HW6: Work and energy PDF

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Work and energy...

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HW6: Work and energy Due: 11:59pm on Wednesday, April 4, 2018 You will receive no credit for items you complete after the assignment is due. Grading Policy

Understanding Work Done by a Constant Force Learning Goal: To explore the definition of work and learn how to find the work done by a force on an object. The word "work" has many meanings when used in everyday life. However, in physics work has a very specific definition. This definition is important to learn and understand. Work and energy are two of the most fundamental and important concepts you will learn in your study of physics. Energy cannot be created or destroyed; it can only be transformed from one form to another. How this energy is transferred affects our daily lives from microscopic processes, such as protein synthesis, to macroscopic processes, such as the expansion of the universe!

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When energy is transferred either to or away from an object by a force F acting over a displacementd , work W is done on that object. The amount of work done by a constant force can be found using the equation

W = F d cos θ , 

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where F is the magnitude of F , d is the magnitude of d , and θ is the angle between F and d .

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The SI unit for work is the joule, J . A single joule of work is not very big. Your heart uses about0.5 J each time it beats, and the 60-watt lightbulb in your desk lamp uses 216, 000 J each hour. A joule is defined as follows:

1J = 1Nm = 1

kg m 2 s2

The net work done on an object is the sum of the work done by each individual force acting on that object. In other words,

Wnet = W1 + W 2 + W3 + ⋯ = ∑i W i . The net work can also be expressed as the work done by the net force acting on an object, which can be represented by the following equation:

Wnet = Fnet d cos θ. Knowing the sign of the work done on an object is a crucial element to understanding work. Positive work indicates that an object has been acted on by a force that tranfers energy to the object, thereby increasing the object's energy. Negative work indicates that an object has been acted on by a force that has reduced the energy of the object. The next few questions will ask you to determine the sign of the work done by the various forces acting on a box that is being pushed across a rough floor. As illustrated in the figure , the box is being acted on by a normal forcen , the force due to

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gravity w ,  the force of kinetic friction f k , and the pushing force F p . The displacement of the box isd .

Part A Which of the following statements accurately describes the sign of the work done on the box by the force of the push?

Hint 1. Find the angle The work done on the box by the pushing force depends on the angle θ between F p  and the displacement d . What is this angle? ANSWER:

0 degrees 45 degrees 90 degrees 180 degrees

ANSWER:

positive negative zero

Correct

Part B

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Which of the following statements accurately decribes the sign of the work done on the box by the normal force?

Hint 1. Finding theta

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 the displacement d . What The work done on the box by the normal force depends on the angle θ between n and is this angle? ANSWER:

0 degrees 45 degrees 90 degrees 180 degrees

ANSWER:

positive negative zero

Correct

Part C Which of the following statements accurately decribes the sign of the work done on the box by the force of kinetic friction?

Hint 1. Finding theta

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The work done on the box by the force of kinetic friction depends on the angle θ between f k and the  displacement d . What is this angle? ANSWER:

0 degrees 45 degrees 90 degrees 180 degrees

ANSWER:

positive negative zero

Correct

Part D Which of the following statements accurately decribes the sign of the work done on the box by the force of gravity?

Hint 1. Finding the angle

 the displacement d .  What is The work done on the box by the weight depends on the angle θ between w and this angle? ANSWER:

0 degrees 45 degrees 90 degrees 180 degrees

ANSWER:

positive negative zero

Correct

Making generalizations You may have noticed that the force due to gravity and normal forces do no work on the box. Any force that is perpendicular to the displacement of the object on which it acts does no work on the object. The force of kinetic friction did negative work on the box. In other words, it took energy away from the box. Typically, this

energy gets transformed into heat, like the heat that radiates from your skin when you get a rug burn due to the friction between your skin and the carpet. A force that acts on an object in a direction opposite to the direction of the object's displacement does negative work on the object. The pushing force acts on the box in the same direction as the object's displacement and does positive work on the box. These generalizations allow physicists to rewrite the equation for work as

W = F∥ d, 

 is either parallel or antiparallel to the displacement. If F is parallel to d , as in the where F|| is the component of F that || 

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case of F p , then the work done is positive. If F || is antiparallel to d , as in the case of f k , then the work done is negative.

Part E You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,500 N across the carpet to a spot 5 m away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser? ANSWER:

W > 1.75 × 104 J W = 1.75 × 104 J 1.75 × 104 J > W > 0 J W = 0J

Correct Remember that to a physicist work means something very specific, and since you were unable to move the dresser, d = 0 and therefore W = 0 . However, you got tired and sweaty trying to move the dresser, just as you do when you go to "work out" at the gym.Your muscles are not static strips of fibrous tissue. They continually contract and expand a slight amount when you exert them. Chemical energy from food is being transformed into the energy needed to move your muscles. Work is being done inside your muscles, but work is not being done on the dresser.

