Work, Energy, Power. Questions

Work, Energy, Power. Prelecture Concept Question Part A A person applies a 50 N force on a crate, causing it to move horizontally at a constant speed through a distance of 10 m. What is the net work done on the crate? ANSWER: The net work done on the crate is 500 J. The net work done on the crate is 0.2 J. The net work done on the crate is 5 J. The net work done on the crate is 250 J. The net work done on the crate is zero joules. Correct Prelecture Concept Question Part A For motion in one dimension, a graph depicts force versus position. What represents the work done by the force over the given displacement? ANSWER: Work cannot be determined from this type of graph. The work done is equal to the area under the curve. The work done is equal to length of the curve. The work done is equal to the slope of the curve. The work done is equal to the product of the maximum force times the maximum position. Correct ± Hooke's Law Learning Goal: To understand the use of Hooke's law for a spring. → Hooke's law states that the restoring force FF_vec on a spring when it has been stretched or compressed is proportional to the displacement → xx_vec of the spring from its equilibrium position. The equilibrium position is the position at which the spring is neither stretched nor compressed. → → Recall that F → x means that FF_vec is equal to a constant times → xx_vec. For a spring, the proportionality constant is called the spring constant and denoted by kk. The spring constant is a property of the spring and must be measured experimentally. The larger the value of kk, the stiffer the spring. In equation form, Hooke's law can be written → F = −k→x. The minus sign indicates that the force is in the opposite direction to that of the spring's displacement from its equilibrium length and is "trying" to restore the spring to its equilibrium position. The magnitude of the force is given by F = kx, where xx is the magnitude of the displacement. In Haiti, public transportation is often by taptaps, small pickup trucks with seats along the sides of the pickup bed and railings to which passengers can hang on. Typically they carry two dozen or more passengers plus an assortment of chickens, goats, luggage, etc. Putting this much into the back of a pickup truck puts quite a large load on the truck springs. A truck has springs for each wheel, but for simplicity assume that the individual springs can be treated as one spring with a spring constant that includes the effect of all the springs. Also for simplicity, assume that all four springs compress equally when weight is added to the truck and that the equilibrium length of the springs is the length they have when they support the load of an empty truck. Part A A 65 kg driver gets into an empty taptap to start the day's work. The springs compress 1.7×10−2 m . What is the effective spring constant of the spring system in the taptap? Enter the spring constant numerically in newtons per meter using two significant figures. Hint 1. How to approach the problem The compression of the springs is governed by Hooke's law. The amount the springs are compressed when the driver climbs into the truck is given in the problem statement. The force that acts to compress the springs is the force caused by the driver getting into the truck. ANSWER: kk = 3.7×104 N / m Correct If you need to use the spring constant in subsequent parts, use the full precision value you calculated, only rounding as a final step before submitting your answer. Part B After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average mass of 65 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed? Enter the compression numerically in meters using two significant figures. Hint 1. How to find the compression of the spring The spring compression is governed by Hooke's law. Use the spring constant you calculated to full precision in Part A prior to rounding your answer. To find the force add the total weight of the load on the truck. Only round as a final step before submitting your answer. ANSWER: xx = 0.49 m Correct Part C Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physically realistic. Think about how far off the ground a typical small truck is. Is the answer to Part B physically realistic? Select the best choice below. ANSWER: No, typical small pickup truck springs are not large enough to compress 0.48 m . Yes, typical small pickup truck springs can easily compress 0.48 m . Correct The answer to Part B is not physically realistic because the springs of a typical light truck will compress their maximum amount (typically about 10 cm) before the total weight of all the passengers and other cargo given in Part B is added to the truck. When this maximum compression is reached, the springs will bottom out, and the ride will be very rough. Part D Now imagine that you are a Haitian taptap driver and want a more comfortable ride. You decide to replace the springs with new springs that can handle the typical heavy load on your vehicle. What spring constant do you want your new spring system to have? ANSWER: substantially larger slightly larger The new springs should have a