Mastering Physics: Ch 06

(1)

Ch 06 HW Due: 11:59am on Sunday, April 6, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy

The WorkEnergy Theorem

Learning Goal: To understand the meaning and possible applications of the workenergy theorem. In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the workenergy theorem. We will start with a special case: a particle of mass moving in the x direction at constant acceleration . During a certain interval of time, the particle accelerates from to , undergoing displacement

given by . Part A Find the acceleration of the particle. Express the acceleration in terms of , , and . Hint 1. Some helpful relationships from kinematics By definition for constant acceleration, . Furthermore, the average speed is , and the displacement is . Combine these relationships to eliminate . ANSWER:

Correct

Part B Find the net force acting on the particle. Express your answer in terms of and . Hint 1. Using Newton's laws Which of Newton's laws may be helpful here?

m

a

vinitial

vfinal

s

s =

xfinal

−

xinitial

a

vinitial

vfinal

s

a =

vfinal−vinitial t

=

v

avg vinitial₂+vfinal

s =

vavg

t

=

a

vfinal2−vinitial2 2s

F

m

a

(2)

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920 2/27 ANSWER:

Correct

Part C Find the net work done on the particle by the external forces during the particle's motion. Express your answer in terms of and . ANSWER:

Correct

Part D

Substitute for from Part B in the expression for work from Part C. Then substitute for from the relation in Part A. This will yield an expression for the net work done on the particle by the external forces during the particle's motion in terms of mass and the initial and final velocities. Give an expression for the work in terms of those quantities. Express your answer in terms of , , and . ANSWER:

Correct

The expression that you obtained can be rearranged as The quantity has the same units as work. It is called the kinetic energy of the moving particle and is denoted by . Therefore, we can write and . Note that like momentum, kinetic energy depends on both the mass and the velocity of the moving object. However, the mathematical expressions for momentum and kinetic energy are different. Also, unlike momentum, kinetic energy is a scalar. That is, it does not depend on the sign (therefore direction) of the velocities. =

F

ma

W

F

s

=

W

Fs

F

a

W

m vinitial

vfinal

=

W

_m

vfinal2−vinitial2 2

W = m

1

− m

.

2

v

final2 12

v

initial2

m

1 2

v

2

K

= m

(3)

Part E Find the net work done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies. Express your answer in terms of and . ANSWER:

Correct

This result is called the workenergy theorem. It states that the net work done on a particle equals the change in kinetic energy of that particle. Also notice that if is zero, then the workenergy theorem reduces to . In other words, kinetic energy can be understood as the amount of work that is done to accelerate the particle from rest to its final velocity. The workenergy theorem can be most easily used if the object is moving in one dimension and is being acted upon by a constant net force directed along the direction of motion. However, the theorem is valid for more general cases as well. Let us now consider a situation in which the particle is still moving along the x axis, but the net force, which is still directed along the x axis, is no longer constant. Let's see how our earlier definition of work, needs to be modified by being replaced by an integral. If the path of the particle is divided into very small displacements , we can assume that over each of these small displacement intervals, the net force remains essentially constant and the work done to move the particle from to is

,

where is the x component of the net force (which remains virtually constant for the small displacement from to ). The net work done on the particle is then given by . Now, using and , it can be shown that . Part F Evaluate the integral .

W

Kinitial

Kfinal

=

W

−

Kinitial

+

Kfinal

K

initial

W = K

final

W = ⋅ ,

F⃗ s⃗

dx

dW

x x + dx

dW = F dx

F

x

x + dx

W

W =

∫

xfinal

dW =

F dx

x_initial

∫

xx_initialfinal

F = ma

a =

dv

=

= v

dt dxdv dxdt dxdv

W =

∫

vfinal

mvdv

v_initial

W =

∫

vfinal

mvdv

v_initial

(4)

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920 4/27 Express your answer in terms of , , and . Hint 1. An integration formula The formula for is . ANSWER:

Correct

The expression that you havejust obtained is equivalent to . Not surprisingly, we are back to the same expression of the workenergy theorem! Let us see how the theorem can be applied to problem solving. Part G

A particle moving in the x direction is being acted upon by a net force , for some constant . The particle moves from to . What is , the change in kinetic energy of the particle during that time? Express your answer in terms of and . Hint 1. Finding the work Integrate to calculate the work done on the particle. Hint 2. An integration formula The formula for is . ANSWER:

m vinitial

vfinal

t dt

∫

_ab

t dt =

∫

_ab b2−_a2 2 =

W

1₂

_m(

_v

_final2

₋

_v

_initial2

₎

W =

K

final

−

K

initial

F(x) = Cx

2

_C

= L

xinitial

xfinal

= 3L

ΔK

C

L

F(x) dx

du

∫

_ab

u

2

du =

∫

_ab

u

2 b3−a3 3 =

ΔK

26₃

_C

_L

3

(5)

Correct

It can also be shown that the workenergy theorem is valid for two and threedimensional motion and for a varying net force that is not necessarily directed along the instantaneous direction of motion of the particle. In that case, the work done by the net force is given by the line integral

where and are the initial and the final positions of the particle, is the vector representing a small displacement, and is the net force acting on the particle.

When Push Comes to Shove

Two forces, of magnitudes and , act in opposite directions on a block, which sits atop a frictionless surface. Initially, the center of the block is at position . At some later

time, the block has moved to the right, so that its center is at position , where .

Part A

Find the work done on the block by the force of magnitude as the block moves from to . Express your answer in terms of some or all of the variables given in the problem introduction.

Hint 1. Formula for the work done by a force

The work done by a force in producing a displacement is given by ,

where and are the magnitudes of and respectively, and is the smaller angle between the two vectors. The scalar that results from the operation is called the scalar product, or dot product, of the vectors and .

W =

∫

Sfinal

⋅

,

S_initial

F ⃗ dL

→

S

initial

S

final

dL

→

F⃗

F1

F2

xi

xf

>

xi

W1

F1

xi

xf

W

F⃗

s⃗

W = ⋅ =

_{F⃗ s⃗ ∣∣F ⃗ ∣∣∣∣s⃗ ∣∣}

cosϕ

∣∣F⃗ ∣∣ ∣∣s⃗ ∣∣

F⃗

s⃗

ϕ

⋅

F⃗ s⃗

F⃗

s⃗

(6)

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920 6/27

ANSWER:

Correct

Part B

Find the work done by the force of magnitude as the block moves from to .

Express your answer in terms of some or all of the variables given in the problem introduction. Hint 1. Is the work positive or negative? The force of magnitude acts in the opposite direction to that of the motion of the block. Therefore, the work done by that force must be negative. ANSWER:

Correct

Part C What is the net work done on the block by the two forces? ANSWER:

Correct

Part D Imagine that the two forces are equal in magnitude, , and that there are no other horizontal forces acting on the block. Determine the change in the kinetic energy of the block as it moves from to .

Hint 1. If the forces are equal, how can the block be moving? Although the net horizontal force acting on the block (and therefore the acceleration of the block) is zero, the block may have had some initial velocity. =

W1

F1

( − )

xf

xi

W2

F2

xi

xf

F2

=

W2

− ( − )

F2

xf

xi

Wnet

=

Wnet

( − )( − )

F1

F2

xf

xi

=

F

1

F

2

−

Kf

Ki

xi

xf

(7)

ANSWER:

Correct

Work from a Constant Force

Learning Goal: To understand how to compute the work done by a constant force acting on a particle that moves in a straight line. In this problem, you will calculate the work done by a constant force. A force is considered constant if is independent of . This is the most frequently encountered situation in elementary Newtonian mechanics. Part A Consider a particle moving in a straight line from initial point B to final point A, acted upon by a constant force . In the figure the force is indicated by a series of identical vectors pointing to the left, parallel to the horizontal axis. The vectors are all identical to reflect the idea that the force is constant everywhere along the path. The magnitude of the force is , and the displacement vector from point B to point A is (of magnitude , making an angle (radians) with the positive x axis). Find , the work that the force performs on the particle as it moves from point B to point A.