Part F A box of mass m is sliding down a frictionless plane that is inclined at an angle ϕ above the horizontal, as shown in the figure . What is the work done on the box by the force due to gravityw , if the box moves a distanced ?

Hint 1. Finding Theta. The work done on the box by the force of gravity depends on the angle between the weight and the displacement; this is the angle θ that goes into the equation

W = F d cos θ . ANSWER:

W = wd cos ϕ W = wd cos(90 − ϕ) W =0 None of these

Correct The angle given to you in a problem is not always the same angle that you use in the equation for work!

Part G The planet Earth travels in a circular orbit at constant speed around the Sun. What is the net work done on the Earth by the gravitational attraction between it and the Sun in one complete orbit? Assume that the mass of the Earth is given by M e , the mass of the Sun is given by Ms , and the Earth-Sun distance is given byr es .

Hint 1. Newton's law of universal gravitation The magnitude of the force of attraction between two objects of masses M 1 and M 2 that are separated by a distance r is given by:

F =G

M1 M 2 . r2

Hint 2. Circumference of a circle The circumference of a circle with radius r is

C = 2πr Hint 3. Finding the angle The work done on the Earth by the gravitational attraction between it and the Sun depends on the angle between the gravitational force and the displacement of the Earth; this is the angle θ that goes into

W = F d cos θ . The force of attraction always points from the Earth toward the Sun along the radius of the Earth's orbit. At any instant in time the displacement of the Earth is considered to be tangent to its orbit; perpendicular to the radius.

ANSWER:

W = 2πG

Me M s res

W = GπMe Ms W =0 None of these.

Correct An object undergoing uniform circular motion experiences a net force that is directed in toward the center of the circle; this net force is called the centripetal force. This force is always perpendicular to the distance the object moves and therefore never does any work on the object.

Part H A block of mass m is pushed up against a spring with spring constant k until the spring has been compressed a distance x from equilibrium. What is the work done on the block by the spring?

Hint 1. Hooke's Law The force exerted by a spring with spring constant k is given by

 −k x,  F = where

x is the spring's displacement from equilibrium position x eq .

ANSWER:

W = kx 2 W = −kx 2 W =0 None of these.

Correct The equation for work presented in this problem requires that the force be constant. Because the force exerted

 kx ) you cannot use W on an object varies with the spring's displacement from equilibrium (F = to find the work done by a spring. In actuality the work done by a spring is given by the equation

= F d cos θ

W spring = − 21 kx2 .

Congratulations! Now that you have the basics down and have been exposed to some tricky situations involving the equation for work, you are ready to apply this knowledge to new situations.

Work and Kinetic Energy Two blocks of ice, one four times as heavy as the other, are at rest on a frozen lake. A person pushes each block the same

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distance d . Ignore friction and assume that an equal force F is exerted on each block.

Part A Which of the following statements is true about the kinetic energy of the heavier block after the push?

Hint 1. How to approach the problem The work-energy theorem states that the change in kinetic energy of an object equals the net work done on that object: Wtotal = ∆K . The work done on an object can also be related to the distance d that the object moves while being acted on by a force F :

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W = F∥ d, where

 F|| is the component of F parallel to the direction of displacement.

Hint 2. Find the work done on each block What can be said about the net work done on the heavier block? ANSWER:

It is greater than the work done on the lighter block. It is equal to the work done on the lighter block. It is less than the work done on the lighter block.

ANSWER:

It is smaller than the kinetic energy of the lighter block. It is equal to the kinetic energy of the lighter block. It is larger than the kinetic energy of the lighter block. It cannot be determined without knowing the force and the mass of each block.

Correct The work-energy theorem states that the change in kinetic energy of an object equals the net work done on that object. The only force doing work on the blocks is the force from the person, which is the same in both cases. Since the initial kinetic energy of each block is zero, both blocks have the same final kinetic energy.

Part B Compared to the speed of the heavier block, what is the speed of the light block after both blocks move the same distance d ?

Hint 1. How to approach the problem In Part A, you determined that the kinetic energy of the heavier block was the same as that of the lighter block. Relate this to the speed of the blocks.

Hint 2. Proportional reasoning Proportional reasoning becomes easier with practice. First relate the kinetic energies of the blocks to each other. To accomplish this, let the subscript h refer to the heavier block and the subscript ℓ to the lighter block. Now

K h = Kℓ can be written as 1 mh (v h ) 2 = 12 mℓ (v ℓ )2 . 2 The problem states that the heavier block is four time as massive as the lighter block. This can be represented by the expression

mh = 4m ℓ . Substituting this expression into the expression for kinetic energy yields

1 2

2

2

(4m ℓ )(v h ) = 12 m ℓ( vℓ )

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How many times larger than v2h is v 2ℓ ? ANSWER:

v 2ℓ = 4 v2h

ANSWER:

one quarter as fast half as fast the same speed twice as fast four times as fast

Correct Since the kinetic energy of the lighter block is equal to the kinetic energy of the heavier block, the lighter block must be moving faster than the heavier block.