spring constant that is slightly smaller substantially smaller springs. than the spring constant of the old Correct A spring constant with a large value is a stiff spring. It will take more force to compress (or stretch) a stiff spring. On a taptap, stiffer springs are less likely to bottom out under a heavy load. However, with a lighter load, for most vehicles, very stiff springs will not compress as much for a bump in the road. Hence very stiff springs will give a better ride with a very heavy load, but less-stiff springs (lower spring constant) will give a smoother ride with a light load. This is why larger vehicles need stiffer springs than smaller vehicles. Prelecture Concept Question 8.01 Part A Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first object. They both slide up the same frictionless incline plane. Which object rises to a greater height? ANSWER: Object 1 rises to the greater height because it weighs less. Object 2 rises to the greater height because it possesses a larger amount of kinetic energy. Object 2 rises to the greater height because it contains more mass. Object 1 rises to the greater height because it possesses a smaller amount of kinetic energy. The two objects rise to the same height. Correct Video Tutor: Chin Basher? First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point. Part A Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g =30m / s 2. When he releases the ball from chin height without giving it a push, how will the ball's behavior differ from its behavior on Earth? Ignore friction and air resistance. (Select all that apply.) Hint 1. How to approach the problem To answer this question, you must evaluate how a change in gg affects the pendulum’s trajectory (choices A and B), period (choices C and D), and mass (choice E). Trajectory: Here, the key is energy conservation. Think about how kinetic and potential energy interchange as a pendulum swings. For the pendulum to return to a position higher than its starting point, the system of the pendulum and planet would have to gain mechanical energy. Where would that energy come from? (The same argument applies to choice B.) Period: You can answer this question by seeing whether the equation for the period of a pendulum includes gg. Or you can use reasoning. A greater downward acceleration gg will cause the pendulum to reach the bottom of its swing more quickly. What does that say about the pendulum’s total period? Mass: The pendulum will weigh more on the massive planet. Will it have more mass? ANSWER: It will take more time to return to the point from which it was released. Its mass will be greater. It will smash his face. It will take less time to return to the point from which it was released. It will stop well short of his face. Correct The pendulum will swing back and forth more quickly (with a shorter period) because it is oscillating in a stronger gravitational field than that on Earth. Fat: The Fuel of Migrating Birds Small birds like that in can migrate over long distances without feeding, storing energy mostly as fat rather than carbohydrate. Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4184 joules. Therefore, in this problem use the conversion factor 1Cal = 4184J. Part A Consider a bird that flies at an average speed of 10.7 m / s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance d bd_b without stopping for feeding. How far will the bird fly before feeding again? Express your answer in kilometers. Hint 1. How to approach the problem From the average speed of the bird, you can calculate how far the bird can fly without stopping if you know the duration of the flight. To determine the duration of the flight, first find the amount of energy available from converting 4 grams of fat, and then use the definition of power. Hint 2. Find the energy used during the flight How much energy E bE_b does the bird have available when it converts 4 grams of fat? Express your answer in kilojoules. Hint 1. Converting fat into energy As stated in the introduction of this problem, 1.00 grams of fat provides about 9.40 (food) Calories. Also keep in mind that 1Calorie = 1000calories = 4184J. ANSWER: E bE_b = 157 kJ Correct Hint 3. Find the duration of the flight If the bird consumes energy at a rate of 3.70 W , how many hours t bt_b can it fly using the energy supply provided by 4.00 grams of fat? Express your answer in hours. Hint 1. Definition of power The average power PP (measured in watts) is the ratio of the energy ΔEDeltaE transformed in the time interval ΔtDeltat: P = ΔE Δt . Note that power measures either the rate at which energy is transferred (or transformed) or the rate at which work is performed. Hint 2. Power: units Power is measured in watts. One watt is equal to 1 joule per second (i.e., 1W = 1J / s). ANSWER: t bt_b = 11.8 hr Correct Hint 4. Find the distance in terms of average velocity Which of the following expressions gives the distance dd traveled in the time interval tt at an average speed vv? ANSWER: v/t d= vt2 vt t/v ANSWER: d bd_b = 455 km Correct Part B How many grams of carbohydrate m carbm_carb would the bird have to consume to travel the same distance d