Ch 07 HW Due: 11:59pm on Sunday, February 14, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy
Work on a Sliding Box A box of mass m is sliding along a horizontal surface.
Part A The box leaves position x
=0
with speed v0 . The box is slowed by a constant frictional force until it comes to rest at position x
= x 1 .
Find F f frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in in your f , the magnitude of the average frictional answer.) Express the frictional force in terms of m , v0 , and x 1 .
Hint 1. How to approach the problem Use the work-energy theorem. As applied to this part, the theorem states that the work done by friction is equal to the change in kinetic ene rgy of the box:
W f = ∆ K = K f
− K i .
Find K i , K f , and W f (which will depend on F f ), then solve for F f .
Hint 2. Find the initial kinetic energy What is K i , the kinetic energy of the box at position x
= 0?
ANSWER:
K i = 1 mv 0 2 2
Hint 3. Find the final kinetic energy What is K f f , the kinetic energy of the box when it reaches position x
= x 1 ?
ANSWER:
K f = 0
Hint 4. Find the work done by friction Find W f work done by friction on the box. Note that the work done by friction is always negative (i.e., friction always always dissipates energy). f , the work Express your answer in terms of F f and other given variables. ANSWER:
W f =
−F f x 1
ANSWER:
F f =
mv 0 2 2x 1
Correct
Part B
After the box comes to rest at position x 1 , a person starts pushing the box, giving it a speed v1 . When the box reaches position x 2 (where x 2 > x 1 ), how much work W p has the person done on the box? Assume that the box reaches x 2 after the person has accelerated it from rest to speed v1 . Express the work in terms of m , v0 , x 1 , x 2 , and v1 .
Hint 1. How to approach the problem Again, use the work-energy theorem. In this part of the problem, both the person and friction are doing work on the box:
W f + W p = ∆ K = K f
− K i .
Hint 2. Find the work done by friction What is W f , the total work done by friction on the box as the person pushes it from position x 1 to position x 2 ? Answer in terms of given variables. (Y (Your our answer should not include F f f .)
Hint 1. Finding the force of friction The normal force on the box is unchanged from partA. Therefore, the force of friction is the same in this part as in part A.
ANSWER:
W f =
(mv 0 2 )(x2 −x 1 )
−
2x 1
Hint 3. Find the change in kinetic energy What is ∆K , the change in kinetic energy of the box from the moment it is at position x 1 to the moment it is at positionx 2 ? ANSWER: ∆K =
1 2
mv 1 2
ANSWER:
p =
(
2 1 mv 0 2 x1
)(x
2 − x1
) + 12 mv1 2
Correct
Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states
K f = K i + W all , where W all is the work done by all all forces forces that act on the object, and K i and K f are the initial and final kinetic energies, respectively respectively.. Part A
After the box comes to rest at position x 1 , a person starts pushing the box, giving it a speed v1 . When the box reaches position x 2 (where x 2 > x 1 ), how much work W p has the person done on the box? Assume that the box reaches x 2 after the person has accelerated it from rest to speed v1 . Express the work in terms of m , v0 , x 1 , x 2 , and v1 .
Hint 1. How to approach the problem Again, use the work-energy theorem. In this part of the problem, both the person and friction are doing work on the box:
W f + W p = ∆ K = K f
− K i .
Hint 2. Find the work done by friction What is W f , the total work done by friction on the box as the person pushes it from position x 1 to position x 2 ? Answer in terms of given variables. (Y (Your our answer should not include F f f .)
Hint 1. Finding the force of friction The normal force on the box is unchanged from partA. Therefore, the force of friction is the same in this part as in part A.
ANSWER:
W f =
(mv 0 2 )(x2 −x 1 )
−
2x 1
Hint 3. Find the change in kinetic energy What is ∆K , the change in kinetic energy of the box from the moment it is at position x 1 to the moment it is at positionx 2 ? ANSWER: ∆K =
1 2
mv 1 2
ANSWER:
p =
(
2 1 mv 0 2 x1
)(x
2 − x1
) + 12 mv1 2
Correct
Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states
K f = K i + W all , where W all is the work done by all all forces forces that act on the object, and K i and K f are the initial and final kinetic energies, respectively respectively.. Part A
The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. Choose the best answer to fill in the blanks above: ANSWER: distance / potential distance / kinetic vertical displacement / potential none of the above
Correct It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball.
Part B To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change. Choose the best answer to fill in the blank above: ANSWER: acceleration work distance potential energy
Correct
Part C To illustrate the work-energy concept, consider the case of a stone falling from x i to x f under the influence of gravity gravity.. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Choose the best answer to fill in the blanks above: ANSWER: force / kinetic potential energy / potential force / potential potential energy / kinetic
Correct
Potential Energy You Energy You should read about potential energy in your text before answering the following questions. Potential energy is is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative negative of of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. nonconservative forces
In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy:
K f + U f = E f = W nc + E i = W nc + K i + U i , where U f and U i are the final and initial potential energies, and W nc is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy.
Part D Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ e nergy. Choose the best answer to fill in the blanks above: ANSWER: work / potential force / kinetic change / potential
Correct
Part E This process happens in such a way that total mechanical energy , equal to the ______ of the kinetic and potential energies, is _______. Choose the best answer to fill in the blanks above: ANSWER: sum / conserved sum / zero sum / not conserved difference / conserved
Correct
Potential Energy Calculations Learning Goal: To understand the relationship between the force and the potential energy changes associated with that force and to be able to calculate the changes in potential energy as definite integrals. Imagine that a conservative force field is defined in a certain region of space. Does this sound too abstract? Well, think of a gravitational field (the one that makes apples fall down and keeps the planets orbiting) or an electrostatic field existing around any electrically charged object. If a particle is moving in such a field, its change in potential energy does not depend on the particle's path and is determined only by the particle's initial and final positions. Recall that, in general, the component of the net force acting on a particle equals the negative derivative of the potential energy fu nction along the corresponding axis:
F x = −
dU (x) dx
.