Part C

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Now assume that both blocks have the same speed after being pushed with the same force F . What can be said about the distances the two blocks are pushed?

Hint 1. How to approach the problem The work-energy theorem states that the change in kinetic energy of an object equals the net work done on that object: Wtotal = ∆K . The work done on an object can also be related to the distance d that the object moves while being acted on by a force F : 

W = F∥ d, where

 F|| is the component of F parallel to the direction of displacement.

Hint 2. Relate the kinetic energies of the blocks Let the subscript h refer to the heavier block and the subscript ℓ to the lighter block. What is the ratio

Kh Kℓ ?

Hint 1. The kinetic energies To relate the kinetic energies of the blocks to each other, recall that

v h = vℓ and

mh = 4m ℓ . ANSWER: Kh Kℓ = 4

Hint 3. Compare the amount of work done on each block In the previous hint, you found that K h work done on the lighter block,

= 4Kℓ . What is the ratio of the work done on the heavy block to the

Wh Wℓ ? ANSWER: Wh Wℓ =

4

ANSWER:

The heavy block must be pushed 16 times farther than the light block. The heavy block must be pushed 4 times farther than the light block. The heavy block must be pushed 2 times farther than the light block. The heavy block must be pushed the same distance as the light block. The heavy block must be pushed half as far as the light block.

Correct Because the heavier block has four times the mass of the lighter block, when the two blocks travel with the same speed, the heavier block will have four times as much kinetic energy. The work-energy theorem implies that four times more work must be done on the heavier block than on the lighter block. Since the same force is applied to both blocks, the heavier block must be pushed through four times the distance as the lighter block.

Problem 7.17

Part A Fleas are agile, wingless insects that feed on the blood of their hosts. Although they are typically 2–3 mm long with a mass of 3.8×10−4 kg , they have an astonishing ability to jump when threatened. Their propulsion, which can briefly produce accelerations more than 100 times that of gravity, comes not from muscles but, in fact, from an elastomeric protein called resilin, which acts as a spring. Given that the typical launch velocity of a flea is about 1 m/s , what total energy must be stored in the resilin just before the flea jumps? Express your answer in Joules to two significant figures. ANSWER: 1.9×10−4

J

Correct

Ups and Downs Learning Goal: To apply the law of conservation of energy to an object launched upward in Earth's gravitational field. In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy. In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K = (1/2)mv2 and its gravitational potential energy U = mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation

K i + U i = K f + Uf

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where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem. First, let us consider an object launched vertically upward with an initial speed v . Neglect air resistance.

Part A

As the projectile goes upward, what energy changes take place? ANSWER:

Both kinetic and potential energy decrease. Both kinetic and potential energy increase. Kinetic energy decreases; potential energy increases. Kinetic energy increases; potential energy decreases.

Correct

Part B At the top point of the flight, what can be said about the projectile's kinetic and potential energy? ANSWER:

Both kinetic and potential energy are at their maximum values. Both kinetic and potential energy are at their minimum values. Kinetic energy is at a maximum; potential energy is at a minimum. Kinetic energy is at a minimum; potential energy is at a maximum.

Correct Strictly speaking, it is not the ball that possesses potential energy; rather, it is the system "Earth-ball." Although we will often talk about "the gravitational potential energy of an elevated object," it is useful to keep in mind that the energy, in fact, is associated with the interactions between Earth and the elevated object.

Part C The potential energy of the object at the moment of launch __________. ANSWER:

is negative is positive is zero depends on the choice of the "zero level" of potential energy

Correct Usually, the zero level is chosen so as to make the relevant calculations simpler. In this case, it makes good sense to assume that U = 0 at the ground level, but this is not by any means the only choice!

Part D Using conservation of energy, find the maximum height h max to which the object will rise. Express your answer in terms of v andg . You may or may not use all of these quantities. ANSWER:

hmax =

v2 2g

Answer Requested You may remember this result from kinematics. It is comforting to know that our new approach yields the same answer.

Part E At what height h above the ground does the projectile have a speed of 0.5v ? Express your answer in terms of v andg . You may or may not use all of these quantities. ANSWER:

h=

3v2 8g

Correct

Part F What is the speed u of the object at the height of (1/2)h max ? Express your answer in terms of v andg . You may or may not use all of these quantities.

Hint 1. How to approach the problem You are being asked for the speed at half of the maximum height. You know that at the initial height (h = 0 ), the speed is v. All of the energy is kinetic energy, and so the total energy is (1/2)mv 2 . At the maximum height, all of the energy is potential energy. Since the gravitational potential energy is proportional to h , half of the initial kinetic energy must have been converted to potential energy when the projectile is at (1/2) hmax . Thus, the 1 1 1 kinetic energy must be half of its original value (i.e., K = 2 ⋅ 2 m v 2 = 4 m...