Ch 04 HW
Short Description
Chapter 4 HW Mastering Physics...
Description
3/6/2016
Ch 04 HW
Ch 04 HW Due: 10:59pm on Sunday, March 6, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy
Newton's 1st Law Learning Goal: To understand Newton's 1st law. Newton's Principia states this first law of motion: An object subject to no net force maintains its state of motion, either at rest or at constant speed in a right line. This law may be stated as follows: If the vector sum of all forces acting on an object is zero, then the acceleration of that object is zero. Mathematically this is just a special case of the 2nd law of motion, F ⃗ = ma⃗ when F ⃗ = 0⃗ , prompting scholars to advance the following reasons (among others) for Newton's spelling it out separately: 1. This expression only holds in an inertial coordinate systemone that is not acceleratingand this law really says you have to use this type of coordinate system (i.e., Newton's laws won't work inside an accelerating rocket ship.) 2. This was a direct challenge to the Impetus theory of motion, described as follows: A mover, while moving a body, impresses on it a certain impetus, a certain power capable of moving this body in the direction in which the mover set it going, whether upwards, downwards, sideways or in a circle. By the same amount that the mover moves the same body swiftly, by that amount is the impetus that is impressed on it powerful. It is by this impetus that the stone is moved after the thrower ceases to move it; but because of the resistance of the air and the gravity of the stone, which inclines it to move in a direction opposite to that towards which the impetus tends to move it, this impetus is continually weakened. Therefore the movement of the stone will become continually slower, and at length, the impetus is so diminished or destroyed that the gravity of the stone prevails over it and moves the stone down towards its natural place. A. C. Crombie, Medieval and Early Modern Science This theory is sometimes called the Animistic theory of motion since it envisions a "life force" being associated with motion. Newton's 1st law is often very difficult to grasp because it contradicts various commonsense ideas of motion that may have been acquired from experience in everyday life. For example, unaccounted for forces like friction might cause a ball rolling on the playground to eventually stop, even though no obvious forces seem to be acting. When studying Newtonian mechanics, it is best to remember this as two laws: 1. If the net force (i.e., vector sum of all forces) acting on an object is zero, the object will keep moving with constant velocity (which may be zero). 2. If an object is moving with constant velocity (not speed), that is, with zero acceleration, then the net force acting on that object must be zero. Complete the following sentences to see if you can apply these ideas.
Part A If a car is moving to the left with constant velocity, one can conclude that ANSWER:
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there must be no forces exerted on the car. the net force exerted on the car is directed to the left. the net force exerted on the car is zero. there is exactly one force exerted on the car.
Correct
Part B An object cannot remain at rest unless ANSWER: there are no forces at all exerted on it. the net force exerted on it is zero. the net force exerted on it is constant. there is only one force exerted on it.
Correct
A Book on a Table A book weighing 5 N rests on top of a table.
Part A A downward force of magnitude 5 N is exerted on the book by the force of ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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the table gravity
.
inertia
Correct
Part B An upward force of magnitude _____ is exerted on the _____ by the table. ANSWER: 6 N / table 5 N / table 5 N / book 6 N / book
Correct
Part C Do the downward force in Part A and the upward force in Part B constitute a 3rd law pair?
Hint 1. The force of gravity The force of gravity is another name for the force exerted by the earth (or any astronomical object) on objects near its surface.
Hint 2. Exploring Newton's 3rd law Indicate whether the following statements about Newton's 3rd law are true, false, or indeterminate. 1. According to Newton's 3rd law, every real force has a unique pair force. 2. The pair force is called a "fictitious force." 3. The force and pair force must act on different point masses. 4. The force and the pair force must always have the same magnitude and must also act in exactly opposite directions. Enter t for true, f for false, or i for indeterminate for each statement, separating the answers with commas (e.g., if all but the first statement were true, you would enter f,t,t,t). ANSWER: , , ,
t f t t
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ANSWER: yes no
Correct
Part D The reaction to the force in Part A is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____ .
