Mastering Physics: Ch 06
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4/9/2014
Ch 06 HW
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 m moving in the x direction at constant acceleration a. During a certain interval of time, the particle accelerates from vinitial to vfinal , undergoing displacement s given by s = xfinal − xinitial .
Part A Find the acceleration a of the particle. Express the acceleration in terms of vinitial , vfinal , and s .
Hint 1. Some helpful relationships from kinematics By definition for constant acceleration, a=
vf inal −vinitial t
.
Furthermore, the average speed is vavg =
vinitial +vf inal 2
,
and the displacement is s = v avg t
.
Combine these relationships to eliminate t.
ANSWER: =
a
vf inal
2
− vinitial
2
2s
Correct
Part B Find the net force F acting on the particle. Express your answer in terms of m and a.
Hint 1. Using Newton's laws Which of Newton's laws may be helpful here? http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW
ANSWER: F
=
ma
Correct
Part C Find the net work W done on the particle by the external forces during the particle's motion. Express your answer in terms of F and s . ANSWER: W
=
Fs
Correct
Part D Substitute for F from Part B in the expression for work from Part C. Then substitute for a from the relation in Part A. This will yield an expression for the net work W 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 W in terms of those quantities. Express your answer in terms of m, vinitial , and vfinal . ANSWER: W
=
m
vf inal
2
− vinitial
2
2
Correct The expression that you obtained can be rearranged as W =
The quantity
1 2
2
mv
1 2
m vf inal
2
−
1 2
m vinitial
2
.
has the same units as work. It is called the kinetic energy of the moving particle and is
denoted by K . Therefore, we can write Kinitial =
1 2
mvinitial
2
and Kf inal
=
1 2
mvf inal
2
.
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.
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Ch 06 HW
Part E Find the net work W 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 K initial and K final . ANSWER: W
=
−K initial + K final
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 K initial is zero, then the workenergy theorem reduces to W = K final
.
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, ⃗ ⃗ W = F ⋅ s,
needs to be modified by being replaced by an integral. If the path of the particle is divided into very small displacements dx, we can assume that over each of these small displacement intervals, the net force remains essentially constant and the work dW done to move the particle from x to x + dx is , where F is the x component of the net force (which remains virtually constant for the small displacement from x to x + dx). The net work W done on the particle is then given by dW = F dx
W =∫
Now, using
xf inal xinitial
dW = ∫
xf inal xinitial
F dx .
F = ma
and a=
it can be shown that
dv dt
=
W = ∫
dv
dx
dx
dt
vf inal vinitial
=v
dv dx
,
mv dv.
Part F Evaluate the integral W
= ∫
vf inal vinitial
mv dv.
http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW
Express your answer in terms of m, vinitial , and vfinal .
Hint 1. An integration formula b
The formula for ∫a
t dt is
∫
b a
2
t dt =
b −a
2
2
.
ANSWER: W
=
1 2
m (vf inal
2
− vinitial
2
)
Correct The expression that you havejust obtained is equivalent to W
= K final − K initial
. 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 F (x) = Cx2 , for some constant C . The particle moves from xinitial
= L
to xfinal
= 3L
. What is ΔK , the change in kinetic energy of the particle during
that time? Express your answer in terms of C and L.
Hint 1. Finding the work Integrate F (x) dx to calculate the work done on the particle.
Hint 2. An integration formula b
The formula for ∫a
u
2
du is
∫
b a
u
2
3
du =
b −a 3
3
.
ANSWER: ΔK
=
26 3
CL
3
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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 W = ∫
S f inal S initial
→ ⃗ F ⋅ dL , →
where S initial and S final are the initial and the final positions of the particle, dL is the vector representing a small displacement, and F ⃗ is the net force acting on the particle.
When Push Comes to Shove Two forces, of magnitudes F1 and F2 , act in opposite directions on a block, which sits atop a frictionless surface. Initially, the center of the block is at position xi . At some later time, the block has moved to the right, so that its center is at position xf , where xf > xi .
Part A Find the work W1 done on the block by the force of magnitude F1 as the block moves from xi to xf . 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 W done by a force F ⃗ in producing a displacement s ⃗ is given by ⃗ ⃗ W = F ⋅ s ⃗ = ∣F ∣∣ s ∣⃗ cos ϕ ∣ ∣∣ ∣
,
where ∣∣F ∣∣⃗ and ∣∣s ∣∣⃗ are the magnitudes of F ⃗ and s ⃗ respectively, and ϕ is the smaller angle between the two vectors. The scalar that results from the operation F ⃗ ⋅ s ⃗ is called the scalar product, or dot product, of the vectors F ⃗ and s ⃗ . http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW
ANSWER: W1
=
F1 (xf − xi )
Correct
Part B Find the work W2 done by the force of magnitude F2 as the block moves from xi to xf . 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 F2 acts in the opposite direction to that of the motion of the block. Therefore, the work done by that force must be negative. ANSWER: W2
=
−F2 (xf − xi )
Correct
Part C What is the net work Wnet done on the block by the two forces? ANSWER: Wnet
=
(F1 − F2 )(xf − xi )
Correct
Part D Imagine that the two forces are equal in magnitude, F1 on the block. Determine the changeK f
− Ki
= F2
, and that there are no other horizontal forces acting
in the kinetic energy of the block as it moves from xi to xf .
