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Chapter 6
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Chapter 6
Due: 11:59pm on Sunday, February 26, 2017
To understand how points are awarded, read the Grading Policy for this assignment.
Work­Energy Theorem Reviewed
Description: Simple questions about the work­energy theorem to test understanding, then one easy analytic question
applying the theorem.
Learning Goal:
Review the work­energy theorem and apply it to a simple problem.
If you push a particle of mass M in the direction in which it is already moving, you expect the particle's speed to increase. If
you push with a constant force F , then the particle will accelerate with acceleration a = F /M (from Newton's 2nd law).
Part A
Enter a one­ or two­word answer that correctly completes the following statement.
If the constant force is applied for a fixed interval of time t, then the _____ of the particle will increase by an amount at.
Hint 1. Kinematic equations recalled
Recall the kinematic equations ­
v = u + at ;
s = ut +
2
v
= u
2
1
2
at
+ 2as
2
; .
ANSWER:
speed
Also accepted: velocity, v
Part B
Enter a one­ or two­word answer that correctly completes the following statement.
If the constant force is applied over a given distance D, along the path of the particle, then the _____ of the particle will
increase by F D.
ANSWER:
kinetic energy
Also accepted: energy, KE, K
The work W done on the particle by the force F over the distance D is F D.
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Part C
If the initial kinetic energy of the particle is Ki , and its final kinetic energy is Kf , express Kf in terms of Ki and the work W done on the particle.
ANSWER:
Kf
= This is the work­energy theorem, often written W = Kf − Ki . It is, essentially, a statement of energy conservation
that does not include potential energy explicitly. All forces­­even conservative forces like gravity­­contribute to the
work.
Part D
⃗ In general, the work done by a force F is written as
W = ∫
i
f
⃗ ⃗ ⋅ dr ⃗ .
F (r )
Now, consider whether the following statements are true or false:
The dot product assures that the integrand is always nonnegative.
The dot product indicates that only the component of the force perpendicular to the path contributes to the
integral.
The dot product indicates that only the component of the force parallel to the path contributes to the integral.
Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,t).
ANSWER:
f,f,t
Part E
Assume that the particle has initial speed vi . Find its final kinetic energy Kf in terms of vi , M , F , and D.
Hint 1. Find the initial kinetic energy
Express the initial kinetic energy Ki in terms of the particle's initial velocity vi and its mass M .
ANSWER:
Ki
= ANSWER:
Kf
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Part F
What is the final speed of the particle?
Express your answer in terms of Kf and M .
ANSWER:
vf
= Work on a Sliding Block
Description: A box is pushed up a frictionless incline. Find the work done by gravity, the pushing force, and the normal
force.
A block of weight w sits on a frictionless inclined plane, which makes an angle θ with respect to the horizontal, as shown. A
force of magnitude F , applied parallel to the incline, pulls the block
up the plane at constant speed.
Part A
The block moves a distance L up the incline. The block does not stop after moving this distance but continues to move
with constant speed. What is the total work Wtot done on the block by all forces? (Include only the work done after the
block has started moving, not the work needed to start the block moving from rest.)
Express your answer in terms of given quantities.
Hint 1. What physical principle to use
To find the total work done on the block, use the work­energy theorem:
Wtot = Kf − Ki .
Hint 2. Find the change in kinetic energy
What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled a
distance L? Remember that the block is pulled at constant speed.
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Hint 1. Consider kinetic energy
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If the block's speed does not change, its kinetic energy cannot change.
ANSWER:
Kf − Ki
= 0
ANSWER:
Wtot
= 0
Part B
What is Wg , the work done on the block by the force of gravity as the block moves a distance L up the incline?
Express the work done by gravity in terms of the weight w and any other quantities given in the problem
introduction.
Hint 1. Force diagram
Hint 2. Force of gravity component
What is the component of the force of gravity in the direction of the block's displacement (along the inclined
plane)?
Express your answer in terms of w and θ.
Hint 1. Relative direction of the force and the motion
Remember that the force of gravity acts down the plane, whereas the block's displacement is directed up
the plane.
ANSWER:
F g||
= Typesetting math: 95%
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ANSWER:
Wg
= Also accepted: Part C
What is WF , the work 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 F and other given quantities.
