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Unformatted text preview: Homework 6: Work and Kinetic Energy Due: 11:59pm on Sunday, February 26, 2017 To understand how points are awarded, read the Grading Policy for this assignment. Understanding Work Done by a Constant Force ­ Copy Learning Goal: To explore the definition of work and learn how to find the work done by a force on an object. The word "work" has many meanings when used in everyday life. However, in physics work has a very specific definition. This definition is important to learn and understand. Work and energy are two of the most fundamental and important concepts you will learn in your study of physics. Energy cannot be created or destroyed; it can only be transformed from one form to another. How this energy is transferred affects our daily lives from microscopic processes, such as protein synthesis, to macroscopic processes, such as the expansion of the universe! ⃗ ⃗ When energy is transferred either to or away from an object by a force F acting over a displacement d , work W is done on that object. The amount of work done by a constant force can be found using the equation , W = F d cos θ ⃗ ⃗ ⃗ ⃗ where F is the magnitude of F , d is the magnitude of d , and θ is the angle between F and d . The SI unit for work is the joule, J. A single joule of work is not very big. Your heart uses about 0.5 J each time it beats, and the 60­watt lightbulb in your desk lamp uses 216, 000 J each hour. A joule is defined as follows: 2 1J = 1Nm = 1 kg m 2 s The net work done on an object is the sum of the work done by each individual force acting on that object. In other words, Wnet = W1 + W2 + W3 + ⋯ = ∑ Wi i . The net work can also be expressed as the work done by the net force acting on an object, which can be represented by the following equation: . Wnet = F net d cos θ Knowing the sign of the work done on an object is a crucial element to understanding work. Positive work indicates that an object has been acted on by a force that tranfers energy to the object, thereby increasing the object's energy. Negative work indicates that an object has been acted on by a force that has reduced the energy of the object. The next few questions will ask you to determine the sign of the work done by the various forces acting on a box that is being pushed across a rough floor. As illustrated in the figure , the box is being acted on by a normal force n⃗ , the force due to gravity w⃗ , the force of kinetic friction f k⃗ , and the pushing force F p⃗ . The displacement of the box is d ⃗ . Part A Which of the following statements accurately describes the sign of the work done on the box by the force of the push? Hint 1. Find the angle The work done on the box by the pushing force depends on the angle θ between F p⃗ and the displacement d ⃗ . What is this angle? ANSWER: 0 degrees 45 degrees 90 degrees 180 degrees ANSWER: positive negative zero Correct Part B Which of the following statements accurately decribes the sign of the work done on the box by the normal force? Hint 1. Finding theta The work done on the box by the normal force depends on the angle θ between n⃗ and the displacement d ⃗ . What is this angle? ANSWER: 0 degrees 45 degrees 90 degrees 180 degrees ANSWER: positive negative zero Correct Part C Which of the following statements accurately decribes the sign of the work done on the box by the force of kinetic friction? Hint 1. Finding theta The work done on the box by the force of kinetic friction depends on the angle θ between f k⃗ and the displacement d ⃗ . What is this angle? ANSWER: 0 degrees 45 degrees 90 degrees 180 degrees ANSWER: positive negative zero Correct Part D Which of the following statements accurately decribes the sign of the work done on the box by the force of gravity? Hint 1. Finding the angle The work done on the box by the weight depends on the angle θ between w⃗ and the displacement d ⃗ . What is this angle? ANSWER: 0 degrees 45 degrees 90 degrees 180 degrees ANSWER: positive negative zero Correct Making generalizations You may have noticed that the force due to gravity and normal forces do no work on the box. Any force that is perpendicular to the displacement of the object on which it acts does no work on the object. The force of kinetic friction did negative work on the box. In other words, it took energy away from the box. Typically, this energy gets transformed into heat, like the heat that radiates from your skin when you get a rug burn due to the friction between your skin and the carpet. A force that acts on an object in a direction opposite to the direction of the object's displacement does negative work on the object. The pushing force acts on the box in the same direction as the