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.
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Ch 06 HW
ANSWER: Kf − Ki
= 0
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 F (⃗ r )⃗ is ⃗ independent of r . 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 F ⃗ . 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 F , and the displacement vector from point B to point A is L⃗ (of magnitude L, making an angle θ (radians) with the positive x axis). Find WBA , the work that the force F ⃗ performs on the particle as it moves from point B to point A. Express the work in terms of L, F , and θ . 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 F ⃗ along a straight path represented by the displacement vector ⃗ L
, the net work done by F ⃗ is F ⃗ ⋅ L⃗ .
Hint 2. Find the angle between F ⃗ and L⃗ You need to find the angle between the vector F ⃗ , which is directed horizontally to the left, and the vector L⃗ in the direction of the particle's motion (at an angle θ (radians) relative to the positive x axis). It may help to visualize F ⃗ directed along the negative x axis at the origin. What is the angle ϕ between F ⃗ and L⃗ ? Express your answer in radians, not degrees. ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW ϕ
=
π−θ
ANSWER: WBA
=
−F cos(θ)L
Correct This result is worth remembering! The work done by a constant force of magnitude F , which acts at an angle of ϕ with respect to the direction of motion along a straight path of length L, is WBA = F L cos(ϕ) . This equation correctly gives the sign in this problem. Since θ is the angle with respect to the positive x axis (in radians), ϕ = π − θ ; hence cos(ϕ) = cos(π − θ) = − cos(θ).
Part B Now consider the same force F ⃗ acting on a particle that travels from point A to point B. The displacement vector L⃗ now points in the opposite direction as it did in Part A. Find the work WAB done by F ⃗ in this case. Express your answer in terms of L, F , and θ .
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, WBA + WAB = 0.
ANSWER: WAB
=
F Lcos(θ)
Correct http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Exercise 6.4 A factory worker pushes a 31.5kg crate a distance of 4.2m 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: F
= 100 N
Correct
Part B How much work is done on the crate by this force when the crate is pushed a distance of 4.2m ? Express your answer using two significant figures. ANSWER: W
= 380 J
Correct
Part C How much work is done on the crate by friction during this displacement? Express your answer using two significant figures. ANSWER: Wf
= 380 J
Correct
Part D How much work is done by the normal force? ANSWER: Wnf
= 0 J
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Ch 06 HW
Correct
Part E How much work is done by gravity? ANSWER: Wg
= 0 J
Correct
Part F What is the total work done on the crate? ANSWER: Wnet
= 0 J
Correct
Exercise 6.6 Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.7×106N , one an angle 15∘ west of north and the other an angle 15∘ east of north, as they pull the tanker a distance 0.86km toward the north.
Part A What is the total work they do on the supertanker? Express your answer using two significant figures. ANSWER: W
= 2.8×109 J
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 Ybot and released. The point marked Y0 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 Ytop . Y0 http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW
For most situations, including this problem, the point Y0 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 Y0 when it was still in contact with the spring. The speed of the ball was decreasing on its way from point Y0 to point Ytop . The speed of the ball was zero at point Ytop . The speed of the ball was the same for all points in its motion between points Y0 and Ytop . Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,f,t). ANSWER: t,t,t,f
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 K
= (1/2)mv
2
, where m is the mass of the object and v is its speed.
ANSWER:
http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW
Ybot Y0
only
only
Ytop
only
Ybot
and Y0
Y0
and Ytop
Ytop
and Ybot
Ybot
and Y0 and Ytop
Correct
Dragging a Board A uniform board of length L and mass M lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is μ1 , and in region 2, the coefficient is μ2 . The positive direction is shown in the figure.
Part A Find the net work W 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 M , g, L, μ1 , and μ2 .
Hint 1. The net force of friction Suppose that the right edge of the board is a distance x from the boundary, as shown. When the board is at this position, what is the magnitude of the force of friction, Fnet (x), acting on the board (assuming that it's moving)?
http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW
Express the force acting on the board in terms of M , g, L, x, μ1 , and μ2 .