Hint 1. How to find the work done by a constant force
Remember that the work done on an object by a particular force is the integral of the dot product of the force and
the instantaneous displacement of the object, over the path followed by the object. In this case, since the force is
constant and the path is a straight segment of length L up the inclined plane, the dot product becomes simple
multiplication.
ANSWER:
WF
= Part D
What is Wnormal , the work done on the block by the normal force as the block moves a distance L up the inclined plane?
Express your answer in terms of given quantities.
Hint 1. First step in computing the work
The work done by the normal force is equal to the dot product of the force vector and the block's displacement
vector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dot
product?
ANSWER:
= 0
⃗ ⃗ N ⋅L
ANSWER:
Wnormal
= 0
± Baby Bounce with a Hooke
Description: ± Includes Math Remediation. Application of Hooke's Law
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One of the pioneers of modern science, Robert Hooke (1635­1703), studied the elastic properties of springs and formulated
⃗ the law that bears his name. Hooke found the relationship among the force a spring exerts, F , the distance from equilibrium
the end of the spring is displaced, x⃗ , and a number k called the spring constant (or, sometimes, the force constant of the
spring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium, or
⃗ F = −kx⃗ .
In its scalar form, this equation is simply
F = kx
.
Where F is the force exerted on the spring and x is the extension of the spring from equilibrium caused by the force F . The
value of k depends on the geometry and the material of the spring; it can be easily determined experimentally using this
scalar equation.
Toy makers have always been interested in springs for the entertainment value of the motion they produce. One well­known
application shown in is a baby bouncer, which consists of a harness
seat for a toddler, attached to a spring. The entire contraption hooks
onto the top of a doorway. The idea is for the baby to hang in the
seat with his or her feet just touching the ground so that a good
push up will get the baby bouncing, providing potentially hours of
entertainment.
Part A
The following chart and accompanying graph shown in depict
an experiment to determine the spring constant for a baby
bouncer.
Displacement from
equilibrium, x (m)
Force exerted on
the spring, F (N)
0
0
0.005
2.5
0.010
5.0
0.015
7.5
0.020
10
What is the spring constant k of the spring being tested for the
baby bouncer?
Express your answer to two significant figures in newtons per meter.
Hint 1. How to approach the problem
Look at the pattern in the data to determine what number must multiply the distance to achieve the force exerted
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on the spring. Look at both the table and the graph.
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Hint 2. Find the spring constant from the graph
The relationship between the displacement and force is linear. This set of data follows the form of y = mx + b ,
where m is the slope of the line and b is the y intercept. For all springs, the force is 0 when the displacement is 0
so b = 0. This leaves the slope of the line to determine the relationship between displacement and force. What is
the slope that you get from the graph?
Express your answer as a fraction in unsimplified form.
Hint 1. Slope equation
Slope is given by the change in y divided by the change in x. In this case, m
.
= ΔF /Δx
ANSWER:
= 500 N/m k
All you need to do now is to convert the fraction to its decimal value.
ANSWER:
= 500 N/m k
Part B
One of the greatest difficulties with setting up the baby bouncer is determining the right height above the floor so that the
child can push off and bounce. Knowledge of physics can be really helpful here.
If the spring constant k = 5.0 × 102 N/m , the baby has a mass m = 11 kg, and the baby's legs reach a distance d = 0.15 m from the bouncer, what should be the height h of the "empty" bouncer above the floor?
Express your answer in meters to two significant figures.
Hint 1. How to approach the problem
Use Hooke's law to determine the displacement x of the spring from equilibrium given the force the spring must
exert to hold up the baby. The displacement must lower the baby toward the floor until the baby's feet can touch.
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Hint 2. Which force to use
The force the spring exerts is equal in magnitude but opposite in direction to the force exerted on it by the weight w
of the baby.
Hint 3. Find the force exerted by the baby
The weight of the baby is equal to the force exerted on the spring. What is the weight of the baby?
Express your answer in newtons to three significant figures.
Hint 1. Formula for weight
Recall that the weight of an object is given by
w = mg ,
where m is the mass of the object and g is the acceleration due to gravity.