bd_b? Express your answer in grams Hint 1. How to approach the problem As stated in the introduction of this problem, 1.00 gram of fat provides about 9.40 Calories, while 1.00 grams of carbohydrate provides 4.20 Calories. ANSWER: m carbm_carb = 8.95 g Correct This is more than twice the amount of fat that was needed! In addition, to store 1 gram of carbohydrate (in the form of glycogen, the most common form of animal carbohydrate) about 3 grams of water are needed. Therefore, if energy were stored as carbohydrates, the bird would need to carry more than eight times the fuel mass to perform the same migratory flight! Part C Field observations suggest that a migrating ruby-throated hummingbird can fly across the Gulf of Mexico on a nonstop flight traveling a distance of about 800 km . Assuming that the bird has an average speed of 40.0 km / hr and an average power consumption of 1.70 W , how many grams of fat m fatm_fat does a ruby-throated hummingbird need to accomplish the nonstop flight across the Gulf of Mexico? Express your answer in grams. Hint 1. How to approach the problem In Part A you were given the amount of fat consumed over the entire flight and were asked to calculate the distance traveled by the migrating bird. Now you need to solve the reverse problem. That is, given the distance traveled, calculate the amount of energy required to perform the flight. Thus, apply the same method as the one used in part A, only in reverse. From the information on distance and average speed, calculate the duration of the nonstop flight. Then use your result and the given power consumption to determine the amount of energy required for the flight. Finally, calculate how many grams are needed to provide that amount of energy. Hint 2. Find the duration of the flight How many hours t ht_h will the ruby-throated hummingbird fly to travel a distance of 800 km at an average speed of 40.0 km / hr ? Express your answer in hours ANSWER: t ht_h = 20.0 hr Hint 3. Find the energy required for the nonstop flight Given that the hummingbird consumes energy at an average rate of 1.70 W , how much energy E hE_h will it require during the nonstop flight over the Gulf of Mexico? Express your answer in joules. Hint 1. How to use power and time Remember that power is energy transferred (or transformed) per unit time. Make sure you are using the correct unit for time. ANSWER: E hE_h = 1.22×105 J ANSWER: m fatm_fat = 3.11 g Correct Considering that in normal conditions the mass of a ruby-throated hummingbird is only 3 or 4 grams, the bird will need to almost double its body mass to store enough fat to perform the nonstop flight. Prelecture Concept Question 8.02 Part A Two identical balls are thrown vertically upward. The second ball is thrown with an initial speed that is twice that of the first ball. How does the maximum height of the two balls compare? ANSWER: The maximum heights of the two balls are equal. The maximum height of the second ball is four times that of the first ball. The maximum height of the second ball is two times that of the first ball. The maximum height of the second ball is 1.41 times that of the first ball. The maximum height of the second ball is eight times that of the first ball. Correct Fun with a Spring Gun A spring-loaded toy gun is used to shoot a ball of mass m = 1.50kg straight up in the air, as shown in . The spring has spring constant k = 667N / m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y =0 and then released, the ball reaches a maximum height h maxh_max (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis. Part A Which of the following statements are true? Check all that apply. Hint 1. Nonconservative forces Dissipative, or nonconservative, forces are those that always oppose the motion of the object on which they act. Forces such as friction and drag are dissipative forces. Hint 2. Forces acting on the ball The ball is acted on by the spring force only when the two are in contact. The force of tension in the spring is a conservative force. Also, the ball is always acted on by gravity, which is also a conservative, or nondissipative, force. ANSWER: Mechanical energy is conserved because no dissipative forces perform work on the ball. The forces of gravity and the spring have potential energies associated with them. No conservative forces act in this problem after the ball is released from the spring gun. Correct Part B Find v mv_m the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position y =0). Hint 1. Determine how to approach the problem What physical relationship can you use to solve this problem? Choose the best answer. ANSWER: kinematics equations Newton's second law law of conservation of energy conservation of momentum Hint 2. Energy equations Recall that kinetic energy is given by the equation 1 2 mv2, where vv is the speed of the object and mm is the object's mass. Gravitational potential energy is given by mgy, where yy is the object's height measured from y =0. The elastic potential energy of a spring is given by 1 2 kx2, where kk is the spring constant and xx is the spring's displacement from equilibrium. Hint 3. Determine which two locations