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Dynamics 2e 747
Problem 4.62
While the stiffness of an elastic cord can be nearly constant over a large range
of deformation, as a bungee cord is stretched, it tends to get less stiff as it gets
longer. Assume a softening force-displacement relation of the form
kı ˇı3
,
where
kD2:58 lb=ft
,
ˇD0:000013 lb=ft3
, and
ı
(measured in ft) is the
displacement of the cord from its unstretched length. For a bungee cord whose
unstretched length is 150 ft, determine
(a) the expression of the cord’s potential energy as a function of ı;
(b)
the velocity at the bottom of a
400 ft
tower of a bungee jumper weighing
170 lb and starting from rest;
(c)
the maximum acceleration, expressed in
g
’s, felt by the bungee jumper in
question.
Solution
Part (a).
To find the potential energy of the cord we need to find the expres-
sion of the work done by the force of the cord as. To do so, it is sufficient to
748 Solutions Manual
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Dynamics 2e 749
Problem 4.63
A crate, initially traveling horizontally with a speed of
18 ft=s
,
is made to slide down a
14 ft
chute inclined at
35ı
. The surface
of the chute has a coefficient of kinetic friction
k
, and at its
lower end, it smoothly lets the crate onto a horizontal trajectory.
The horizontal surface at the end of the chute has a coefficient of
kinetic friction
k2
. Model the crate as a particle, and assume that
gravity and the contact forces between the crate and the sliding
surface are the only relevant forces.
If kD0:35, what is the speed with which the crate reaches
the bottom of the chute (immediately before the crate’s trajectory
becomes horizontal)?
Solution
750 Solutions Manual
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Dynamics 2e 751
Problem 4.64
A crate, initially traveling horizontally with a speed of
18 ft=s
,
is made to slide down a
14 ft
chute inclined at
35ı
. The surface
of the chute has a coefficient of kinetic friction
k
, and at its
lower end, it smoothly lets the crate onto a horizontal trajectory.
The horizontal surface at the end of the chute has a coefficient of
kinetic friction
k2
. Model the crate as a particle, and assume that
gravity and the contact forces between the crate and the sliding
surface are the only relevant forces.
Find
k
such that the crate’s speed at the bottom of the chute
(immediately before the crate’s trajectory becomes horizontal) is
15 ft=s.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
752 Solutions Manual
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Dynamics 2e 753
Problem 4.65
A crate, initially traveling horizontally with a speed of
18 ft=s
,
is made to slide down a
14 ft
chute inclined at
35ı
. The surface
of the chute has a coefficient of kinetic friction
k
, and at its
lower end, it smoothly lets the crate onto a horizontal trajectory.
The horizontal surface at the end of the chute has a coefficient of
kinetic friction
k2
. Model the crate as a particle, and assume that
gravity and the contact forces between the crate and the sliding
surface are the only relevant forces.
Let
kD0:5
and suppose that, once the crate reaches the
bottom of the chute and after sliding horizontally for
5ft
, the
crate runs into a bumper. If the weight of the crate is
WD110 lb
,
k2 D0:33
, modeling the bumper as a linear spring with constant
k
, and neglecting the mass of the bumper, determine the value of
k
so that the crate comes to a stop
2ft
after impacting with the
bumper.
Solution
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
754 Solutions Manual
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Dynamics 2e 755
Problem 4.66
The Lennard-Jones force law between two atoms can be represented as
fij D24✏✓6
r7
ij 212
r13
ij ◆;
where
rij
is the distance between atoms
i
and
j
, and
✏
and
are material-specific parameters with
dimensions of energy and length, respectively. Using this equation, determine the potential between two
Ni atoms. Assume the potential between the two atoms is zero when the distance between those two atoms
is infinite.
Solution
We consider atoms
i
and
j
. We regard
i
as being fixed and we sketch the
FBD of atom
j
subject to the interatomic force
fij
directed from
j
to
i
.
756 Solutions Manual
Problem 4.67
Spring scales work by measuring the displacement of a spring that supports
both the platform and the object, of mass
m
, whose weight is being measured.
Neglect the mass of the platform on which the mass sits, and assume that the
spring is uncompressed before the mass is placed on the platform. In addition,
assume that the spring is linear elastic with spring constant
k
. You may have
solved these same problems using Newton’s second law when doing Prob. 3.27
and 3.28 — here use the work-energy principle to solve them.
If the mass
m
is gently placed on the spring scale (i.e., it is dropped from
zero height above the scale), determine the maximum reading on the scale after
the mass is released.
Solution
We model the object placed on the platform as a particle subject to
its own weight
mg
and the force of the spring
Fs
. Both these forces