Hint 1. The force of gravity The force of gravity is another name for the force exerted by the earth (or any astronomical object) on objects near its surface. ANSWER: 5 N / earth / book / upward 5 N / book / table / upward 5 N / book / earth / upward 5 N / earth / book / downward
Correct
Part E The reaction to the force in Part B is a force of magnitude _____, exerted on the _____ by the _____. Its direction is _____. ANSWER: 5 N / table / book / upward 5 N / table / earth / upward 5 N / book / table / upward 5 N / table / book / downward 5 N / earth / book / downward
Correct
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Part F Which of Newton's laws could we have used to predict that the forces in Parts A and B are equal and opposite? Check all that apply. ANSWER: Newton's 1st law Newton's 2nd law Newton's 3rd law
Correct Since the book is at rest, either Newton's 1st or 2nd law can tell us that the net force on it must be zero. This means that the force exerted on it by the earth must be equal and opposite to the force exerted on it by the table.
Part G Which of Newton's laws could we have used to predict that the forces in Parts B and E are equal and opposite? Check all that apply. ANSWER: Newton's 1st law Newton's 2nd law Newton's 3rd law
Correct
Video Tutor: Suspended Balls: Which String Breaks? 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.
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Part A A heavy crate is attached to the wall by a light rope, as shown in the figure. Another rope hangs off the opposite edge of the box. If you slowly increase the force on the free rope by pulling on it in a horizontal direction, which rope will break? Ignore friction and the mass of the ropes.
Hint 1. How to approach the problem Because you slowly increase the force you exert on the rope, the block’s inertia does not affect the outcome. Why is that the case? As you pull harder, the ropes stretch a bit, so the block slides slightly toward you. But the changes are so gradual that the accelerations of the block and ropes are practically zero at any instant. Shown here are freebody diagrams for the ropes and block:
What does Newton’s third law say about the tension forces exerted by the block on the two ropes?
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ANSWER: The rope that you are pulling on will break. The rope attached to the wall will break. Both ropes are equally likely to break.
Correct Since the attached rope doesn’t have to support any weight (as it did in the vertical case), the tension is the same in both ropes.
Tension in a Hanging Massive Rope Consider a rope with length l, mass per unit length λ , experiencing a gravitational acceleration g and hanging vertically as shown. Let y refer to the height of a point P above the bottom of the rope.
Part A The force exerted on the rope by the ceiling is in the _____ direction.
Hint 1. Consider the weight of the rope To support the rope, the ceiling must exert a force opposite to the weight of the rope. ANSWER: upward downward
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Correct
Part B Find F , the magnitude of the force exerted on the rope by the ceiling. Express F in terms of quantities given in the problem introduction.
Hint 1. Use Newton's 2nd law Take W to be the weight of the rope. Apply Newton's 2nd law to find an expression for W
−F
.
ANSWER: W −F
= 0
Hint 2. Find the weight of the rope What is the weight of the rope, W ? Express your answer in terms of λ , l, and g .
Hint 1. Find the mass of the rope What is the mass M of the rope? ANSWER: M
=
λl
ANSWER: W
=
λlg
ANSWER: F
=
lλg
Correct
Part C What is the tension TP at point P in the rope? Express TP in terms of quantitites given in the problem introduction. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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Hint 1. How to approach this problem The tension at point P is the magnitude of the force exerted by the part of the rope above P on the part of the rope below P. Remember that point P is located a distance y above the bottom of the rope.
Hint 2. Find the weight of the rope below point P What is the weight W P of the part of the rope below point P? Express your answer in terms of quantities given in the problem introduction. ANSWER: WP
=
λyg
ANSWER: TP
=
yλg
Correct
Pulling Two Blocks In the situation shown in the figure, a person is pulling with a constant, nonzero force F ⃗ on string 1, which is attached to block A. Block A is also attached to block B via string 2, as shown. For this problem, assume that neither string stretches and that friction is negligible. Both blocks have finite (nonzero) mass.