Therefore, the change in potential energy can be found as the integral
2 ,⃗ ∆U = − ∫ 1 F ⃗ ⋅ d s
⃗ is the net force acting on the particle at a given point of its path, where ∆U is the change in potential energy for a particle moving from point 1 to point 2, F and ds ⃗ is a small displacement of the particle along its path from 1 to 2.
Evaluating such an integral in a general case can be a tedious and lengthy task. However, two circumstances make it easier: 1. Because the result is path-independent , it is always possible to consider the most straightforward way to reach point 2 from point 1.
2. The most common real-world fields are rather simply defined.
In this problem, you will practice calculating the change in poten tial energy for a particle moving in three common force fields.
^ ^ ^ is the unit vector in the radial Note that, in the equations for the forces, i is the unit vector in the x direction, j is the unit vector in the y direction, and r direction in case of a spherically symmetrical force field. Part A Consider a uniform gravitational field (a fair approximation near the surface of a planet). Find y f
⃗ d s U (y f ) − U (y 0 ) = − ∫ y 0 F g ⋅ ,⃗ where ⃗ ^ ^. F ⃗ = dy j g = − mg j and d s
Express your answer in terms of m , g , y 0 , and y f . ⃗ and d s ⃗ Hint 1. Relative directions of F g
Note that F g⃗ and d s ⃗ are parallel, and their dot product is simply the product of their magnitudes. That is, ⃗ ⃗ = − mgdy . F g ⋅ d s
ANSWER:
(
U (y f ) − U (y 0 ) = mg y f − y 0
)
Correct
Part B Consider the force exerted by a spring that obeys Hooke's law. Find
x
f U (xf ) − U (x0 ) = − ∫ x0 F s ⋅ d s ,⃗
where ⃗ ^ F ⃗ = dx ^ i, s = − kx i , d s
and the spring constant k is positive. Express your answer in terms of k , x 0 , and x f . ⃗ and d s ⃗ Hint 1. Relative directions of F s
Note that F s⃗ and d s ⃗ are parallel, and their dot product is simply the product of their magnitudes. That is, ⃗ F ⃗ = − kxdx . s ⋅ d s
ANSWER:
U (x f ) − U (x 0 ) = .5k(x f 2
− x0
2
)
Correct
Part C Finally, consider the gravitational force generated by a spherically symmetrical massive object. The magnitude and direction of such a force are given
by Newton's law of gravity:
F G = − ^; G , m 1 , and m 2 are constants; and r where d s ⃗ = dr r
> 0 . Find
Gm 1 m 2 r2
r^ ,
rf
U (r f ) − U (r0 ) = − ∫ r0 F G ⋅ d s .⃗ Express your answer in terms of G , m 1 , m 2 , r 0 , and r f . ⃗ and d s ⃗ Hint 1. Relative directions of F G
⃗ and d s ⃗ are parallel, and their dot product is simply the product of their magnitudes. That is, Note that F G
⃗ ⃗ = − F G ⋅ d s Hint 2. Integrating
Gm 1 m 2 r2
dr.
1/r2
Recall that
= − 1r . ∫ dr r2
Carefully account for all the negative signs in your calculations.
ANSWER:
U (r f ) − U (r 0 ) = −Gm1 m2
(
1
1
rf − r0
)
Correct As you can see, the change in potential energy of the particle can be found by integrating the force along the particle's path. However, this method, as we mentioned before, does have an important restriction: It can only be applied to a conservative force field. For conservative forces such as gravity or tension the work done on the particle does not depend on the particle's path, and the potential energy is the function of the particle's position. In case of a nonconservative force--such as a frictional or magnetic force--the potential energy can no longer be defined as a function of th e particle's position, and the method that you used in this problem would not be applicable.
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 g 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 ene rgy. 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 g . Or you can use reasoning. A greater downward acceleration g 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: 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 take more 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.
Exercise 7.32 While a roofer is working on a roof that slants at 39.0 ∘ above the horizontal, he accidentally nudges his 87.0 N toolbox, causing it to start sliding downward, starting from rest.
Part A If it starts 4.90 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N ? ANSWER:
v = 6.01 m/s
Correct
Exercise 7.35 A force parallel to the x -axis acts on a particle moving along the x -axis. This force produces a potential energy U (x) given by U (x) 4 . α = 1.44 J/m
Part A What is the force when the particle is at position x = -0.990 m ? ANSWER:
F x = 5.59
Correct
N
= α x 4
where
Energy Required to Lift a Heavy Box As you are trying to move a heavy box of mass m , you realize that it is too heavy for you to lift by yourself. There is no one around to help, so you attach an ideal pulley to the box and a massless rope to the ceiling, which you wrap around the pulley. You pull up on the rope to lift the box. Use g for the magnitude of the acceleration due to gravity and neglect friction forces.
Part A Once you have pulled hard enough to start the box moving upward, what is the magnitude F of the upward force you must apply to the rope to start raising the box with constant velocity? Express the magnitude of the force in terms of m , the mass of the box.
Hint 1. What force must be applied to the box to keep it moving at a constant speed? Once you have pulled hard enough to start the box moving upward, what is the magnitude of the force that the pulley must exert on the box so that it moves at a constant speed? Express your answer in terms of the mass of the box. ANSWER:
F p = mg
Hint 2. What force does the pulley exert on the box? If you take the tension in the rope to be T , what is p , the magnitude of the net upward force that the pulley exerts on the box? Express your answer in terms of T . ANSWER: p =
2T
Hint 3. Find the tension in the rope Find the tension in the rope in terms of F , the force with which you are pulling upward. ANSWER:
T = F
Hint 4. Putting it all together On your own or using the previous hints, you should have found equations for he following: 1. the force needed to lift the box at constant velocity, in terms of its mass, 2. the relationship between the force on the box due to the pulley and the tension in the rope, and 3. the relationship between the force applied to the rope and the tension in the rope.
Use two of these equations to eliminate the force applied by the pulley and the tension in the rope. You should then be able to express the force applied on the rope in terms of the mass of the box.