Express the work in terms of , , and . Remember to use radians, not degrees, for any angles that appear in your answer. Hint 1. Formula for work done by a constant force For a particle subjected to a constant force along a straight path represented by the displacement vector , the net work done by is . Hint 2. Find the angle between and You need to find the angle between the vector , which is directed horizontally to the left, and the vector in the direction of the particle's motion (at an angle (radians) relative to the positive x axis). It may help to visualize directed along the negative x axis at the origin. What is the angle between and ?

Express your answer in radians, not degrees. ANSWER: = 0

−

Kf

Ki

( )

F⃗ r⃗

r⃗

F⃗

F

L⃗

L

θ

WBA

F⃗

L F

θ

F⃗

L⃗

F⃗

F⃗ L⃗

⋅

F⃗

L⃗

F⃗

L⃗

θ

F⃗

ϕ

F⃗

L⃗

(8)

ANSWER:

Correct

This result is worth remembering! The work done by a constant force of magnitude , which acts at an angle of with respect to the direction of motion along a straight path of length , is . This equation correctly gives the sign in this problem. Since is the angle with respect to the positive x axis (in radians), ; hence .

Part B

Now consider the same force acting on a particle that travels from point A to point B. The displacement vector now points in the opposite direction as it did in Part A.

Find the work done by in this case. Express your answer in terms of , , and . Hint 1. A physical argument The easiest argument to make is a physical one: If the particle were to go straight from point A to point B and then back from point B to point A with the same force acting, the total work done would be zero (i.e., the gain in energy on the way to point B due to this force would be lost on the way back, and vice versa). This holds for all conservative forces (but does not hold for nonconservative forces), and a constant force is indeed conservative. Therefore, ANSWER:

Correct

=

ϕ π − θ

=

WBA

−Fcos(θ)L

F

ϕ

L

W

BA

= FL cos(ϕ)

θ

ϕ = π − θ

cos(ϕ) = cos(π − θ) = −cos(θ)

F⃗

L⃗

WAB

F⃗

L F

θ

+

= 0.

WBA

WAB

=

WAB

FLcos(θ)

(9)

Exercise 6.4

A factory worker pushes a 31.5 crate a distance of 4.2 along a level floor at constant velocity by pushing downward at an angle of 29 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25. Part A What magnitude of force must the worker apply to move the crate at constant velocity? Express your answer using two significant figures. ANSWER:

Correct

Part B How much work is done on the crate by this force when the crate is pushed a distance of 4.2 ? Express your answer using two significant figures. ANSWER:

Correct

Part C How much work is done on the crate by friction during this displacement? Express your answer using two significant figures. ANSWER:

Correct

Part D How much work is done by the normal force? ANSWER:

kg

m

∘ = 100

F

N

m

= 380

W

J

= 380

Wf

J

= 0

W

nf

J

(10)

Correct

Part E How much work is done by gravity? ANSWER:

Correct

Part F What is the total work done on the crate? ANSWER:

Correct

Exercise 6.6

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.7×106 , one an angle 15 west of north and the other an angle 15 east of north, as they pull the tanker a distance 0.86 toward the north.