Part A https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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Which one of the following statements correctly descibes the relationship between the accelerations of blocks A and B?
Hint 1. Relative movement of the blocks The two masses are connected (by string 2), which means, if they are being pulled, they must move together. ANSWER: Block A has a larger acceleration than block B. Block B has a larger acceleration than block A. Both blocks have the same acceleration. More information is needed to determine the relationship between the accelerations.
Correct Since the two blocks are connected, they won't move independently when string 1 is pulled. As block A is accelerated, its motion will impart the same acceleration to block B.
Part B How does the magnitude of the tension in string 1, T1 , compare with the tension in string 2, T2 ?
Hint 1. How to approach the problem Suppose that block A has a mass m. Draw a freebody diagram for block A, then write down Newton's 2nd law for block A's horizontal motion. What is the tension T1 ? Express the tension in terms of T2 , m, and the block's acceleration a .
Hint 1. Freebody diagram for block A Taking the positive direction to be to the right, what is the net horizontal force acting on block A? (Note that, in the figure, forces are not drawn to scale.) Express your answer in terms of T1 , T2 , n A , and w A .
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ANSWER: F = ma
=
T1 − T2
ANSWER: T1
=
ma + T2
ANSWER: T1 > T2
T1 = T2
T1 < T2
More information is needed to determine the relationship between T1 and T2 .
Correct The force transmitted through string 1 (proportional to T1 ) must be enough to accelerate both blocks, but the force transmitted through string 2 only needs to accelerate block B. Consider the case where block A is very heavy and block B is very light: In this case, string 2 would only need to supply a tiny amount of tension to keep the blocks connected as block A is pulled around.
Enhanced EOC: Exercise 4.10 A dockworker applies a constant horizontal force of 72.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 10.0 m in a time of 5.20 s . You may want to review (
pages 112 117) .
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For general problemsolving tips and strategies for this topic, you may want to view a Video Tutor Solution of Determining force from acceleration.
Part A What is the mass of the block of ice?
Hint 1. How to approach the problem Start by drawing a sketch of the block of ice and its motion across the smooth horizontal floor. Create a standard coordinate system, and draw all the forces on the block. Do you need to worry about the vertical forces (gravity and the surface’s normal force) on the block? Why? You know the force applied horizontally on the block. To determine its mass, what other quantity must you find? How can you use the information provided about the block’s motion to help you determine that other quantity? ANSWER: m
= 97.3 kg
Correct
Part B If the worker stops pushing after 5.20 s , how far does the block move in the next 5.50 s ?
Hint 1. How to approach the problem To find out how far something moves in a time period, you need to know its initial velocity and acceleration. How can you determine these quantities for the block in the second time interval, when the worker is no longer pushing? During that second interval of motion, there is no force on the block of ice from the dockworker. What can be said about the acceleration on the block in this time? Use this information to find the distance travelled during the second time interval. ANSWER: x
= 21.2 m
Correct
Applying Newton's 2nd Law https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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Learning Goal: To learn a systematic approach to solving Newton's 2nd law problems using a simple example. Once you have decided to solve a problem using Newton's 2nd law, there are steps that will lead you to a solution. One such prescription is the following: Visualize the problem and identify special cases. Isolate each body and draw the forces acting on it. Choose a coordinate system for each body. Apply Newton's 2nd law to each body. Write equations for the constraints and other given information. Solve the resulting equations symbolically. Check that your answer has the correct dimensions and satisfies special cases. If numbers are given in the problem, plug them in and check that the answer makes sense. Think about generalizations or simplfications of the problem. As an example, we will apply this procedure to find the acceleration of a block of mass m2 that is pulled up a frictionless plane inclined at angle θ with respect to the horizontal by a perfect string that passes over a perfect pulley to a block of mass m1 that is hanging vertically.