ANSWER:
= 108 N w
Hint 4. Find the displacement of the spring
Use Hooke's law to determine how far the spring would stretch downward once the baby is placed in the seat.
How far does the bottom end of the spring move?
Express your answer in meters to two significant figures.
ANSWER:
x
= ­0.22 m To finish the problem, you must consider the length of the baby's legs.
ANSWER:
h
= 0.37 m A displacement of −0.22 m for the spring holding up a baby may not seem very large but you must consider how
small babies are. Also, once the baby begins jumping up and down, the extra energy allows the spring to stretch
further than 0.22 m and a resonant frequency may be achieved. At resonance the bouncing may become too violent,
leading to a potentially dangerous situation for the little bouncer.
Exercise 6.10
Description: An ##­kg package in a mail­sorting room slides ## m down a chute that is inclined at ## degree(s) below
the horizontal. The coefficient of kinetic friction between the package and the chute's surface is ##. (a) Calculate the...
An 11.0­kg package in a mail­sorting room slides 3.00 m down a chute that is inclined at 54.0 ∘ below the horizontal. The
coefficient of kinetic friction between the package and the chute's surface is 0.10.
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Part A
Calculate the work done on the package by friction.
Express your answer with the appropriate units.
ANSWER:
Wf riction
= ­19
= Also accepted: = ­19.0 , = ­19.0 , = ­19
Part B
Calculate the work done on the package by gravity.
Express your answer with the appropriate units.
ANSWER:
Wgravity
= Also accepted: = 260
= 262 , = 262 , = 260
Part C
Calculate the work done on the package by the normal force.
Express your answer with the appropriate units.
ANSWER:
WN
= 0
Part D
What is the net work done on the package?
Express your answer with the appropriate units.
ANSWER:
Wnet
= 240
= Also accepted: = 243 , = 243 , = 240
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Exercise 6.20
Description: A m­kg watermelon is dropped from rest from the roof of a s­m­tall building and feels no appreciable air
resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground.
(b) Just...
A 4.11­kg watermelon is dropped from rest from the roof of a 19.9­m­tall building and feels no appreciable air resistance.
Part A
Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground.
Express your answer with the appropriate units.
ANSWER:
W
= 802
= Also accepted: = 802 , = 802
Part B
Just before it strikes the ground, what is the watermelon's kinetic energy?
Express your answer with the appropriate units.
ANSWER:
K
= 802
= Also accepted: = 802 , = 802
Part C
Just before it strikes the ground, what is the watermelon's speed?
Express your answer with the appropriate units.
ANSWER:
= 19.7
= v
Also accepted: = 19.8
, = 19.7
Part D
Would the answer in part A be different if there were appreciable air resistance?
ANSWER:
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yes
no
The work done by gravity would be the same.
Part E
Would the answer in part B be different if there were appreciable air resistance?
ANSWER:
yes
no
Air resistance would do negative work and Wtot would be less than Wgrav .
Part F
Would the answer in part C be different if there were appreciable air resistance?
ANSWER:
yes
no
Air resistance would do negative work and Wtot would be less than Wgrav .
Exercise 6.28
Description: A soccer ball with mass m is initially moving with speed vi. A soccer player kicks the ball, exerting a
constant force of magnitude F in the same direction as the ball's motion. (a) Over what distance must the player's foot be
in contact with the...
A soccer ball with mass 0.420 kg is initially moving with speed 2.10 m/s . A soccer player kicks the ball, exerting a constant
force of magnitude 45.0 N in the same direction as the ball's motion.
Part A
Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 {\rm{ m/s}}?
ANSWER:
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= 0.147 \rm m s = Exercise 6.34
Description: To stretch a spring l from its unstretched length, W of work must be done. (a) What is the force constant of
this spring? (b) What magnitude force is needed to stretch the spring l from its unstretched length? (c) How much work
must be done to...
To stretch a spring 5.00 {\rm cm} from its unstretched length, 12.0 {\rm J} of work must be done.
Part A
What is the force constant of this spring?
ANSWER:
k = = 9600 \rm N/m If you need to use the value of the spring constant 'k' in subsequent parts, please use the unrounded full precision
value and not the one you submitted for this part rounded using three significant figures.
Part B
What magnitude force is needed to stretch the spring 5.00 {\rm cm} from its unstretched length?