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 x 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; f raction of the board in region 2 =
length of the board in region 2 total length of the board
.
ANSWER: Fraction of board in region 2 =
x L
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 M , g, L, x, and μ1 .
Hint 1. Fraction of the board in region 1 When the right edge of the board is a distance x from the boundary, what fraction of the board lies in region 1? f raction of the board in region 1 =
length of the board in region 1 total length of the board
.
ANSWER: Fraction of board in region 1 =
L−x L
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Ch 06 HW
ANSWER: Fregion1
=
M g μ1(L−x) L
ANSWER: Fnet (x)
=
Mg L
( μ1 ( L − x) + μ2 x)
Hint 2. Work as integral of force After you find the net force of friction F (x) that acts on the board, as a function of x, to find the net work done by this force, you will need to perform the appropriate work integral, W = ∫ F (x) dx
The lower limit of this integral will be at x = 0. What will be the upper limit? ANSWER: Upper limit at x =
L
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 ∫
b
a
x dx
∫
b a
2
x dx =
b −a
2
2
ANSWER: W
=
L⋅ − gM
μ 2 +μ 1 2
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 M , g, L, μ1 , and μ2 . http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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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: Wext
=
LgM
μ +μ 2
1
2
Correct
Exercise 6.20 You throw a 20N rock vertically into the air from ground level. You observe that when it is a height 14.6m above the ground, it is traveling at a speed of 24.1m/s upward.
Part A Use the workenergy theorem to find its speed just as it left the ground. ANSWER: v0
= 29.4 m/s
Correct
Part B Use the workenergy theorem to find its maximum height. ANSWER: = 44.2 m
h
Correct
Work Done by a Spring Consider a spring, with spring constant k, one end of which is attached to a wall. The spring is initially unstretched, with the unconstrained end of the spring at position x = 0.
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Part A The spring is now compressed so that the unconstrained end moves from x = 0 to x = L. Using the work integral W = ∫
xf xi
⃗ F (x⃗ ) ⋅ d x⃗ ,
find the work done by the spring as it is compressed. Express the work done by the spring in terms of k and L.
Hint 1. Spring force as a function of position The spring force vector F ⃗ as a function of displacement x from the spring's equilibrium position, is given by ⃗ ^ F = −kx i
where k is the spring constant and ^ i 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: W = ∫
xf xi
⃗ F (x⃗ ) ⋅ d x⃗ = ∫
xf xi
⃗ ^ F (x) ⋅ i dx ,
⃗ where the infinitesmal displacement vector dx⃗ has been written as ^ i dx . Write F (x) in terms of given
quantities, and then compute the dot product to find an expression for the integrand. (Note, ^ i ⋅^ i
.)
= 1
Express your answer in terms of k, x, and dx. ANSWER: ⃗ ^ F (x) ⋅ i dx
=
−kxdx
Hint 3. Upper limit of the work integral http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920 i = 0
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Ch 06 HW
The lower limit of the work integral will be at xi
= 0
. What will be the integral's upper limit?
ANSWER: xf
=
L
ANSWER: W
=
−
1 2
kL
2
Correct
± Holding Force of a Nail A hammer of mass M is moving at speed v0 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 L deeper into the block.
Part A Find the magnitude F 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 v0 ≫ √2gL , 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 M , v0 , and L.
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 F 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 F .
Hint 2. Find the work done in terms of F The workenergy theorem connects the work needed to stop the hammer with the change in the hammer's kinetic energy. Find the work W done on the hammer by the nail. Don't forget to consider the sign of your answer. Express your answer in terms of F and L. ANSWER: W
=
−F L
Hint 3. Find the change in kinetic energy of the hammer Kf − K i http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW
What is K f
− Ki
, the change in kinetic energy of the hammer?
Express your answer in terms of M and v0 . ANSWER: Kf − K i
=
−0.5M v0
2
ANSWER: 1
F
=
2
M v0
2
L
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 M = 0.5 kg hammer (about 18 ounces), moving with speed v0 = 10 m/s. (If such a hammer were swung this hard upward and released, it would rise 5 m). Take the penetration depth L to be 2 cm, which is appropriate for one hit on a relatively heavy construction nail. Express your answer to the nearest pound. (Note: 1
lb = 4.45 N
.)
ANSWER: ⃗ |F |
= 281 lb
Correct
Pulling a Block on an Incline with Friction A block of weight mg sits on an inclined plane as shown. A force of magnitude F is applied to pull the block up the incline at constant speed. The coefficient of kinetic friction between the plane and the block is μ.
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Ch 06 HW
Part A What is the total work Wfric done on the block by the force of friction as the block moves a distance L up the incline? Express the work done by friction in terms of any or all of the variables μ, m, g, θ , L, and F .