object's displacement and does positive work on the box. These generalizations allow physicists to rewrite the equation for work as , W = F∥ d ⃗ ⃗ where F|| is the component of F that is either parallel or antiparallel to the displacement. If F || is parallel to d , as in the case of F p⃗ , then the work done is positive. If F|| is antiparallel to d ⃗ , as in the case of f k⃗ , then the work done is negative. Part E You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,500 N across the carpet to a spot 5 m away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser? ANSWER: W W > = 1.75 × 10 1.75 × 10 1.75 × 10 W = 4 J > 4 J 4 J W > 0 J 0 J Correct Remember that to a physicist work means something very specific, and since you were unable to move the dresser, d = 0 and therefore W = 0. However, you got tired and sweaty trying to move the dresser, just as you do when you go to "work out" at the gym.Your muscles are not static strips of fibrous tissue. They continually contract and expand a slight amount when you exert them. Chemical energy from food is being transformed into the energy needed to move your muscles. Work is being done inside your muscles, but work is not being done on the dresser. Part F A box of mass m is sliding down a frictionless plane that is inclined at an angle ϕ above the horizontal, as shown in the figure . What is the work done on the box by the force due to gravity w, if the box moves a distance d? Hint 1. Finding Theta. The work done on the box by the force of gravity depends on the angle between the weight and the displacement; this is the angle θ that goes into the equationW = F d cos θ. ANSWER: W = wd cos ϕ W = wd cos(90 − ϕ) W = 0 None of these Correct The angle given to you in a problem is not always the same angle that you use in the equation for work! ± All Work and No Play Learning Goal: To be able tocalculate work done by a constant force directed at different angles relative to displacement If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as ⃗ ⃗ ∣ ∣ ⃗ cos θ W = F ⋅ s ⃗ = F ∣ ∣s ∣ ∣ ∣ ∣ , ⃗ ⃗ ⃗ where W is the work done by force F on the object that undergoes displacement s directed at angle θ relative to F . Note that depending on the value of cos θ, the work done can be positive, negative, or zero. In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. Part A What can be said about the sign of the work done by the force F 1⃗ ? ANSWER: It is positive. It is negative. It is zero. There is not enough information to answer the question. Correct When θ = 90 ∘ , the cosine of θ is zero, and therefore the work done is zero. Part B What can be said about the work done by force F 2⃗ ? ANSWER: It is positive. It is negative. It is zero. Correct When 0∘ < θ < 90 ∘ , cos θ is positive, and so the work done is positive. Part C The work done by force F 3⃗ is ANSWER: positive negative zero Correct When 90 ∘ ∘ < θ < 180 , cos θ is negative, and so the work done is negative. Part D The work done by force F 4⃗ is ANSWER: positive negative zero Correct Part E The work done by force F 5⃗ is ANSWER: positive negative zero Correct Part F The work done by force F 6⃗ is ANSWER: positive negative zero Correct Part G The work done by force F 7⃗ is ANSWER: positive negative zero Correct In the next series of questions, you will use the formula W ⃗ ⃗ ∣ ∣ ⃗ cos θ = F ⋅ s ⃗ = F ∣ ∣s ∣ ∣ ∣ ∣ to calculate the work done by various forces on an object that moves 160 meters to the right. Part H Find the work W done by the 18­newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: W = 2900 J Correct Part I Find the work W done by the 30­newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: W = 4200 J Correct Part J Find the work W done by the 12­newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: W = ­1900 J Correct Part K Find the work W done by the 15­newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: W = ­1800 J Correct Exercise 6.6 Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.2×106 N , one an angle 11 ∘ west of north and the other an angle 11 ∘ east of north, as they pull the tanker a distance 0.90 km toward the north. Part A What is the total work they do on the supertanker? Express your answer using two significant figures. ANSWER: W = 3.9×109 J Correct Exercise 6.8 A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force F ⃗ = ( 35 N )^i − ( 42 N )^ j to the cart as it undergoes a displacement s ⃗ = 8.7 m )^i − ( 3.7 m )^ j. (− Part A How much work does the force you apply do on the grocery cart? Express your answer using three significant figures. ANSWER: W = ­149 J Correct When Push Comes to Shove Two forces, of magnitudes F1 = 85.0 N and F2 = 45.0 N , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position xi = ­5.00 cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 1.00 cm . Part A Find the work W1 done on the block by the force of magnitude F1 = 85.0 N as the block moves from xi = ­5.00 cm to xf = 1.00 cm . Express your answer numerically, in joules. Hint 1. Formula for the work done by a force ⃗ ⃗ The work W done by a force F in producing a displacement s is given by ⃗ ∣ ∣ ⃗ ⃗ cos ϕ W = F ⋅ s ⃗ = F ∣ s∣ ∣ ∣∣ ∣ , ⃗ ⃗ where ∣∣F ∣∣⃗ and ∣∣s ∣∣⃗ are the magnitudes of F and s respectively, and ϕ is the smaller angle between the two vectors. ⃗ ⃗ The scalar that results from the operation F ⃗ ⋅ s is called the scalar product, or the dot product, of the vectors F and s .