed into thermal energy the change is irreversible; the energy isn’t lost, but it is lost to our use QuickCheck Question 11.1 When you walk at a constant speed on level ground, what energy transformation is taking place? A. Echem → Ug B. Ug → Eth C. Echem → K D. Echem → Eth E. K → Eth QuickCheck Question 11.1 When you walk at a constant speed on level ground, what energy transformation is taking place? A. Echem → Ug B. Ug → Eth C. Echem → K D. Echem → Eth E. K → Eth Efficiency This woman climbs the stairs, converting her body’s chemical energy into gravitational potential energy Efficiency This woman climbs the stairs, converting her body’s chemical energy into gravitational potential energy How efficiently will this woman climb to the top of the stairs? Efficiency The larger the energy losses in a system, the lower its efficiency Efficiency The larger the energy losses in a system, the lower its efficiency Reductions in efficiency are caused by: Efficiency The larger the energy losses in a system, the lower its efficiency Reductions in efficiency are caused by: 1. Process limitations: energy loss due to practical details such as the design of the mechanism Efficiency The larger the energy losses in a system, the lower its efficiency Reductions in efficiency are caused by: 1. Process limitations: energy loss due to practical details such as the design of the mechanism 2. Fundamental limitations: energy loss due to physical laws (transformation of energy into thermal energy) Efficiency 35% efficiency is close to the theoretical maximum for power plants due to fundamental limitations Efficiency Example 1.1: Lightbulb Efficiency A 15 W compact fluorescent bulb and a 75 W incandescent bulb each produce 3.0 W of visible-light energy. What are the efficiencies of these two types of bulbs for converting electric energy into light? Example 1.1: Lightbulb Efficiency PREPARE The problem statement gives us values for power. But 15 W is 15 J/s, so we will consider the value for the power to be the energy in 1 second. For each of the bulbs, “what you get” is the visible light output—3.0 J of light every second for each bulb. “What you had to pay” is the electric energy to run the bulb. This is how the bulbs are rated. A bulb labeled “15 W” uses 15 J of electric energy each second. A 75 W bulb uses 75 J each second. Example 1.1: Lightbulb Efficiency SOLVE The efficiencies of the two bulbs are computed using the energy in 1 second: Example 1.1: Lightbulb Efficiency SOLVE The efficiencies of the two bulbs are computed using the energy in 1 second: Both bulbs produce the same visible-light output, but the compact fluorescent bulb does so with a significantly lower energy input, so it is more efficient. Compact fluorescent bulbs are more efficient than incandescent bulbs, but their efficiency is still relatively low—only 20%. ASSESS Getting Energy From Food: Energy Inputs The chemical energy in food provides the necessary energy input for your body to function Getting Energy From Food: Energy Inputs Chemical energy in food is broken down into simpler molecules such as glucose and glycogen and are metabolized in the cells by combining with oxygen Getting Energy From Food: Energy Inputs Chemical energy in food is broken down into simpler molecules such as glucose and glycogen and are metabolized in the cells by combining with oxygen Oxidation reactions “burn” the fuel you obtain by eating Getting Energy From Food: Energy Inputs Burning food transforms all of its chemical energy into thermal energy Getting Energy From Food: Energy Inputs Burning food transforms all of its chemical energy into thermal energy Thermal energy is measured in units of calories (cal) 1.00 calorie = 4.19 joules Getting Energy From Food: Energy Inputs Getting Energy From Food: Energy Inputs Example 11.2: Energy in Food A 12 oz can of soda contains approximately 40 g (or a bit less than 1/4 cup) of sugar, a simple carbohydrate. What is the chemical energy in joules? How many Calories is this? Example 11.2: Energy in Food A 12 oz can of soda contains approximately 40 g (or a bit less than 1/4 cup) of sugar, a simple carbohydrate. What is the chemical energy in joules? How many Calories is this? SOLVE From Table 11.1, 1 g of sugar contains 17 kJ of energy, so 40 g contains Example 11.2: Energy in Food A 12 oz can of soda contains approximately 40 g (or a bit less than 1/4 cup) of sugar, a simple carbohydrate. What is the chemical energy in joules? How many Calories is this? SOLVE From Table 11.1, 1 g of sugar contains 17 kJ of energy, so 40 g contains Example 11.2: Energy in Food A 12 oz can of soda contains approximately 40 g (or a bit less than 1/4 cup) of sugar, a simple carbohydrate. What is the chemical energy in joules? How many Calories is this? SOLVE From Table 11.1, 1 g of sugar contains 17 kJ of energy, so 40 g contains Converting to Calories, we get Example 11.2: Energy in Food A 12 oz can of soda contains approximately 40 g (or a bit less than 1/4 cup) of sugar, a simple carbohydrate. What is the chemical energy in joules? How many Calories is