ANSWER:
F =
mg
2
Correct
Part B Consider lifting a box of mass m to a height h using two different methods: lifting the box directly or lifting the box using a pulley (as in the previous part). What is W d /W p , the ratio of the work done lifting the box directly to the work done lifting the box with a pulley? Express the ratio numerically.
Hint 1. Definition of work In each case, the amount of work W you do is equal to the force F you apply times the distance d over which you apply the force:
W = F d . Hint 2. Ratio of the forces What is the ratio of the force needed to lift the box directly to the force needed to lift the box using the pulley? Express your answer numerically. ANSWER: F d F p =
2
Answer Requested
Hint 3. Ratio of the distances What is the ratio of the distance over which force is applied when lifting the box directly to the distance over which force is applied when lifting the box with the pulley? Express the ratio of distances numerically.
Hint 1. Find the distance when using the pulley Find
p , the distance over which you must apply force when lifting the box using the pulley.
Express your answer in terms of h , the total height that the box is lifted.
Hint 1. Pulling the rope a short distance Using the pully, imagine that you pull the end of the rope a short distance dx upward. The box will actually rise a distance dx/2 . (Draw a picture if you have trouble visualizing this.)
ANSWER: p =
2h
Answer Requested
Hint 2. Find the distance when lifting directly When lifting the box directly, the distance over which force is applied, D d , is equal to the vertical distance h that the box is raised.
ANSWER: Dd Dp = 0.500
ANSWER: W d = 1 W p
Correct No matter which method you use to lift the box, its gravitational potential energy will increase by mgh . So, neglecting friction, you will always need to do an amount of work equal to mgh to lift it.
Potential Energy Graphs and Motion Learning Goal: To be able to interpret potential energy diagrams and predict the corresponding motion of a particle. Potential energy diagrams for a particle are useful in predicting the motion of that particle. These diagrams allow one to determine the direction of the force acting on the particle at any point, the points of stable and unstable equilibrium, the particle's kinetic energy, etc. Consider the potential energy diagram shown. The curve represents the value of potential energy U as a function of the particle's coordinate x . The horizontal line above the curve represents the constant value of the total energy of the particle E . The total energy E is the sum of kinetic ( K ) and potential ( U ) energies of the particle. The key idea in interpreting the graph can be expressed in the equation
F x (x) = −
dU (x)
,
dx where F x (x) is the x component of the net force as function of the particle's coordinate x . Note the negative sign: It means that the x component of the net force is negative when the derivative is positive and vice versa. For instance, if the particle is moving to the right, and its potential energy is increasing, the net force would be pulling the particle to the left.
If you are still having trouble visualizing this, consider the following: If a massive particle is increasing its gravitational potential energy (that is, moving upward), the force of gravity is pulling in the opposite direction (that is, downward). If the x component of the net force is zero, the particle is said to be in equilibrium. There are two kinds of equilibrium: Stable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle back toward the equilibrium point (think of a ball rolling between two hills). Unstable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle further away from the equilibrium point (think of a ball on top of a hill).
In answering the following questions, we will assume that there is a single varying force F acting on the particle along the x axis. Therefore, we will use the term force instead of the cumbersome x component of the net force.
Part A The force acting on the particle at point A is __________.
Hint 1. Sign of the derivative If a function increases (as x increases) in a certain region, then the derivative of the function in that region is positive.
Hint 2. Sign of the component If x increases to the right, as in the graph shown, then a (one-dimensional) vector with a positive x component points to the right, and vice versa.
ANSWER: directed to the right directed to the left equal to zero
Correct Consider the graph in the region of point A. If the particle is moving to the right, it would be "climbing the hill," and the force would "pull it down," that is, pull the particle back to the left. Another, more abstract way of thinking about this is to say that the slope of the graph at point A is positive;
therefore, the direction of F ⃗ is negative.
Part B The force acting on the particle at point C is __________.
Hint 1. Sign of the derivative If a function increases (as x increases) in a certain region, then the derivative of the function in that region is positive, and vice versa.
Hint 2. Sign of the component If x increases to the right, as in the graph shown, then a (one-dimensional) vector with a positive x component points to the right, and vice versa.
ANSWER: directed to the right directed to the left equal to zero
Correct
Part C The force acting on the particle at point B is __________.
Hint 1. Derivative of a function at a local maximum At a local maximum, the derivative of a function is equal to zero.
ANSWER: directed to the right directed to the left equal to zero
Correct The slope of the graph is zero; therefore, the derivative dU /dx
|⃗ = 0 = 0 , and |F .
Part D The acceleration of the particle at point B is __________.
Hint 1. Relation between acceleration and force The relation between acceleration and force is given by Newton's 2nd law,
F = ma . ANSWER: directed to the right directed to the left equal to zero
Correct If the net force is zero, so is the acceleration. The particle is said to be in a state of equilibrium.
Part E If the particle is located slightly to the left of point B, its acceleration is __________.
Hint 1. The force on such a particle To the left of B, U (x) is an increasing function and so its derivative is positive. This implies that the x component of the force on a particle at this location is negative, or that the force is directed to the left, just like at A. What can you say now about the acceleration?
ANSWER: directed to the right directed to the left equal to zero
Correct
Part F If the particle is located slightly to the right of point B, its acceleration is __________.
Hint 1. The force on such a particle To the right of B, U (x) is a decreasing function and so its derivative is negative. This implies that the x component of the force on a particle at this location is positive, or that the force is directed to the right, just like at C. What can you now say about the acceleration?
ANSWER: directed to the right directed to the left equal to zero
Correct As you can see, small deviations from equilibrium at point B cause a force that accelerates the particle further away; hence the particle is in unstable equilibrium.
Part G Name all labeled points on the graph corresponding to unstable equilibrium. List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE.
Hint 1. Definition of unstable equilibrium Unstable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle further away from the equilibrium point (think of a ball on top of a hill).
ANSWER: BF
Correct
Part H Name all labeled points on the graph corresponding to stable equilibrium. List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE.
Hint 1. Definition of stable equilibrium Stable equilibrium means that small deviations from the equilibrium point create a net force that accelerates the particle back toward the equilibrium point. (Think of a ball rolling between two hills.)