Visualize the problem and identify special cases First examine the problem by drawing a picture and visualizing the motion. Apply Newton's 2nd law, ∑ F ⃗ = ma⃗ , to each body in your mind. Don't worry about which quantities are given. Think about the forces on each body: How are these consistent with the direction of the acceleration for that body? Can you think of any special cases that you can solve quickly now and use to test your understanding later? One special case in this problem is if m2 gravity: a⃗ 1
^ = −g j
= 0
, in which case block 1 would simply fall freely under the acceleration of
.
Part A Consider another special case in which the inclined plane is vertical (θ would the acceleration of the two blocks be equal to zero?
= π/2
). In this case, for what value of m1
Express your answer in terms of some or all of the variables m2 and g . ANSWER:
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m1
=
m 2 sin(
π 2
)
Correct
Isolate each body and draw the forces acting on it A force diagram should include only real forces that act on the body and satisfy Newton's 3rd law. One way to check if the forces are real is to detrmine whether they are part of a Newton's 3rd law pair, that is, whether they result from a physical interaction that also causes an opposite force on some other body, which may not be part of the problem. Do not decompose the forces into components, and do not include resultant forces that are combinations of other real forces like centripetal force or fictitious forces like the "centrifugal" force. Assign each force a symbol, but don't start to solve the problem at this point.
Part B Which of the four drawings is a correct force diagram for this problem?
ANSWER: a b c d
Correct
Choose a coordinate system for each body Newton's 2nd law, ∑ F ⃗ =
, is a vector equation. To add or subtract vectors it is often easiest to decompose each
ma⃗
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vector into components. Whereas a particular set of vector components is only valid in a particular coordinate system, the vector equality holds in any coordinate system, giving you freedom to pick a coordinate system that most simplifies the equations that result from the component equations. It's generally best to pick a coordinate system where the acceleration of the system lies directly on one of the coordinate axes. If there is no acceleration, then pick a coordinate system with as many unknowns as possible along the coordinate axes. Vectors that lie along the axes appear in only one of the equations for each component, rather than in two equations with trigonometric prefactors. Note that it is sometimes advantageous to use different coordinate systems for each body in the problem. In this problem, you should use Cartesian coordinates and your axes should be stationary with respect to the inclined plane.
Part C Given the criteria just described, what orientation of the coordinate axes would be best to use in this problem? In the answer options, "tilted" means with the x axis oriented parallel to the plane (i.e., at angle θ to the horizontal), and "level" means with the x axis horizontal. ANSWER: tilted for both block 1 and block 2 tilted for block 1 and level for block 2 level for block 1 and tilted for block 2 level for both block 1 and block 2
Correct
Apply Newton's 2nd law to each body Part D What is ∑ F2x , the sum of the x components of the forces acting on block 2? Take forces acting up the incline to be positive. Express your answer in terms of some or all of the variables tension T , m2 , the magnitude of the acceleration of gravity g , and θ .
Hint 1. Decompose the force of gravity on block 2 In this problem, the hardest force vector to express in terms of its coordinates is the force of gravity on block 2. The magnitude of the weight is m2 g. Find the force of gravity in terms of its components, using a tilted coordinate system whose x axis is parallel to and pointing up the inclined plane. ⃗ Express the force of gravity on block 2, F g2 , in terms of some or all of the variables m2 , g , and θ . ^ Express your answer as a vector in terms of the unit vectors ^ i and j .
ANSWER:
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⃗ F g2
=
^ ^ −m 2 g(sin(θ) i + cos(θ) j )
ANSWER: m 2 a 2x =
∑ F 2x
=
T − m 2 gsin(θ)
Correct
Part E Now determine m1 a 1y = ∑ F1y , the sum of the y components of the forces acting on block 1. Take forces acting upward as positive. Express your answer in terms of some or all of the variables T , m1 , and g . ANSWER: m 1 a 1y =
∑ F 1y
=
T − m1 g
Correct
Part F Write equations for the constraints and other given information In this problem, the fact that the length of the string does not change imposes a constraint on relative accelerations of the two blocks. Find a relationship between the x component of the acceleration of block 2, a 2x , and the acceleration of block 1. Pay careful attention to signs. Express a 2x in terms of a 1x and/or a 1y , the components of the acceleration vector of block 1.