you should examine Pick the two points along the ball's path that would be most useful to compare in order to find the solution to this problem. Choose from among the following three points: Check all that apply. ANSWER: y = −25cm, the location of the ball when the spring is compressed. y =0, the equilibrium position of the spring. y = h max, the maximum height that the ball reaches above the point y =0. Hint 4. Find the initial energy of the system A useful statement of mechanical energy conservation relating the initial and final kinetic (KK) and potential (UU) energies is K initial+ U initial = K final+ U final. In this situation, the initial position is y = −25.0cm and the final position is y =0, which is the equilibrium position of the spring. What kind(s) of energy does the system "spring-ball" have at the initial position? ANSWER: kinetic only elastic potential only gravitational potential only kinetic and gravitational potential kinetic and elastic potential elastic and gravitational potentials Hint 5. Determine the final energy A useful statement of mechanical energy conservation relating the initial and final kinetic (KK) and potential (UU) energies is K initial+ U initial = K final+ U final. In this situation, the initial position is y = −25.0cm and the final position is y =0, which is the equilibrium position of the spring. What kind(s) of energy does the system "spring-ball" have at the final position? ANSWER: kinetic only elastic potential only gravitational potential only kinetic and gravitational potential kinetic and elastic potential elastic and gravitational potentials Hint 6. Creating an equation From the hints you now know what kinds of energy are present at the initial and final positions chosen for the ball in this part of the problem. You also know that K initial+ U initial = K final+ U final. It has been determined that K initialK_initial is zero and U initialU_initial consists of two terms: gravitational potential energy and elastic potential energy. In addition, U finalU_final is zero. ANSWER: v m = 4.78 m / s Correct Part C Find the maximum height h maxh_max of the ball. Express your answer numerically, in meters. Hint 1. Choose two locations to examine Pick the two points along the ball's movement that would be most useful to compare in order to find a solution to this problem. Choose from among the following three points: Check all that apply. ANSWER: y = −25cm, the location of the ball when the spring is compressed. y =0, the equilibrium position of the spring. y = h max, the maximum height that the ball reaches measured from y =0. Hint 2. Find the initial energy A useful statement of mechanical energy conservation is K inital+ U initial = K final+ U final. Recall that in the problem statement, y =0 is set to correspond to the equilibrium position of the spring. Therefore, in this situation, the initial location is at y = −25cm and the final position should be taken as y = h max. What kind(s) of energy does the ball have at the initial location? ANSWER: kinetic only elastic potential only gravitational potential only kinetic and gravitational potential kinetic and elastic potential elastic and gravitational potentials Hint 3. Determine the final energy A useful statement of mechanical energy conservation is K initial+ U initial = K final+ U final. In this situation, the initial location is at y = −25cm, and the final position should be taken as y = h max. What kind(s) of energy does the ball have at y = h max? Hint 1. Find the speed of the ball at the top of its trajectory What is the speed v topv_top of the ball at the top of its trajectory? Express your answer numerically, in meters per second. Hint 1. Motion in the vertical direction Recall from kinematics that a ball travels upward until its speed decreases to zero, at which point it starts falling back to Earth. ANSWER: v topv_top = 0 m / s ANSWER: kinetic only elastic potential only gravitational potential only kinetic and gravitational potential kinetic and elastic potential elastic and gravitational potentials Hint 4. Creating an equation From the above hints, you now know what kind of energy is present at the inital and final positions chosen for the ball in this part of the problem. You know that K initial+ U initial = K final+ U final. It was determined that K initialK_initial is zero and that U initialU_initial consists of two terms: gravitational potential energy and elastic potential energy. In addition, K finalK_final is zero. ANSWER: h max = 1.17 m Correct In this problem you practiced applying the law of conservation of mechanical energy to a physical situation to find the muzzle velocity and the maximum height reached by the ball. Part D Which of the following actions, if done independently, would increase the maximum height reached by the ball? Check all that apply. ANSWER: reducing the spring constant kk increasing the spring constant kk decreasing the distance the spring is compressed increasing the distance the spring is compressed decreasing the mass of the ball increasing the mass of the ball tilting the spring gun so that it is at an angle θ <90 degrees from the horizontal Correct