ANSWER: DH
Correct
Part I Name all labeled points on the graph where the acceleration of the particle is zero. List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE.
Hint 1. Relation between acceleration and force The relation between acceleration and force is given by Newton's 2nd law,
F = ma . ANSWER: BDFH
Correct Your answer, of course, includes the locations of both stable and unstable equilibrium.
Part J Name all labeled points such that when a particle is released from rest there, it would accelerate to the left. List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE.
You did not open hints for this part.
ANSWER: BDFGH
Incorrect; Try Again; 2 attempts remaining
Part K Consider points A, E, and G. Of these three points, which one corresponds to the greatest magnitude of acceleration of the particle?
Hint 1. Acceleration and force The greatest acceleration corresponds to the greatest magnitude of the net force, represented on the graph by the magnitude of the slope.
ANSWER: A E G
Correct
Kinetic energy If the total energy E of the particle is known, one can also use the graph of U (t) to draw conclusions about the kinetic energy of the particle since
K = E − U . As a reminder, on this graph, the total energy E is shown by the horizontal line.
Part L What point on the graph corresponds to the maximum kinetic energy of the moving particle?
Hint 1. K , U , and E Since the total energy does not change, the maximum kinetic energy corresponds to the minimum potential energy.
ANSWER: D
Correct It makes sense that the kinetic energy of the particle is maximum at one of the (force) equilibrium points. For example, think of a pendulum (which has only one force equilibrium point--at the very bottom).
Part M
At what point on the graph does the particle have the lowest speed? ANSWER: B
Correct As you can see, many different conclusions can be made about the particle's motion merely by looking at the graph. It is helpful to understand the character of motion qualitatively before you attempt quantitative problems. This problem should prove useful in improving such an understanding.
Loop the Loop A roller coaster car may be approximated by a block of mass m . The car, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R , as shown. Assume that the initial height h is great enough so that the car never loses contact with the track.
Part A Find an expression for the kinetic energy of the car at the top of the loop. Express the kinetic energy in terms of m , g , h , and R .
Hint 1. Find the potential energy at the top of the loop What is the potential energy of the car when it is at the top of the loop? Define the gravitational potential energy to be zero ath
=0.
Express your answer in terms of R and other given quantities. ANSWER:
U top = mg(2R)
ANSWER:
K = mg(h − 2R)
Correct
Part B Find the minimum initial height h at which the car can be released that still allows the car to stay in contact with the track at the top of the loop. Express the minimum height in terms of R .
Hint 1. How to approach this part
Meaning of "stay in contact" For the car to just stay in contact through the loop, without falling, the normal force that acts on th e car when it's at the top of the loop must be zero (i.e., N = 0 ). Find the velocity at the top such that the remaining force on the car i.e. its weight provides the necessary centripetal acceleration. If the velocity were any greater, you would additionally require some force from the track to provide the necessary centripetal acceleration. If the velocity were any less, the car would fall off the track. Use the above described condition to find the velocity and then the result from the above part to find the required height.
Hint 2. Acceleration at the top of the loop Assuming that the speed of the car at the top of the loop is vtop , and that the car stays on the track, find the acceleration of the car. Take the positive y direction to be upward. Express your answer in terms of vtop and any other quantities given in the problem introduction. ANSWER:
atop =
−v top
2
R
Hint 3. Normal force at the top of the loop Suppose the car stays on the track and has speed vtop at the top of the loop. Use Newton's 2nd law to find an expression for N , the magnitude of the normal force that the loop exerts on the car when the car is at the top of the loop. Express your answer in terms of m , g , R , and vtop .
Hint 1. Find the sum of forces at the top of the loop Find the sum of the forces acting on the car at the top of the loop. Remember that the positivey direction is upward . Express your answer in terms of N , m , and g . ANSWER:
∑ F top =
−N − mg
ANSWER: \texttip{N}{N} = \large{m \left(\frac{{v_{\rm{top}}}^{2}}{R}-g\right)}
Hint 4. Solving for \texttip{h}{h} The requirement to stay in contact results in an expression for v_{top}^2 in terms of \texttip{R}{R} and \texttip{g}{g}. Substitute this into your expression for kinetic energy, found in Part A, to determine a relation between \texttip{h}{h} and \texttip{R}{R}.
ANSWER: \texttip{h_{\rm min}}{h_min} = \large{\frac{5R}{2}}
Correct For h > 2.5 \; R the car will still complete the loop, though it will require some normal reaction even at the very top. For h < R the car will just oscillate. Do you see this? For R < h < 2.5 \; R, the cart will lose contact with the track at some earlier point. That is why roller coasters must have a lot of safety features. If you like, you can check that the angle at which the cart loses contact with the track is given by \large{\theta = \arcsin\left(\frac{2}{3} \left( \frac{h}{R} - 1\right)\right)}. Where \theta is the angle measured counterclockwise from the horizontal positive x -axis, where the origin of the x -axis is at the center of the loop.
Drag on a Skydiver A skydiver of mass \texttip{m}{m} jumps from a hot air balloon and falls a distance \texttip{d}{d} before reaching a terminal velocity of magnitude \texttip{v}{v}. Assume that the magnitude of the acceleration due to gravity is \texttip{g}{g}.
Part A What is the work \texttip{W_{\rm d}}{W_d} done on the skydiver, over the distance \texttip{d}{d}, by the drag force of the air? Express the work in terms of \texttip{d}{d}, \texttip{v}{v}, \texttip{m}{m}, and the magnitude of the acceleration due to gravity \texttip{g}{g}.
Hint 1. How to approach the problem If no nonconservative forces were acting, then the total mechanical energy (kinetic plus potential) of the skydiver upon leaving the plane would be equal to the total mechanical energy of the skydiver after falling a distance \texttip{d}{d}. Now consider the drag force, which is nonconservative. The drag force opposes the motion of the skydiver, which means that it does negative work on the skydiver. Thus, the final mechanical energy of the skydiver will be smaller than the initial mechanical energy by an amount equal to the work done by the drag force.