Hint 1. Visualize the motion If block 2 has an acceleration a 2x up the incline, must the acceleration of block 1 be upward or downward to keep the string taut? ANSWER: a 2x
=
−a 1y
Correct
Part G https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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Solve and check In the previous parts, you obtained the following equations using Newton's 2nd law and the constraint on the motion of the two blocks: m 2 a 2x = T − m 2 g sin(θ), m 1 a 1y = T − m 1 g,
(1)
(2)
and a 2x = −a 1y .
(3)
Solve these equations to find a 1y . Before you enter your answer, make sure it satisfies the special cases you already identified: if m2 = 0 and 0 if m 1 = m 2 and θ
a 1y = −g a 1y =
= π/2
.
Also make sure that your answer has dimensions of acceleration. Express a 1y in terms of some or all of the variables m1 , m2 , θ , and g .
Hint 1. How to solve the equations Substitute for T from equation (1) into equation (2) and then use a 2x from equation (3) in the new equation (2). This will yield a linear equation in a 1y that is easy to solve. ANSWER:
a 1y
=
(m 2 sin(θ)−m 1 )g m 1 +m 2
Correct Can you see how a simple generalization of the problem could be solved with a little extra work or how you could solve a nontrivial problem that is a subset of this one? For example, imagine that there is friction in this problem between the plane and block 2. This would lead to an additional force on block 2: Ff 2 = μN , where the normal force N is given by N = m2 g cos(θ). This additional force would lead to a new term in the expression for the acceleration of block 1: a1y =
Now, by choosing whether or not μ frictionless or not!
= 0
m 2 sin(θ)−μm 2 cos(θ)−m 1 m 1 +m 2
g.
, you have a result that can be applied whether the plane is
Enhanced EOC: Exercise 4.5 Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is 56.0 ∘ . Dog A exerts a force of 290 N , and dog B exerts a force of 306 N . You may want to review (
pages 105 108) .
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For help with math skills, you may want to review: Vector Magnitudes Determining the Angle of a Vector Vector Addition For general problemsolving tips and strategies for this topic, you may want to view a Video Tutor Solution of Superposition of forces.
Part A Find the magnitude of the resultant force.
Hint 1. How to approach the problem Imagine you were sitting in a tree directly above the post, so that you were looking down on the dogs and the ropes tying them to the post. Start by drawing a sketch of the two dogs pulling on their ropes as seen by you from above, and choose a coordinate system so that the post is at the origin and the rope tied to dog A is along the xaxis. In what direction would dog B be seen pulling its rope? Now, find the resultant force. How is the magnitude of a vector related to its components? ANSWER: F
= 526 N
Correct
Part B Find the angle the resultant force makes with the rope of dog A.
Hint 1. How to approach the problem How can you find the angle the resultant force vector makes with the xaxis (the rope of dog A) from the components of that vector? ANSWER: θ
= 28.8 ∘
Correct
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Newton's 3rd Law Discussed Learning Goal: To understand Newton's 3rd law, which states that a physical interaction always generates a pair of forces on the two interacting bodies. In Principia, Newton wrote: To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts. (translation by Cajori) The phrase after the colon (often omitted from textbooks) makes it clear that this is a statement about the nature of force. The central idea is that physical interactions (e.g., due to gravity, bodies touching, or electric forces) cause forces to arise between pairs of bodies. Each pairwise interaction produces a pair of opposite forces, one acting on each body. In summary, each physical interaction between two bodies generates a pair of forces. Whatever the physical cause of the interaction, the force on body A from body B is equal in magnitude and opposite in direction to the force on body B from body A. Incidentally, Newton states that the word "action" denotes both (a) the force due to an interaction and (b) the changes in momentum that it imparts to the two interacting bodies. If you haven't learned about momentum, don't worry; for now this is just a statement about the origin of forces. Mark each of the following statements as true or false. If a statement refers to "two bodies" interacting via some force, you are not to assume that these two bodies have the same mass.