Part A What is the total work they do on the supertanker? Express your answer using two significant figures. ANSWER:

Correct

Vertical Spring Gun: Speed and Kinetic Energy

The figure represents a multiflash photograph of a ball being shot straight up by a spring. The spring, with the ball atop, was initially compressed to the point marked and released. The point marked is the point where the ball would remain at rest if it were placed gently on the spring, and the ball reaches its highest point at the point marked . = 0

Wg

J

= 0

Wnet

J

N

∘ ∘

_km

= 2.8×109

W

J

Ybot

Y0

Ytop

Y0

(11)

For most situations, including this problem, the point may be taken to be at the top of the spring, where the ball loses contact with the spring. Part A Indicate whether the following statements are true or false. Assume that air resistance is negligible. The speed of the ball was greatest at point when it was still in contact with the spring. The speed of the ball was decreasing on its way from point to point . The speed of the ball was zero at point . The speed of the ball was the same for all points in its motion between points and . Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,f,t). ANSWER:

Correct

Part B Consider the kinetic energy of the ball. At what point or points is the ball's kinetic energy greatest? Hint 1. What equation to use

The kinetic energy is given by , where is the mass of the object and is its speed.

ANSWER:

Y0

Ytop

Y0

Ytop

t,t,t,f

K = (1/2)mv

2

_m

_v

(12)

Correct

Dragging a Board

A uniform board of length and mass lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is , and in region 2, the coefficient is . The positive direction is shown in the figure. Part A Find the net work done by friction in pulling the board directly from region 1 to region 2. Assume that the board moves at constant velocity. Express the net work in terms of , , , , and . Hint 1. The net force of friction Suppose that the right edge of the board is a distance from the boundary, as shown. When the board is at this position, what is the magnitude of the force of friction, , acting on the board (assuming that it's moving)? only only only and and and and and

Ybot

Y0

Ytop

Ybot

Y0

Ytop

Ybot

Y0

Ytop

L

M

μ

1

μ

2

W

M g L μ

₁

μ

₂

x

(x)

Fnet

(13)

Express the force acting on the board in terms of , , , , , and . Hint 1. Fraction of board in region 2 Consider the part of the board in region 2 when the right edge of the board is a distance from the boundary. The magnitude of the force of friction acting on the board (only considering the friction from region 2) will be the coefficient of friction, multiplied by the magnitude of the normal force that acts on the board. Since the ground is horizontal, and the board is not accelerating in the vertical direction, the normal force should equal the board's weight. But, only a fraction of the board's total mass is in region 2. Find the fraction of the board in region 2 in terms of the given lengths; . ANSWER: Hint 2. Force of friction in region 1 Now consider that part of the board in region 1. Again, only a fraction of the board's mass is in region 1. Using this fact, find the magnitude of the force of friction acting on the board, just due to friction in region 1. Express your answer in terms of , , , , and . Hint 1. Fraction of the board in region 1 When the right edge of the board is a distance from the boundary, what fraction of the board lies in region 1? . ANSWER:

M g L x μ

₁

μ

₂

x

fraction of the board in region 2 =

length of the board in region 2_{total length of the board}

Fraction of board in region 2 = x L

M g L x

μ

₁

x

fraction of the board in region 1 =

length of the board in region 1_{total length of the board}

Fraction of board in region 1 = L−x L

(14)

ANSWER:

Hint 2. Work as integral of force

After you find the net force of friction that acts on the board, as a function of , to find the net work done by this force, you will need to perform the appropriate work integral, The lower limit of this integral will be at . What will be the upper limit? ANSWER: Hint 3. Direction of force of friction Don't forget that the force of friction is directed opposite to the direction of the board's motion. Hint 4. Formula for ANSWER:

Correct

This answer makes sense because it is as if the board spent half its time in region 1, and half in region 2, which on average, it in fact did. Part B What is the total work done by the external force in pulling the board from region 1 to region 2? (Again, assume that the board moves at constant velocity.) Express your answer in terms of , , , , and . =

Fregion1

Mg (L−x)μ1 L =

(x)

F

net Mg_L

( (L−x)+ x)

μ

₁

μ

₂

F(x)

x

W = ∫ F(x) dx

x = 0

Upper limit at =

x L

x dx

∫

_ab

x dx =

∫

_ab b2−a2 2 =

W

_L⋅−gM

μ2+μ1 2

M g L μ

1

μ

2

(15)