Hint 2. Find the change in potential energy Find the change in the skydiver's gravitational potential energy, after falling a distance \texttip{d}{d}. Express your answer in terms of given quantities. ANSWER: \texttip{\Delta U_{\rm g}}{DeltaU_g} =
-m g d
Hint 3. Find the change in kinetic energy Find the change in the skydiver's kinetic energy, after falling a distance \texttip{d}{d}. Express your answer in terms of given quantities. ANSWER: \texttip{\Delta K}{DeltaK} = \large{\frac{1}{2} m v^{2}}
ANSWER: \texttip{W_{\rm d}}{W_d} =
\large{{\frac{1}{2}}mv^{2}-mgd}
Correct
Part B Find the power \texttip{P_{\rm d}}{P_d} supplied by the drag force after the skydiver has reached terminal velocity \texttip{v}{v}. Express the power in terms of quantities given in the problem introduction.
Hint 1. How to approach the problem One way to approach this problem would be to apply the definition of power as the time derivative of the work done. A simpler method that works for this problem is to use the formula for the power delivered by a force \texttip{\vec{F}}{F_vec} acting on an object moving with velocity \texttip{\vec{v}}{v_vec}: P=\vec{F}\cdot\vec{v}.
Hint 2. Magnitude of the drag force Find the magnitude \texttip{F_{\rm d}}{F_d} of the drag force after the skydiver has reached terminal velocity.
Express your answer in terms of the skydiver's mass m and other given quantities.
Hint 1. Definition of terminal velocity Once terminal velocity is reached, the skydiver's acceleration goes to zero. (The drag force exactly balances the downward acceleration due to gravity.)
ANSWER: \texttip{F_{\rm d}}{F_d} = m g
Hint 3. Relative direction of the drag force and velocity When you find \vec{F_d}\cdot\vec{v}, remember that the direction of t he drag force is opposite to the d irection of the skydiver's velocity.
ANSWER: \texttip{P_{\rm d}}{P_d} = -mgv
Correct
Problem 7.72 If a fish is attached to a vertical spring and slowly lowered to its equilibrium position, it is found to stretch the spring by an amount d.
Part A If the same fish is attached to the end of the un stretched spring and then allowed to fall from rest, through what maximum distance does it stretch the spring? (Hint: Calculate the force constant of the spring in terms of the distance d and the mass m of the fish.) Express your answer in terms of d. ANSWER: y = 2d
Correct
Problem 7.65 In a truck-loading station at a post o ffice, a small 0.200-{\rm kg} package is released from rest at point \texttip{A}{A} on a track that is one-quarter of a circle with radius 1.60 {\rm m} (the figure ). The size of the package is much less than 1.60 {\rm m}, so the package can be treated as a particle. It slides down the track and reaches point \texttip{B}{B} with a speed of 4.20 {\rm m/s} . From point \texttip{B}{B}, it slides on a level surface a distance o f 3.00 {\rm m} to point {C}, where it comes to rest.
Part A What is the coefficient of kinetic friction on the horizontal surface? ANSWER: \mu = 0.300
Correct
Part B How much work is done on the package by friction as it slides down the circular arc from \texttip{A}{A} to \texttip{B}{B}? ANSWER: W=
-1.37 {\rm J}
Correct
Conservation of Energy Ranking Task Six pendulums of various masses \texttip{m}{m} are released from various heights \texttip{h}{h} above a tabletop, as shown in the figures below. All the pendulums have the same length a nd are mounted such that at the vertical position their lowest points are the height of the tabletop and just do not strike the tabletop when released. Assume that the size of each bob is negligible.
Part A Rank each pendulum on the basis of its initial gravitational potential energy (before being released) relative to the tabletop. Rank from largest to smallest To rank items as equivalent, overlap them.
Hint 1. Gravitational potential energy Gravitational potential energy \texttip{U}{U} is defined as the product of the mass of the object, the acceleration due to gravity, and the height of the object above a reference level, summarized as U = mgh.
ANSWER:
Correct
Part B Rank each pendulum on the basis of the maximum kinetic energy it attains after release. Rank from largest to smallest To rank items as equivalent, overlap them.
Hint 1. Kinetic energy Each pendulum begins at rest (i.e., kinetic energy of zero). The maximum kinetic energy for a pendulum will occur when it is at the lowest point in its motion. At this point, all of the pendulums will be the same height above the tabletop. The kinetic energy then will equal the change in potential energy from the initial point to the point just above the tabletop.
ANSWER:
Correct
Part C Rank each pendulum on the basis of its maximum speed. Rank from largest to smallest To rank items as equivalent, overlap them.
Hint 1. The role of mass Both kinetic energy and gravitational potential energy are proportional to mass. Thus, a pendulum with larger mass has a larger potential energy upon release, and a larger kinetic energy at its lowest point. Since the kinetic energy is equal to the change in potential energy, you may write down the equation mg\Delta h = ^1/_2 m v^2. Notice that mass may be canceled from both sides. Thus, the final speed depends upon the change in height but is independent of mass. The situation is similar to that of an object in free fall. In free fall, although larger masses are acted upon by a larger gravitational force, a larger mass also has more inertia. These two effects cancel out. All ob jects fall with the same acceleration, and therefore reach the same velocity after falling equal distances.
ANSWER:
Correct
Exercise 7.10 A 26.0 {\rm {\rm kg}} child plays on a swing having support ropes that are 1.80 {\rm {\rm m}} long. A friend pulls her back until the ropes are 45.0 {\rm ^\circ} from the vertical and releases her from rest.
Part A What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing? ANSWER: U=
134
{\rm J}
Correct
Part B How fast will she be moving at the bottom of the swing? ANSWER: v=
3.21
{\rm m/s}
Correct
Part C How much work does the tension in the ropes do as the child swings from the initial position to the bottom? ANSWER:
W =
0
{\rm J}
Correct
Exercise 7.11 You are testing a new amusement park roller coaster with an empty car with a mass of 110 {\rm kg} . One part of the track is a vertical loop with a radius of 12.0 {\rm m} . At the bottom of the loop (point A) the car has a speed of 25.0 {\rm m/s} and at the top of the loop (point B) it has speed of 8.00 {\rm m/s} .