ANSWER:
F = = 480 \rm N Part C
How much work must be done to compress this spring 4.00 {\rm cm} from its unstretched length?
ANSWER:
W = = 7.68 \rm J Part D
What force is needed to compress it this distance?
ANSWER:
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F = = 384 \rm N Exercise 6.56
Description: When its engine of power P is generating full power, a small single­engine airplane with mass m gains
altitude at a rate of v. (a) What fraction of the engine power is being used to make the airplane climb? (The remainder is
used to overcome the...
When its engine of power 71 {\rm kW} is generating full power, a small single­engine airplane with mass 720 {\rm kg} gains
altitude at a rate of 2.6 {\rm m/s} .
Part A
What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the
effects of air resistance and of inefficiencies in the propeller and engine.)
Express your answer as a percentage using two significant figures.
ANSWER:
= 26 \% Alternative Exercise 6.113
Description: The space shuttle Endeavour, with mass 86400 kg, is in a circular orbit of radius 6.66* 10^6 ( m) around the
earth. It takes 90.1 min for the shuttle to complete each orbit. On a repair mission, the shuttle is cautiously moving x
closer to a disabled...
The space shuttle Endeavour, with mass 86400 {\rm kg}, is in a circular orbit of radius 6.66\times 10^6 {\rm{ m}} around the
earth. It takes 90.1 {\rm min} for the shuttle to complete each orbit. On a repair mission, the shuttle is cautiously moving 4.50
{\rm m} closer to a disabled satellite every 13.5 {\rm s} .
Part A
Calculate the shuttle's kinetic energy relative to the earth.
ANSWER:
K = 2.59×1012 \rm J Part B
Calculate the shuttle's kinetic energy relative to the satellite.
ANSWER:
K = = 4800 \rm J Typesetting math: 95%
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Problem 6.66
Description: A m­kg package slides s down a long ramp that is inclined at theta below the horizontal. The coefficient of
kinetic friction between the package and the ramp is mu_k=##. (a) Calculate the work done on the package by friction. (...
A 7.00­\rm kg package slides 2.60 {\rm \; m} down a long ramp that is inclined at 26.0 {\rm ^\circ} below the horizontal. The
coefficient of kinetic friction between the package and the ramp is \texttip{\mu _{\rm k}}{mu_k} = 0.310.
Part A
Calculate the work done on the package by friction.
Express your answer with the appropriate units.
ANSWER:
= ­49.7
W_{\rm f \hspace{1 pt}} = Also accepted: = ­49.7 , = ­49.7
Part B
Calculate the work done on the package by gravity.
Express your answer with the appropriate units.
ANSWER:
W_{\rm g} = Also accepted: = 78.2
= 78.3 , = 78.2
Part C
Calculate the work done on the package by the normal force.
Express your answer with the appropriate units.
ANSWER:
W_{\rm n} = 0
Part D
Calculate the total work done on the package.
Express your answer with the appropriate units.
ANSWER:
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W_{\rm net} = Also accepted: = 28.5
= 28.5 , = 28.5
Part E
If the package has a speed of 2.30 {\rm \; m/s} at the top of the ramp, what is its speed after it has slid 2.60 {\rm \; m}
down the ramp?
Express your answer with the appropriate units.
ANSWER:
= 3.66
v = Also accepted: = 3.67
, = 3.66
Problem 6.71
Description: A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless,
horizontal surface . The block is originally revolving at a distance of r_0 from the hole with a speed of v_0. The cord is
then...
A small block with a mass of 0.0600 \rm kg is attached to a cord passing through a hole in a frictionless, horizontal surface .
The block is originally revolving at a distance of 0.37 {\rm \; m} from
the hole with a speed of 0.77 {\rm \; m/s} . The cord is then pulled
from below, shortening the radius of the circle in which the block
revolves to 0.19 {\rm \; m} . At this new distance, the speed of the
block is 1.50 {\rm \; m/s} .
Part A
What is the tension in the cord in the original situation when the block has speed
\texttip{v_{\rm 0}}{v_0} = 0.77 {\rm \; m/s} ?
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Express your answer with the appropriate units.