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 Ffric of the friction force. Express your answer in terms of any or all of the variables μ, m, g, and θ .
Hint 1. Find the magnitude of the normal force What is the magnitude N of the normal force? Express your answer in terms of m, g, and θ . ANSWER: N
=
mgcos(θ)
ANSWER: Ffric
=
μmgcos(θ)
ANSWER: http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW Wfric
=
−(F − mgsin(θ))L
Correct
Part B What is the total work WF done on the block by the applied force F ⃗ as the block moves a distance L up the incline? Express your answer in terms of any or all of the variables μ, m, g, θ , L, and F . ANSWER: WF
=
FL
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 Wfric done on the block by the force of friction as the block moves a distance L down the incline? Express your answer in terms of any or all of the variables μ, m, g, θ , L, and F . ANSWER: Wfric
=
−(mgsin(θ) + F )L
Correct
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Ch 06 HW
Part D What is the total work WF done on the box by the appled force in this case? Express your answer in terms of any or all of the variables μ, m, g, θ , L, and F . ANSWER: WF
=
LF
Correct
Exercise 6.47 A force in the +xdirection with magnitude F (x) = 18.0 N − (0.530 N/m)x is applied to a 7.90kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F (x) is the only horizontal force on the box.
Part A If the box is initially at rest at x = 0, what is its speed after it has traveled 17.0m ? Express your answer to three significant figures and include the appropriate units. ANSWER: v
= 7.62
m s
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 30.0mph in time 1.20s .
Part A At full power, how long would it take for the car to accelerate from 0 to 60.0mph ? 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 K1 http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
mph
K2
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Ch 06 HW
Find the (numerical) ratio of the car's kinetic energy K 1 at time 60.0mph to K 2 , the kinetic energy at time 30.0mph . ANSWER: K1 K2
= 4
ANSWER: 4.80 s
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 30.0mph in time 1.20s , how long would it take to go from zero to 60.0mph ? 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: 2.40 s
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 760kg of water per minute from a well of depth 14.0m and eject it with a speed of 17.0m/s .
Part A How much work is done per minute in lifting the water? http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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ANSWER: W
= 1.04×105 J
Correct
Part B How much in giving the water the kinetic energy it has when ejected? ANSWER: K
= 1.10×105 J
Correct
Part C What must be the power output of the pump? ANSWER: P
= 3570 W
Correct
Exercise 6.37 A 7.0kg box moving at 3.0m/s on a horizontal, frictionless surface runs into a light spring of force constant 70N/cm .
Part A Use the workenergy theorem to find the maximum compression of the spring. Express your answer using two significant figures. ANSWER: = 9.5 cm
x
Correct
Problem 6.77 kg http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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A block of ice with mass 6.10kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force F ⃗ to it. As a result, the block moves along the xaxis such that its position as a function of time is given by x(t) = αt2 + βt3 , where α = 0.190m/s2 and β = 1.95×10−2m/s3 .
Part A Calculate the velocity of the object at time t = 4.40s . Express your answer to three significant figures. ANSWER: v
= 2.80 m/s
Correct
Part B Calculate the magnitude of F ⃗ at time t = 4.40s . Express your answer to three significant figures. ANSWER: F
= 5.46 N
Correct
Part C Calculate the work done by the force F ⃗ during the first time interval of 4.40s of the motion. Express your answer to three significant figures. ANSWER: W
= 24.0 J
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 http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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A branch falls from the top of a 95.0mtall redwood tree, starting from rest. How fast is it moving when it reaches the ground? ANSWER: v
= 43.2 m/s
Correct
Part B A volcano ejects a boulder directly upward 527m into the air. How fast was the boulder moving just as it left the volcano? ANSWER: v
= 102 m/s
Correct
Part C A skier moving at 5.50m/s 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: s
= 7.02 m
Correct
Part D Suppose the rough patch in part C was only 2.90m long? How fast would the skier be moving when she reached the end of the patch? ANSWER: v
= 4.21 m/s
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/s http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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toward the hill. How high vertically above the base will it go before stopping? ANSWER: = 7.35 m
h
Correct
Problem 6.85 A 5.00kg block is moving at 6.00 m/s along a frictionless, horizontal surface toward a spring with force constant k=500 N/m 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: = 0.600 m
x
Correct
Part B If the spring is to compress by no more than 0.350m , what should be the maximum value of v0 ? ANSWER: v0
= 3.50 m/s
http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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Ch 06 HW
Correct Score Summary: Your score on this assignment is 96.5%. You received 30.88 out of a possible total of 32 points.
http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2832920
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