Part A Every force has one and only one 3rd law pair force. ANSWER: true false
Correct
Part B The two forces in each pair act in opposite directions. ANSWER: true false
Correct
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Part C The two forces in each pair can either both act on the same body or they can act on different bodies. ANSWER: true false
Correct
Part D The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be due to friction or electric charge). ANSWER: true false
Correct
Part E The two forces of a 3rd law pair always act on different bodies. ANSWER: true false
Correct
Part F Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.)
Hint 1. F ⃗ = ma⃗ Remember F ⃗ = magnitude?
: If the forces are equal in magnitude, must the accelerations also be of equal
ma⃗
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ANSWER: true false
Correct Newton's 3rd law can be summarixed as follows: A physical interaction (e.g., gravity) operates between two interacting bodies and generates a pair of opposite forces, one on each body. It offers you a way to test for real forces (i.e., those that belong on the force side of ΣF ⃗ = ma⃗ )there should be a 3rd law pair force operating on some other body for each real force that acts on the body whose acceleration is under consideration.
Part G According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is ANSWER: greater in magnitude and antiparallel to the force on the earth due to the moon. greater in magnitude and parallel to the force on the earth due to the moon. equal in magnitude but antiparallel to the force on the earth due to the moon. equal in magnitude and parallel to the force on the earth due to the moon. smaller in magnitude and antiparallel to the force on the earth due to the moon. smaller in magnitude and parallel to the force on the earth due to the moon.
Correct
Enhanced EOC: Exercise 4.24 The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N. You may want to review (
pages 112 117) .
For general problemsolving tips and strategies for this topic, you may want to view a Video Tutor Solution of Determining acceleration from force.
Part A What is the magnitude of the acceleration?
Hint 1. How to approach the problem Start by drawing a sketch with an appropriate coordinate system and including all of the forces acting on the elevator passenger. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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Why in this problem is the normal force not equal to the passenger’s weight? What must be happening? Find the resultant net force acting on the elevator passenger, and using the appropriate equations, calculate the passenger’s acceleration. What additional quantity do you need to find to calculate the acceleration? ANSWER: a
= 0.452 m/s 2
Correct
Part B What is the direction of the acceleration?
Hint 1. How to approach the problem Suppose you were in this elevator. What would it feel like if the normal force (from the elevator floor) acting on you in the upward direction were less than the force of gravity acting on you in the downward direction? Suppose the normal force was larger than gravity – which way would you move? In which direction is the resultant force acting on the elevator passenger? ANSWER: upward downward
Correct
Video Tutor: Tension in String between Hanging Weights 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.
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Part A Consider the video tutorial you just watched. Suppose that we duplicate this experimental setup in an elevator. What will the spring scale read if the elevator is moving upward at constant speed?
Hint 1. How to approach the problem What does the phrase "at constant speed" imply about the acceleration of the system? ANSWER: 0 N 18 N More than 18 N Less than 18 N but greater than 0 N
Correct Since the elevator is not accelerating, the reading on the scale is the same as in the video.
± Motion of a Block with Three Forces The diagram below shows a block of mass m = 2.00 kg on a frictionless horizontal surface, as seen from above. Three forces of magnitudes F1 = 4.00 N , F2 = 6.00 N , and F3 = 8.00 N are applied to the block, initially at rest on the surface, at angles shown on the diagram. In this problem, you will determine the resultant (total) force vector from the combination of the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).
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Part A Calculate the magnitude of the total resultant force F r⃗
⃗ ⃗ ⃗ = F1 + F2 + F3
acting on the mass.