ANSWER:
T = = 9.6×10−2
Also accepted: = 9.61×10−2 , = 9.6×10−2
Part B
What is the tension in the cord in the final situation when the block has speed \texttip{v_{\rm 1}}{v_1} = 1.50 {\rm \; m/s} ?
Express your answer with the appropriate units.
ANSWER:
T = = 0.71
Also accepted: = 0.710 , = 0.71
Part C
How much work was done by the person who pulled on the cord?
Express your answer with the appropriate units.
ANSWER:
W = Also accepted: = 5.0×10−2
= 4.97×10−2 , = 5.0×10−2
Alternative Exercise 6.116
Description: A block of ice with mass m is initially at rest on a frictionless, horizontal surface. A worker then applies a
horizontal force F_vec to it. As a result, the block moves along the x­axis such that its position as a function of time is
given by...
A block of ice with mass 5.70 {\rm kg} is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal
force \vec{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)=
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\alpha t^2+\beta t^3, where \texttip{\alpha }{alpha} = 0.200 {\rm m/s^2} and \texttip{\beta }{beta} = 2.01×10−2 {\rm m/s^3} .
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Part A
Calculate the velocity of the object at time \texttip{t}{t} = 4.20 {\rm s} .
Express your answer to three significant figures.
ANSWER:
v = = 2.74 {\rm m/s} Part B
Calculate the magnitude of \vec{F} at time \texttip{t}{t} = 4.20 {\rm s} .
Express your answer to three significant figures.
ANSWER:
F = = 5.17 {\rm N} Part C
Calculate the work done by the force \vec{F} during the first time interval of 4.20 {\rm s} of the motion.
Express your answer to three significant figures.
ANSWER:
W = = 21.5 {\rm J} Problem 6.76
Description: The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed
6.00 cm and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring is
then released,...
The spring of a spring gun has force constant k = 400 {\rm N/m} and negligible mass. The spring is compressed 6.00 {\rm cm}
and a ball with mass 0.0300 {\rm kg} is placed in the horizontal barrel against the compressed spring. The spring is then
released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 {\rm cm} long, so the ball leaves the barrel at
the same point that it loses contact with the spring. The gun is held so the barrel is horizontal.
Part A
Calculate the speed with which the ball leaves the barrel if you can ignore friction.
ANSWER:
v = 6.93 {\rm m/s} Typesetting math: 95%
Part B
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Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 {\rm N} acts on the ball as it
moves along the barrel.
ANSWER:
v = 4.90 {\rm m/s} Part C
For the situation in part B, at what position along the barrel does the ball have the greatest speed? (In this case, the
maximum speed does not occur at the end of the barrel.)
ANSWER:
x = 4.50×10−2 {\rm m} from start point Part D
What is that greatest speed?
ANSWER:
v_{\rm max} = 5.20 {\rm m/s} Problem 6.82
Description: Consider the system shown in the figure. The rope and pulley have negligible mass, and the pulley is
frictionless. The coefficient of kinetic friction between the 8.00­kg block and the tabletop is mu_k=0.250. The blocks are
released from rest. (a)...
Consider the system shown in the figure. The rope and pulley have negligible mass, and the pulley is frictionless. The
coefficient of kinetic friction between the 8.00­kg block and the tabletop is \mu_k=0.250. The blocks are released from rest.
Part A
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Use energy methods to calculate the speed of the 6.00­kg block after it has descended 1.50 {\rm m}.
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ANSWER:
v = 2.90 {\rm m/s} Problem 6.81
Description: Consider the system shown in the figure . The rope and pulley have negligible mass, and the pulley is
frictionless. Initially the 6.00­kg block is moving downward and the 8.00­kg block is moving to the right, both with a speed
of v. The blocks come to ...
Consider the system shown in the figure . The rope and pulley have
negligible mass, and the pulley is frictionless. Initially the 6.00­{\rm
kg} block is moving downward and the 8.00­{\rm kg} block is moving
to the right, both with a speed of 0.400 {\rm m/s} . The blocks come
to rest after moving 2.00 {\rm m} .
Part A
Use the work­energy theorem to calculate the coefficient of kinetic friction between the 8.00­{\rm kg} block and the
tabletop.
ANSWER:
\mu_{\rm k} = = 0.757
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