Part A As the car rolls from point A to point B, how much work is done by friction? Use 9.81 {\rm m/s^2} for the acceleration due to gravity. ANSWER: -4960 \rm J
Correct
Energy in a Spring Graphing Question A toy car is held at rest against a compressed spring, as shown in the figure. When released, the car slides across the room. Let x = 0 be the initial position of the car. Assume that friction is negligible.
Part A Sketch a graph of the total energy of the spring and car system. There is no scale given, so your graph should simply reflect the qualitative shape of the energy vs. time plot. ANSWER:
Correct
Part B Sketch a plot of the elastic potential energy of the spring from the point at which the car is released to the equilibrium position of the spring. Make your graph consistent with the given plot of total energy (the gray line given in the graphing window).
Hint 1. Determine the sign of the initial elastic potential energy At the instant the car is released, the spring is compressed. Therefore, is the spring's initial elastic potential energy positive, negative, or zero? ANSWER: positive negative zero
Correct
Hint 2. Determine the sign of the initial kinetic energy Is the initial kinetic energy of the cart positive, negative, or zero? ANSWER: positive negative zero
Correct
Hint 3. Determine the sign of the final elastic potential energy When the car reaches the equilibrium position of the spring, is the elastic potential energy positive, negative, or zero? ANSWER:
positive negative zero
Correct
Hint 4. The shape of the elastic potential energy graph The elastic potential energy of a spring with spring constant \texttip{k}{k} that is stretched or compressed to position \texttip{x}{x} is given by \large{U = \frac{1}{2}k(x-x_0)^2}, where \texttip{x_{\rm 0}}{x_0} is the equilibrium position of the spring.
ANSWER:
All attempts used; correct answer displayed
Part C Sketch a graph of the car's kinetic energy from the moment it is released until it passes the equilibrium position of the spring. Your graph should be consistent with the given plots of total energy (gray line in graphing window) and potential energy (gray parabola in graphing window). ANSWER:
Answer Requested
Bungee Jumping Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass\texttip{m}{m}, and the surface of the bridge is a height \texttip{h}{h} above the water. The bungee cord, which has length \texttip{L}{L} when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant \texttip{k}{k}. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use \texttip{g}{g} for the magnitude of the acceleration due to gravity.
Part A How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn't touch the water. Express the distance in terms of quantities given in the problem introduction.
Hint 1. Decide how to approach the problem Here are three possible methods for solving this problem: a. No nonconservative forces are acting, so mechanical energy is conserved. Set Kate's gravitational potential energy at the top of the bridge equal to the spring potential ene rgy in the bungee cord (which depends on the cord's final length \texttip{d}{d}) and solve for \texttip{d}{d}. b. Since nonconservative forces are acting, mechanical energy is not conserved. Set the spring potential energy in the bungee cord (which depends on \texttip{d}{d}) equal to Kate's gravitational potential energy plus the work done by dissipative forces. Eliminate the unknown work, and solve for \texttip{d}{d}. c. When Kate comes to rest she has zero acceleration, so the net force acting on her must be zero. Set the spring force due to the bungee cord (which depends on \texttip{d}{d}) equal to the force of gravity and solve for \texttip{d}{d}.
Which of these options is the simplest, most accurate way to find \texttip{d}{d} given the information available? ANSWER:
a b c
Hint 2. Compute the force due to the bungee cord When Kate is at rest, what is the magnitude \texttip{F_{\rm b}}{F_b} of the upward force the bungee cord exerts on her? Express your answer in terms of the cord's final stretched length \texttip{d}{d} and quantities given in the problem introduction. Your answer should not depend on Kate's mass \texttip{m}{m}.
Hint 1. Find the extension of the bungee cord The upward force on Kate is due to the extension of the bungee cord. What is this extension? Express your answer in terms of the cord's final (stretched) length \texttip{d}{d} and \texttip{L}{L}. ANSWER: Extension = d-L
Hint 2. Formula for the force due to a stretched cord The formula for the force due to a stretched cord is F = - k \Delta x, where \texttip{k}{k} is the spring constant of the cord and \texttip{\Delta x}{Deltax} is the extension of the cord.
ANSWER: \texttip{F_{\rm b}}{F_b} = k \left(d-L\right)
ANSWER: \texttip{d}{d} = \large{{\frac{mg}{k}}+L}
Correct
Part B If Kate just touches the surface of the river on her first downward trip (i.e., before the first bounce), what is t he spring constant \texttip{k}{k}? Ignore all dissipative forces. Express \texttip{k}{k} in terms of \texttip{L}{L}, \texttip{h}{h}, \texttip{m}{m}, and \texttip{g}{g}.
Hint 1. Decide how to approach the problem Here are three possible methods for solving this problem: a. Since nonconservative forces are ignored, mechanical energy is conserved. Set Kate's gravitational potential energy at the top of the bridge equal to the spring potential energy in the bungee cord at the lowest point (which depends on \texttip{k}{k}) and solve for \texttip{k}{k}. b. Nonconservative forces can be ignored, so mechanical energy is conserved. Set the spring potential energy in the bungee cord (which depends on \texttip{k}{k}) equal to Kate's gravitational potential energy at the top of the bridge plus the work done by gravity as Kate falls. Compute the work done by gravity, then solve for \texttip{k}{k}. c. When Kate is being held just above the water she has zero acceleration, so the net force acting on her must be zero. Set the spring force due to the bungee cord (which depends on \texttip{k}{k}) equal to the force of g ravity and solve for \texttip{k}{k}.
Which of these options is the simplest, most accurate way to find \texttip{k}{k} given the information available?