Hint 1. No acceleration Since the board is not accelerating, the sum of the external forces on it must be zero. Therefore the external force must be oppositely directed to that of friction. ANSWER:

Correct

Exercise 6.20

You throw a 20 rock vertically into the air from ground level. You observe that when it is a height 14.6 above the ground, it is traveling at a speed of 24.1 upward. Part A Use the workenergy theorem to find its speed just as it left the ground. ANSWER:

Correct

Part B Use the workenergy theorem to find its maximum height. ANSWER:

Correct

Work Done by a Spring

Consider a spring, with spring constant , one end of which is attached to a wall. The spring is initially unstretched, with the unconstrained end of the spring at position . =

Wext

_LgM

μ2+μ1 2

N

m

m/s

= 29.4

v0

m/s

= 44.2

h

m

k

x = 0

(16)

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920 16/27 Part A

The spring is now compressed so that the unconstrained end moves from to . Using the work integral ,

find the work done by the spring as it is compressed.

Express the work done by the spring in terms of and .

Hint 1. Spring force as a function of position

The spring force vector as a function of displacement from the spring's equilibrium position, is given by

where is the spring constant and is a unit vector in the direction of the displacement of the spring (in this case, towards the right).

Hint 2. Integrand of the work integral

The work done by the spring is given by the integral of the dot product of the spring force and an infinitesimal displacement of the end of the spring:

,

where the infinitesmal displacement vector has been written as . Write in terms of given quantities, and then compute the dot product to find an expression for the integrand. (Note, .) Express your answer in terms of , , and . ANSWER: Hint 3. Upper limit of the work integral

x = 0 x = L

W =

∫

xf

( )⋅d

x_i

F ⃗ x⃗ x⃗

k

L

F⃗

x

= −kx

F⃗

i^

k

i^

W =

∫

xf

( )⋅d =

(x)⋅ dx

x_i

F ⃗ x⃗ x⃗ ∫

x_f x_i

F ⃗

i^

dx⃗

i^

dx

F⃗

(x)

⋅ = 1

i^ i^

k x

dx

=

(x) ⋅ dx

F⃗

i^

−kxdx

= 0

i

(17)

The lower limit of the work integral will be at . What will be the integral's upper limit? ANSWER:

ANSWER:

Correct

± Holding Force of a Nail

A hammer of mass is moving at speed when it strikes a nail of negligible mass that is stuck in a wooden block. The hammer is observed to drive the nail a distance deeper into the block. Part A Find the magnitude of the force that the wooden block exerts on the nail, assuming that this force is independent of the depth of penetration of the nail into the wood. You may also assume that , so that the change in the hammer's gravitational potential energy, as it drives the nail into the block, is insignificant. Express the magnitude of the force in terms of , , and . Hint 1. How to approach the problem One way to solve this problem is to use the workenergy theorem. To stop the hammer from moving, the wooden blocknail system must do a certain amount of work on the hammer. One expression for this amount of work involves and the displacement of the hammer. In addition, the workenergy theorem implies that the initial kinetic energy of the hammer plus the work done on the hammer must equal the final kinetic energy of the hammer. This gives another expression for the work done that involves only the change in kinetic energy of the hammer. Equate the two expressions for the work done and solve for . Hint 2. Find the work done in terms of The workenergy theorem connects the work needed to stop the hammer with the change in the hammer's kinetic energy. Find the work done on the hammer by the nail. Don't forget to consider the sign of your answer. Express your answer in terms of and . ANSWER: Hint 3. Find the change in kinetic energy of the hammer

= 0

xi

=

xf

L

=

W

_{− k}

1₂

_L

2

M

v0

L

F

≫

v0

√

− −

2gL

−−

M v0

L

F

W

F

L

=

W

−FL

(18)