this? SOLVE From Table 11.1, 1 g of sugar contains 17 kJ of energy, so 40 g contains Converting to Calories, we get Example 11.2: Energy in Food 160 Calories is a typical value for the energy content of a 12 oz can of soda (check the nutrition label on one to see), so this result seems reasonable. ASSESS Using Energy in the Body: Energy Outputs At rest, your body uses energy to build and repair tissue, digest food and keep warm Using Energy in the Body: Energy Outputs At rest, your body uses energy to build and repair tissue, digest food and keep warm Your body uses approximately 100 W of energy at rest which is converted almost entirely into thermal energy, which is transferred as heat to the environment Using Energy in the Body: Energy Outputs At rest, your body uses energy to build and repair tissue, digest food and keep warm Your body uses approximately 100 W of energy at rest which is converted almost entirely into thermal energy, which is transferred as heat to the environment When you feel cold, your body is releasing its heat to the environment Using Energy in the Body: Energy Outputs Using Energy in the Body: Energy Outputs When you’re active, your cells require oxygen to metabolize carbohydrates Using Energy in the Body: Energy Outputs When you’re active, your cells require oxygen to metabolize carbohydrates Your body’s energy use, or total metabolic energy use, can be measured by how much oxygen it is using Using Energy in the Body: Energy Outputs Efficiency of the Human Body As you climb stairs, you change your potential energy. Efficiency of the Human Body As you climb stairs, you change your potential energy. If you climb 2.7 m and your mass is 68 kg then your potential energy is Efficiency of the Human Body As you climb stairs, you change your potential energy. If you climb 2.7 m and your mass is 68 kg then your potential energy is On average, you use 7200 J to climb the stairs: Efficiency of the Human Body We compute the efficiency of climbing the stairs: Efficiency of the Human Body We compute the efficiency of climbing the stairs: We typically consider the body’s efficiency to be 25% Energy Storage Energy from food that is not used will be stored Energy Storage Energy from food that is not used will be stored If energy input from food continuously exceeds the energy outputs of the body, the energy is stored as fat under the skin and around the organs Energy Storage Energy from food that is not used will be stored If energy input from food continuously exceeds the energy outputs of the body, the energy is stored as fat under the skin and around the organs So get out there and move! Unless winter is coming, then stay put and get ready for hibernation 😜 Example 11.6: Running out of fuel The body stores about 400 g of carbohydrates. Approximately how far could a 68 kg runner travel on this stored energy? Example 11.6: Running out of fuel The body stores about 400 g of carbohydrates. Approximately how far could a 68 kg runner travel on this stored energy? PREPARE Table 11.1 gives a value of 17 kJ per g of carbohydrate. The 400 g of carbohydrates in the body contain an energy of Example 11.6: Running out of fuel The body stores about 400 g of carbohydrates. Approximately how far could a 68 kg runner travel on this stored energy? PREPARE Table 11.1 gives a value of 17 kJ per g of carbohydrate. The 400 g of carbohydrates in the body contain an energy of SOLVE Table 11.4 gives the power used in running at 15 km/h as 1150 W. The time that the stored chemical energy will last at this rate is Example 11.6: Running out of fuel And the distance that can be covered during this time at 15 km/h is to two significant figures. Example 11.6: Running out of fuel And the distance that can be covered during this time at 15 km/h is to two significant figures. ASSESS A marathon is longer than this—just over 42 km. Even with “carbo loading” before the event (eating highcarbohydrate meals), many marathon runners “hit the wall” before the end of the race as they reach the point where they have exhausted their store of carbohydrates. The body has other energy stores (in fats, for instance), but the rate that they can be drawn on is much lower. Energy and Locomotion When you walk at a constant speed on level ground, your kinetic and potential energy are constant Energy and Locomotion When you walk at a constant speed on level ground, your kinetic and potential energy are constant You need energy to walk because the kinetic energy of your leg and foot is transformed into thermal energy in your muscles and shoes, which is lost Summary Summary Summary Summary Summary Summary Summary Summary Summary Summary Summary Summary Summary Summary Summary Things that are due Reading Quiz #11 Due October 27, 2015 by 4:59 pm Homework #7 Due November 2, 2015 at 11:59 pm Exam #3 October 29, 2015 at 5:00 pm EXAM #3 Covers Chapters 8-11 & Lectures 14-18 Thursday, October 29, 2015 Bring a calculator and cheat sheet (turn in with exam, one side of 8.5”x11” piece of paper) Practice exam will be available on website along with solutions QUESTIONS?