ANSWER: a b c
Hint 2. Find the initial gravitational potential energy What is Kate's gravitational potential energy \texttip{U_{\rm g}}{U_g} at the moment she steps off the bridge? (Define the zero of gravitational potential to be at the surface of the water.) Express your answer in terms of quantities given in the problem introduction. ANSWER: \texttip{U_{\rm g}}{U_g} = m g h
Hint 3. Find the elastic potential energy in the bungee cord What is the elastic potential energy \texttip{U_{\rm el}}{U_el} stored in the bungee cord when Kate is at the lowest point of her first downward trip? Express your answer in terms of quantities given in the problem introduction.
Hint 1. Formula for elastic potential energy The elastic potential energy of the bungee cord (which we are treating as an ideal spring) is \large{U_{\rm el} = \frac{1}{2}k (\Delta x)^2}, where \texttip{\Delta x}{Deltax} is the amount by which the cord is stretched beyond its unstretched length.
Hint 2. How much is the bungee cord stretched? By how much is the bungee cord stretched when Kate is at a depth \texttip{d_{\rm 1}}{d_1} below the bridge? Express your answer in terms of \texttip{d_{\rm 1}}{d_1} and \texttip{L}{L}. ANSWER: \texttip{\Delta x}{Deltax} = d_{1}-L
ANSWER: \texttip{U_{\rm el}}{U_el} =
\large{\frac{1}{2} k \left(h-L\right)^{2}}
ANSWER: \texttip{k}{k} = \large{\frac{2mgh}{\left(h-L\right)^{2}}}
Correct
Exercise 7.15 A force of 541 {\rm \; N} keeps a certain spring stretched a distance of 0.500 {\rm \; m} .
Part A What is the potential energy of the spring when it is stretched of 0.500 {\rm \; m} ?
Express your answer with the appropriate units. ANSWER: U_1 = 135 {\rm J}
Correct
Part B What is its potential energy when it is compressed of 9.00 {\rm \; cm} ? Express your answer with the appropriate units. ANSWER: U_2 = 4.38 {\rm J}
Correct
PSS 7.2 Problems Using Mechanical Energy II Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-{\rm kg} circus performer who is shot from a cannon (actually a spring gun). You don’t find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 {\rm N/m} that he will compress with a force of 4400 {\rm N}. The inside of the gun barrel is coated with Teflon, so the average friction force will be only 40.0 {\rm N} during the 4.00 {\rm m} he moves in the barrel. At what speed will he emerge from the end of the barrel, 2.50 {\rm m} above his initial rest position? Problem-Solving Strategy: Problems using mechanical energy II IDENTIFY the relevant concepts: The energy approach is useful in solving problems that involve elastic forces as well as gravitational forces, provided the additional elastic potential energy i included in the potential energy \texttip{U}{U}. SET UP the problem using the following steps: 1. Decide what the initial and final states of the system are. Use the subscript 1 for the initial state and the subscript 2 for the final state. It helps to draw sketches. 2. Define your coordinate system, particularly the level at which y=0. We suggest that you always choose the positive y direction to be upward because this is what U_{\rm grav} = mgy assumes. 3. Identify all forces that do work, including those that can’t be described in terms of potential energy. A free-body diagram is always helpful. 4. List the unknown and known quantities, including the coordinates and velocities at each point. Decide which unknowns are your target variables. EXECUTE the solution as follows: Write expressions for the initial and final kinetic and potential energies—that is, \texttip{K_{\rm 1}}{K_1}, \texttip{K_{\rm 2}}{K_2}, \texttip{U_{\rm 1}}{U_1}, and \texttip{U_{\rm 2}}{U_2}. The potential energy \texttip{U}{U} now includes both the gravitational potential energy \texttip{U_{\rm grav}}{U_grav} and the elastic potential energy \large{U_{\rm el}=\frac{1}{2} kx^2}, where \texttip{x}{x} is the displacement of the spring from its unstretched length. Then, relate the kinetic and potential energies and the work done by other forces, \texttip{W_{\rm other}}{W_other}, using K_1 + U_1 + W_{\rm other} = K_2 + U_ 2. If no other forces do work, this expression becomes K _1 + U_1 = K_2 + U_2. It’s helpful to draw bar graphs showing the initial and final values of \texttip{K}{K}, \texttip{U}{U}, and E=K +U. Then, solve to find whatever unknown quantity is required. EVALUATE your answer: Check whether your answer makes physical sense. Keep in mind that the work done by the gravitational and elastic forces is accounted for by their potential energies; the work of the other forces, \texttip{W_{\rm other}}{W_other}, has to be included separately.
IDENTIFY the relevant concepts The problem involves a spring gun. Therefore, to account for the potential energy associated with its elastic force, the energy approach might be the best method to solve this problem.
SET UP the problem using the following steps
Part A Below is a sketch of the initial state of the situation described in this problem. Draw the most suitable set of coordinate axes for this problem. Note that even though you can choose the y=0 level to be wherever you like, in most situations it is best to set the zero height to coincide with either the initial or final position, so that the calculations for the gravitational potential energy become easier. For this reason, in this particular problem place the origin of your coordinate axes on the black dot marking the performer's initial position. Draw only the positive portion of the coordinate axes. Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vec tors will not be graded. ANSWER:
Part B
This question will be shown after you complete previous question(s).
EXECUTE the solution as follows
Part C
This question will be shown after you complete previous question(s).
EVALUATE your answer
Part D
This question will be shown after you complete previous question(s).
Exercise 7.27 A 0.60-\rm kg book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.4 {\rm \; N} .
Part A How much work is done on the book by friction during a displacement of 2.9 {\rm \; m} to the left? Express your answer with the appropriate units. ANSWER: W_1 = -4.1 {\rm J}
Correct
Part B The book now slides 2.9 {\rm \; m} to the right, returning to its starting point. During this second 2.9-\rm m displacement, how much work is done on the book by friction? Express your answer with the appropriate units. ANSWER: W_2 = -4.1 {\rm J}
Correct
Part C What is the total work done on the book by friction during the complete round trip? Express your answer with the appropriate units. ANSWER: W_3 = -8.1 {\rm J}
Correct
Part D On the basis of your answer to part C, would you say that the friction force is conservative or nonconservative? ANSWER: conservative nonconservative
Correct The net work done by friction for the round trip is not zero, and friction is not a conservative force.