Express your answer in Newtons to three significant figures.
Hint 1. Definition of resultant force When several forces are applied to an object, the vector sum is often called the resultant or the resultant force.
Hint 2. How to find the resultant When working with vectors, the general rule is to think geometrically but to calculate using components. Thus to add vectors one estimates the sum by imagining the tail of the second vector to be placed at the point of the first, the tail of the third to be placed at the second, etc. But to calculate the vector sum each vector is represented by components in a convenient coordinate system and these components are added to find the components of the sum.
Hint 3. Find the components of F 1⃗ What are the x component and y component of F 1⃗ ? Express your answer as an ordered pair of numbers, separated by a comma, to three significant figures.
Hint 1. x component of F 1⃗ The x component of F 1⃗ is F1 cos(θ), where θ is the angle between the positive x axis and the vector's direction. ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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Ch 04 HW F 1x
, F1y = 3.63,1.69 N
Hint 4. Find the components of F 2⃗ Find the x and y components of the vector F 2⃗ . Express your answer as an ordered pair to three significant figures. ANSWER: F 2x
, F2y = 4.91,3.44
Hint 5. Find the components of F r⃗ Now find the x and y components of the resultant (sum) vector, F r⃗ . (Don't forget to include F 3⃗ .) Express your answer as an ordered pair to three significant figures. ANSWER: F rx
, Fry = 0.540,1.75
Hint 6. Magnitude of F r⃗ The magnitude of F r⃗ in terms of its x and y components Frx and Fry is given by −−−−−−−−− 2 2 ⃗ |F r | = √ F rx + F ry
.
ANSWER: ⃗ |F r |
= 1.83 N
Correct
Part B What angle does F r⃗ make with the positive x axis? Express your answer in degrees to two significant figures.
Hint 1. Find the angle symbolically The angle that F r⃗ makes with the x axis can be determined if you know its x and y components, which you should have from your calculation for Part A. What is the angle that F r⃗ makes with the positive x axis? https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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Ch 04 HW
Answer symbolically in terms of Frx and Fry .
Hint 1. How to approach this problem When developing a general formula for the angle, you can choose the vector to lie in any quadrant. It is easiest to have the vector lie in the first quadrant.
ANSWER:
atan(
Fry Frx
)
ANSWER: 73 degrees
Correct
Part C What is the magnitude of the mass's acceleration vector, a⃗ ? Express your answer to two significant figures.
Hint 1. Newton's 2nd law Recall that F ⃗ =
, so you should be able to find |a⃗ | fairly easily here.
ma⃗
ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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Ch 04 HW |a⃗ |
= 0.92 m/s 2
Correct
Part D What is the direction of a⃗ ? In other words, what angle does this vector make with respect to the positive x axis? Express your answer in degrees to two significant figures.
Hint 1. Relation between the direction of a⃗ and F ⃗ Is there any reason why the direction of a⃗ should be different from F ⃗ ? ANSWER: 73 degrees
Correct
Part E How far (in meters) will the mass move in 5.0 s? Express the distance d in meters to two significant figures.
Hint 1. Displacement with constant acceleration Remember that we have constant acceleration here, so you can use the equation: d(t) = d0 + v 0 t +
1 2
at
2
,
where d(t) is the displacement at time t. ANSWER: d
= 11 m
Correct
Part F What is the magnitude of the velocity vector of the block at t https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
= 5.0 s
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Ch 04 HW
Express your answer in meters per second to two significant figures.
Hint 1. Velocity with constant acceleration Remember, we have constant acceleration, and because the object starts from rest, the velocity vector will be parallel to the acceleration vector. Therefore, v f = v i + at . ANSWER: ⃗ |v (5)|
= 4.6 m/s
Correct
Part G In what direction is the mass moving at time t respect to the positive x axis?
= 5.0 s
? That is, what angle does the velocity vector make with
Express your answer in degrees to two significant figures.