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920 18/27 What is , the change in kinetic energy of the hammer? Express your answer in terms of and . ANSWER: ANSWER:

Correct

Part B Now evaluate the magnitude of the holding force of the wooden block on the nail by assuming that the force necessary to pull the nail out is the same as that needed to drive it in, which we just derived. Assume a relatively heavy hammer (about 18 ounces), moving with speed . (If such a hammer were swung this hard upward and released, it would rise 5 m). Take the penetration depth to be 2 cm, which is appropriate for one hit on a relatively heavy construction nail. Express your answer to the nearest pound. (Note: .) ANSWER:

Correct

Pulling a Block on an Incline with Friction

A block of weight sits on an inclined plane as shown. A force of magnitude is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is .

− K

Kf

i

M

v0

=

− K

Kf

i

−0.5Mv0

2 =

F

12Mv02 L

M = 0.5 kg

v0

= 10 m/s

L

1 lb = 4.45 N

= 281 lb

| |

F⃗

mg

F

μ

(19)

Part A

What is the total work done on the block by the force of friction as the block moves a distance up the incline? Express the work done by friction in terms of any or all of the variables , , , , , and . Hint 1. How to start Draw a freebody force diagram showing all real forces acting on the block. Hint 2. Find the magnitude of the friction force Write an expression for the magnitude of the friction force. Express your answer in terms of any or all of the variables , , , and . Hint 1. Find the magnitude of the normal force What is the magnitude of the normal force? Express your answer in terms of , , and . ANSWER: ANSWER: ANSWER:

Wfric

L

μ m g θ L

F

Ffric

μ m g

θ

N

m g

θ

=

N

mgcos(θ)

=

Ffric

μmgcos(θ)

(20)

Correct

Part B

What is the total work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER:

Correct

Now the applied force is changed so that instead of pulling the block up the incline, the force pulls the block down the incline at a constant speed. Part C

What is the total work done on the block by the force of friction as the block moves a distance down the incline? Express your answer in terms of any or all of the variables , , , , , and . ANSWER:

Correct

=

Wfric

−(F − mgsin(θ))L

WF

F⃗

L

μ m g θ L

F

=

WF

FL

W

fric

L

μ m g θ L

F

=

Wfric

−(mgsin(θ) + F)L

(21)

Part D What is the total work done on the box by the appled force in this case? Express your answer in terms of any or all of the variables , , , , , and . ANSWER:

Correct

Exercise 6.47

A force in the direction with magnitude is applied to a 7.90 box that is sitting on the horizontal, frictionless surface of a frozen lake. is the only horizontal force on the box. Part A If the box is initially at rest at , what is its speed after it has traveled 17.0 ? Express your answer to three significant figures and include the appropriate units. ANSWER:

Correct

A Car with Constant Power

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph). At full power, the car can accelerate from zero to 30.0 in time 1.20 . Part A At full power, how long would it take for the car to accelerate from 0 to 60.0 ? Neglect friction and air resistance. Express your answer in seconds. Hint 1. Energy and power In the absence of friction, the constant power of the engine implies that the kinetic energy of the car increases linearly with time. Hint 2. Find the ratio of kinetic energies

WF

μ m g θ L

F

=

WF

LF

+x

F(x) = 18.0 N − (0.530 N/m)x

kg

F(x)

x = 0

m

= 7.62

v

m_s

mph

s

mph

(22)

Find the (numerical) ratio of the car's kinetic energy at time 60.0 to , the kinetic energy at time 30.0 . ANSWER: ANSWER:

Correct

Of course, neglecting friction, especially air friction, is completely unrealistic at such speeds. Part B A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 30.0 in time 1.20 , how long would it take to go from zero to 60.0 ? Express your answer numerically, in seconds. Hint 1. How to approach the problem Constant force means constant acceleration. Use this fact to find how the speed increases with time. ANSWER:

Correct

This is probably the first and last time you will come across an imaginary car that goes slower than the real one!