⃗ ANSWER: W1 = 5.10 J Correct Part B Find the work W2 done by the force of magnitude F2 = 45.0 N as the block moves from xi = ­5.00 cm to xf = 1.00 cm . Express your answer numerically, in joules. Hint 1. Is the work positive or negative? The force of magnitude F2 acts in the opposite direction to that of the motion of the block. Therefore, the work done by that force must be negative. ANSWER: W2 = ­2.70 J Correct Part C What is the net work Wnet done on the block by the two forces? Express your answer numerically, in joules. ANSWER: Wnet = 2.40 J Correct Part D Determine the changeKf − Ki in the kinetic energy of the block as it moves from xi = ­5.00 cm to xf = 1.00 cm . Express your answer numerically, in joules. Hint 1. Conservation of energy The work done on the block goes into changing its kinetic energy. Thus the net work done by the two forces is equal to the change in the kinetic energy. ANSWER: Kf − Ki = 2.40 J Correct Work and Kinetic Energy Two blocks of ice, one four times as heavy as the other, are at rest on a frozen lake. A person pushes each block the same ⃗ distance d. Ignore friction and assume that an equal force F is exerted on each block. Part A Which of the following statements is true about the kinetic energy of the heavier block after the push? Hint 1. How to approach the problem The work­energy theorem states that the change in kinetic energy of an object equals the net work done on that object: Wtotal = ΔK. The work done on an object can also be related to the distance d that the object moves while being acted on by a force F :⃗ , W = F∥ d ⃗ where F|| is the component of F parallel to the direction of displacement. Hint 2. Find the work done on each block What can be said about the net work done on the heavier block? ANSWER: It is greater than the work done on the lighter block. It is equal to the work done on the lighter block. It is less than the work done on the lighter block. ANSWER: It is smaller than the kinetic energy of the lighter block. It is equal to the kinetic energy of the lighter block. It is larger than the kinetic energy of the lighter block. It cannot be determined without knowing the force and the mass of each block. Correct The work­energy theorem states that the change in kinetic energy of an object equals the net work done on that object. The only force doing work on the blocks is the force from the person, which is the same in both cases. Since the initial kinetic energy of each block is zero, both blocks have the same final kinetic energy. Part B Compared to the speed of the heavier block, what is the speed of the light block after both blocks move the same distance d? Hint 1. How to approach the problem In Part A, you determined that the kinetic energy of the heavier block was the same as that of the lighter block. Relate this to the speed of the blocks. Hint 2. Proportional reasoning Proportional reasoning becomes easier with practice. First relate the kinetic energies of the blocks to each other. To accomplish this, let the subscript h refer to the heavier block and the subscript ℓ to the lighter block. Now Kh = Kℓ can be written as 1 2 2 mh (vh ) = 1 2 2 mℓ (vℓ ) . The problem states that the heavier block is four time as massive as the lighter block. This can be represented by the expression m h = 4m ℓ . Substituting this expression into the expression for kinetic energy yields 1 2 2 (4mℓ )(vh ) = 1 2 2 mℓ (vℓ ) . How many times larger than v 2h is v 2ℓ ? ANSWER: v 2 ℓ = 4 v 2h ANSWER: one quarter as fast half as fast the same speed twice as fast four times as fast Correct Since the kinetic energy of the lighter block is equal to the kinetic energy of the heavier block, the lighter block must be moving faster than the heavier block. Part C ⃗ Now assume that both blocks have the same speed after being pushed with the same force F . What can be said about the distances the two blocks are pushed? Hint 1. How to approach the problem The work­energy theorem states that the change in kinetic energy of an object equals the net work done on that object: Wtotal = ΔK. The work done on an object can also be related to the distance d that the object moves while being acted on by a force F :⃗ = , W = F∥ d ⃗ where F|| is the component of F parallel to the direction of displacement. Hint 2. Relate the kinetic energies of the blocks Let the subscript h refer to the heavier block and the subscript ℓ to the lighter block. What is the ratio Kh Kℓ ? Hint 1. The kinetic energies To relate the kinetic energies of the blocks to each other, recall that vh = vℓ and m h = 4m ℓ . ANSWER: Kh Kℓ = 4 Hint 3. Compare the amount of work done on each block In the previous hint, you found that Kh done on the lighter block, Wh Wℓ = 4Kℓ . What is the ratio of the work done on the heavy block to the work ? ANSWER: Wh Wℓ 4 = ANSWER: The heavy block must be pushed 16 times farther than the light block. The heavy block must be pushed 4 times farther than the light block. The heavy block must be pushed 2 times farther than the light block. The heavy block must be pushed the same distance as the light block. The heavy block must be pushed half as far as the light block. Correct Because the heavier block has four times the mass of the lighter block, when the two blocks travel with the same speed, the heavier block will have four times as much kinetic energy. The work­energy theorem implies that four times more work must be done on the heavier block than on the lighter block. Since the same force is applied to both blocks, the heavier block must be pushed through four times the distance as the lighter block. Exercise 6.18 A 4.10­kg watermelon is dropped from rest from the roof of a 21.0­m ­tall buildingand feels no appreciable air resistance. Part A Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. ANSWER: W = 844 J Correct Part B Just before it strikes the ground, what is the watermelon's kinetic energy? ANSWER: K = 844 J Correct Part C Just before it strikes the ground, what is the watermelon's speed? ANSWER: v = 20.3 m/s Correct Part D Would the answer in part A be different if there were appreciable air resistance? ANSWER: Yes No Correct Part E Would the answer in part B be different if there were appreciable air resistance? ANSWER: Yes No Correct Part F Would the answer in part C be different if there were appreciable air resistance? ANSWER: Yes No Correct Exercise 6.34 ⃗ A child applies a force F parallel to the x ­axis to a 10.0­kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x ­component of the force she applies varies with the x ­coordinate of the sled as shown in the figure . Part A ⃗ Calculate the work done by the force F when the sled moves from x=0 to x=8.0m. Express your answer using two significant figurs. ANSWER: W = 40 J Correct Part B ⃗ Calculate the work done by the force F when the sled moves from x=8.0m to x =12.0m. Express your answer using two significant figurs. ANSWER: W = 20 J Correct Part C ⃗ Calculate the work done by the force F when the sled moves from x=0 to x =12.0m. . Express your answer using two significant figurs. ANSWER: W = 60 J Correct Exercise 6.55 Your job is to lift 30­kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck. Part A How many crates would you have to load onto the truck in one minute for the average power output you use to lift the crates to equal 0.50 hp? Express your answer using two significant figures. ANSWER: 85 1/min Correct Part B How many crates for an average power output of 100 W? Express your answer using two significant figures. ANSWER: 23 1/min Correct Problem 6.68 ­ Work Done on an Inclined Plane. A 5.10 kg package slides 1.42 m down a long ramp that is inclined at 12.4 ∘ below the horizontal. The coefficient of kinetic friction between the package and the ramp is μk = 0.315. Hint: Before starting this problem, draw a proper force diagram for the package and use it to solve for the magnitude of the normal force on the package. This will also allow you to solve for the magnitude of the friction force on the package. Part A Calculate the work done on the package by friction. Express your answer to three significant figures. ANSWER: Wf = ­21.8 J Correct Part B Calculate the work done on the package by gravity. Express your answer to three significant figures. ANSWER: Wg = 15.2 J Correct Part C Calculate the work done on the package by the normal force. Express your answer to three significant figures. ANSWER: Wn = 0 J Correct Part D Calculate the total work done on the package. Express your answer to three significant figures. ANSWER: Wnet = ­6.59 J Correct Part E If the package has a speed of 2.29 m/s at the top of the ramp, what is its speed after sliding the distance 1.42 m down the ramp? Express your answer to three si...
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