⃗ Hint 1. Relationship between the direction of v and a⃗
The mass starts at rest and is accelerated in one direction. Therefore, it must have velocity in the direction of the acceleration, and that direction only. ANSWER: 73 degrees
Correct
Enhanced EOC: Exercise 4.28 A person pulls horizontally on block B in the figure , causing both blocks to move together as a unit. You may want to review (
pages 124 125) .
For general problemsolving tips and strategies for this topic, you may want to view a Video Tutor Solution of A Newton’s third law paradox.
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Ch 04 HW
Part A While this system is moving, make a carefully labeled freebody diagram of block A if the table is frictionless. Draw all relevant force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded, but the relative length of one to the other will be graded.
Hint 1. How to approach the problem First, picture what is happening. What force(s) does block B exert on block A? Are there any other forces acting on block A? If the table is frictionless, what happens to block B as it is pulled? Since block A and block B move as a unit, what can be said about their accelerations? So is there a net force on block A? Think about what causes this resultant force on block A. In what direction must it act? ANSWER:
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Ch 04 HW
Correct
Part B While this system is moving, make a carefully labeled freebody diagram of block A if there is friction between block B and the table and the pull is equal to the friction force on block B due to the table. Draw all relevant force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded, but the relative length of one to the other will be graded.
Hint 1. How to approach the problem How has friction with the table changed this problem from part A? What is the acceleration of block B now? Since block A and block B move as a unit, what can be said about their accelerations? So is there a net force on block A? ANSWER:
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Ch 04 HW
Correct
Problem 4.58 The position of a 2.75 × 10 5 r ⃗ = (0.020 m/s
3
)t
3
N
training helicopter under test is given by
2 2 ^ ^ ^ i + (2.2 m/s)t j − (0.060 m/s )t k
.
Part A Find the net force on the helicopter at t
= 5.0 s
.
^ ^ Express your answer in terms of ^ i , j , k . Use the 'unit vector' button to denote unit vectors in your answer. Express your coefficient using two significant figures.
ANSWER:
F
⃗
=
^ ^ 17000 i − 3400k
N
Correct
Exercise 4.31 A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force ∘
N https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4266217
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Ch 04 HW
= 35.0 N that is directed at an angle of 43.0 below the horizontal and the chair slides along the floor. ∘
Part A Use Newton's laws to calculate the normal force that the floor exerts on the chair. ANSWER: n
= 141 N
Correct
Problem 4.36 You have just landed on Planet X. You take out a ball of mass 100 g , release it from rest from a height of 13.0 m and measure that it takes a time of 2.10 s to reach the ground. You can ignore any force on the ball from the atmosphere of the planet.
Part A How much does the ball weigh on the surface of Planet X? ANSWER: W
= 0.590 N
Correct
Problem 4.40
Part A An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 910 kg traveling initially at a speed of 49.0 km/h in a distance equal to the diameter of a dime, which is 1.8 cm ? Express your answer using two significant figures. ANSWER: F = 4.7×106 {\rm N}
Correct
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Ch 04 HW
Problem 4.60 An object with mass m initially at rest is acted on by a force \vec F=k_1 \hat i + k_2 t^3 \hat j where k_1 and k_2 are constants.
Part A Calculate the velocity \vec v\left( t \right) of the object as a function of time. ANSWER: \large{\vec v(t)=m\left({k_1t\hat i+\frac{k_2}{3}t^3\hat j}\right)} \large{\vec v(t)=\frac{1}{m}\left({k_1\hat i+\frac{k_2}{3}t^3\hat j}\right)} \large{\vec v(t)=\frac{1}{m}\left({k_1t\hat i+\frac{k_2}{4}t^4\hat j}\right)} \large{\vec v(t)=m\left({k_1t\hat i+\frac{k_2}{4}t^4\hat j}\right)}
Correct Score Summary: Your score on this assignment is 99.4%. You received 26.83 out of a possible total of 27 points.
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