Problem 6.91

A pump is required to lift a mass of 760 of water per minute from a well of depth 14.0 and eject it with a speed of 17.0 . Part A How much work is done per minute in lifting the water?

K1

mph K2

mph

= 4 K1 K₂ 4.80

s

mph

s

mph

2.40

s

kg

m

m/s

(23)

ANSWER:

Correct

Part B How much in giving the water the kinetic energy it has when ejected? ANSWER:

Correct

Part C What must be the power output of the pump? ANSWER:

Correct

Exercise 6.37

A 7.0 box moving at 3.0 on a horizontal, frictionless surface runs into a light spring of force constant 70 .

Part A Use the workenergy theorem to find the maximum compression of the spring. Express your answer using two significant figures. ANSWER:

Correct

Problem 6.77

= 1.04×105

W

J

= 1.10×105

K

J

= 3570

P

W

kg

m/s

N/cm

= 9.5

x

cm

(24)

A block of ice with mass 6.10 is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force to it. As a result, the block moves along the xaxis such that its position as a function of time is given by , where = 0.190 and = 1.95×10−2 .

Part A Calculate the velocity of the object at time = 4.40 . Express your answer to three significant figures. ANSWER:

Correct

Part B Calculate the magnitude of at time = 4.40 . Express your answer to three significant figures. ANSWER:

Correct

Part C

Calculate the work done by the force during the first time interval of 4.40 of the motion. Express your answer to three significant figures. ANSWER:

All attempts used; correct answer displayed

Exercise 6.19

Use the workenergy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. Part A

kg

F⃗

x(t) = α + β

t

2

_t

3

_α

_m/s

2

_β

_m/s

3

t

s

= 2.80

v

m/s

F⃗

t

s

= 5.46

F

N

F⃗

s

= 24.0

W

J

(25)

A branch falls from the top of a 95.0 tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? ANSWER:

Correct

Part B A volcano ejects a boulder directly upward 527 into the air. How fast was the boulder moving just as it left the volcano? ANSWER:

Correct

Part C A skier moving at 5.50 encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. How far does she travel on this patch before stopping? ANSWER:

Correct

Part D Suppose the rough patch in part C was only 2.90 long? How fast would the skier be moving when she reached the end of the patch? ANSWER:

Correct

Part E At the base of a frictionless icy hill that rises at 30.0 above the horizontal, a toboggan has a speed of 12.0

m

= 43.2

v

m/s

m

= 102

v

m/s

= 7.02

s

m

= 4.21

v

m/s

∘

_m/s

(26)

toward the hill. How high vertically above the base will it go before stopping? ANSWER:

Correct

Problem 6.85

A 5.00 block is moving at 6.00 along a frictionless, horizontal surface toward a spring with force constant =500 that is attached to a wall (the figure ). The spring has negligible mass. Part A Find the maximum distance the spring will be compressed. ANSWER:

Correct

Part B If the spring is to compress by no more than 0.350 , what should be the maximum value of ? ANSWER: = 7.35

Mastering Physics: Ch 06

The Work­Energy Theorem

Correct

m

a

vinitial

vfinal

s

s =

xfinal

−

xinitial

a

vinitial

vfinal

s

a =

=

v

s =

vavg

t

t

a

F

m

a

Correct

Correct

Correct

F

ma

W

F

s

W

Fs

F

a

W

W

m vinitial

vfinal

W

m

W = m

− m

.

v

v

m

v

K

= m

Correct

W

Kinitial

Kfinal

W

−

Kinitial

+

Kfinal

K

W = K

W = ⋅ ,

F⃗ s⃗

dx

dW

x x + dx

dW = F dx

F

x

x + dx

W

W =

∫

dW =

F dx

∫

The WorkEnergy Theorem

_m

_m(

_v

₋

_v

₎

_C

_C

_L