Exercise 7.33 A small block with mass 0.0400 {\rm kg} is moving in the xy -plane. The net force on the block is described by the potential- energy function U(x,y)= (5.55 {\rm J/m^2} )x^2-(3.85 {\rm J/m^3} )y^3.
Part A What is the magnitude of the acceleration of the block when it is at the point x= 0.37 {\rm m} , y= 0.50 {\rm m} ?
Express your answer with the appropriate units. ANSWER: a=
126 \large{{\rm \frac{m}{s^{2}}}}
Correct
Part B What is the direction of the acceleration of the block when it is at the point x= 0.37 {\rm m} , y= 0.50 {\rm m} ? ANSWER: \theta =
145
^\circ counterclockwise from the +x-axis
Correct
Springs in Two Dimensions The ends of two identical springs are connected. Their unstretched lengths \ell are negligibly small and each has spring constant \texttip{k}{k}. After being connected, both springs are stretched an amount \texttip{L}{L} and their free ends are anchored at y = 0 and x= \pm L as shown . The point where the springs are connected to each other is now pulled to the position (\texttip{x}{x}, \texttip{y}{y}). Assume that (\texttip{x}{x}, \texttip{y}{y}) lies in the first quadrant.
Part A What is the potential energy of the two-spring system after the point of connection has been moved to position (\texttip{x}{x}, \texttip{y}{y})? Keep in mind that the unstretched length of each spring \ell is much less than \texttip{L}{L} and can be ignored (i.e., \ell \ll L). Express the potential in terms of \texttip{k}{k}, \texttip{x}{x}, \texttip{y}{y}, and \texttip{L}{L}.
Hint 1. An important property of the potential energy Recall that potential energy is a scalar and hence can be added like a number. You should find the potential energies of each of the springs, then add these two energies together to get the total potential energy of the system.
Hint 2. Potential energy of the left-hand spring Find the potential energy of the spring on the left side of the figure when the point where the two springs are connected is pulled to the position (\texttip{x}{x}, \texttip{y}{y}).
Hint 1. Stretched length of the left-hand spring Find the total elongation of the spring on the left when its end (the point at which it connects to the other spring) is at the point ( \texttip{x} {x}, \texttip{y}{y}).
ANSWER: {\rm elongation}_{\rm left} = \sqrt{\left(\left(L+x\right)^{2} + y^{2}\right)}
ANSWER: \texttip{U_{\rm left}}{U_left} = .5 k \left(\left(L+x\right)^{2}+\left(y^{2}\right)\right)
ANSWER: \texttip{U}{U} = k\left(x^{2}+y^{2}+L^{2}\right)
Correct
Part B Find the force \texttip{\vec{F}}{F_vec} on the junction point, the point where the two springs are attached to each other. Express \texttip{\vec{F}}{F_vec} as a vector in terms of the unit vectors \texttip{\hat{i}}{i_unit} and \texttip{\hat{j}}{j_unit}.
Hint 1. How to approach the problem Once you have obtained the potential energy you can calculate the corresponding force as the gradient of the potential energy.
Hint 2. Components of the force vector The components of the force are \large{F_{x} = -\frac{dU}{dx}\; {\rm and}\; F_{y} = -\frac{dU}{dy}}.
ANSWER: \texttip{\vec{F}}{F_vec} = -2kx \hat{i}-2ky \hat{j}
Correct Notice how much more difficult it would have been to obtain the force via vector addition (computing the two component s of the force from each spring, then adding them). This is the power of scalar potential functions: They allow you to simply add up the contributions, without having to worry about vectors or coordinate axes. By taking the gradient of the potential, you automatically obtain the desired vector quantities.
Exercise 7.35 The potential energy of two atoms in a diatomic molecule is approximated by U\,( r ) = a/r^{12} - b/r^6, where \texttip{r}{r} is the spacing between atoms and \texttip{a}{a} and \texttip{b}{b} are positive constants.
Part A Find the force F(r)on one atom as a function of r . Express your answer in terms of the variables a, b, and r . ANSWER: F(r) = \large{{\frac{12a}{r^{13}}}-{\frac{6b}{r^{7}}}}
Correct
Part B Find the equilibrium distance between the two atoms. Express your answer in terms of the variables a and b. ANSWER: r_{min} =
\large{\sqrt[6]{2} \sqrt[6]{\frac{a}{b}}}
Correct
Part C Is this equilibrium stable? ANSWER: yes no
Correct
Part D Suppose the distance between the two atoms is equal to the equilibrium distance found in part A. What minimum energy must be added to the molecule to dissociate it-that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. Express your answer in terms of the variables a and b. ANSWER: E_0 = \large{\frac{b^{2}}{4a}}
Correct
Part E For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is 1.13\times 10^{ - 10}\; {\rm{ m}} and the dissociation energy is 1.54\times 10^{ - 18} \;{\rm{ J}} per molecule. Find the value of the constant \texttip{a}{a}. ANSWER: a =
6.68!10 138 "
{\rm J \times m^{12}}
Correct
Part F Find the value of the constant \texttip{b}{b}. ANSWER: b =
6.41!10 78
Correct
"
{\rm J \times m^{6}}
Problem 7.39 A block with mass 0.50 {\rm kg} is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 {\rm m} (the figure ). When released, the block moves on a horizontal tabletop for 1.00 {\rm m} before coming to rest. The spring constant \texttip{k}{k} is 100 {\rm{ N/m}}.
Part A What is the coefficient of kinetic friction \mu _{\rm{k}} between the block and the tabletop? Express your answer using two significant figures. ANSWER: \mu _{\rm{k}} = 0.41
Correct
Problem 7.53 A 0.275 {\rm kg} potato is tied to a string with length 2.00 {\rm m} , and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released.
Part A What is the speed of the potato at the lowest point of its motion? Take free fall acceleration to be 9.80 {\rm m/s^2} . ANSWER: v=
6.26
{\rm m}/{\rm s}
Correct
Part B What is the tension in the string at this point? ANSWER: T = 8.09
{\rm N}
Correct Score Summary:
