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P207 - Fall 2007
Solutions Assignment 7
1. Chapter 8, Problem 28
(a) The total mechanical energy of the box is the sum of the potential energy
of the spring, the potential energy of gravity and the kinetic energy. Let
x
= 0 when the spring is in the unstretched position. The change in height
is
h
=
x
sin
θ
.
E
mech
=
U
spring
+
U
gravity
+
K
0 =
1
2
kx
2
−
mgx
sin
θ
+
1
2
mv
2
When
x
= 0,
E
mech
= 0 and mechanical energy is conserved so we can
solve for
v
as a function of
x
v
=
p
2
m
(
mgx
sin
θ
−
1
2
kx
2
)
P
1
2
(1)
When
x
= 0
.
1m,
v
=
p
2
2 kg
((2 kg)(0
.
1 m)(9
.
8 m
/
s
2
) sin(40
◦
)
−
1
2
(120N
/
m)(0
.
1 m)
2
)
P
1
2
= 0
.
81 m
/
s
(b) We need to determine the values of
x
in Equation 1 such that
v
= 0. We
have that
mg
sin
θ
=
1
2
kx
→
x
=
2
mg
sin
θ
k
=
2(2 kg)(9
.
8 m
/
s
2
) sin 40
◦
120N
/
m
= 0
.
21 m
(c) We have that when the box stops when
x
=
x
0
= 2
mg
sin
θ/k
. Then the
force on the box is
F
=
−
kx
+
mg
sin
θ
and the acceleration is
a
=
F/m
1

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−
kx/m
+
g
sin
θ
=
k
p
−
2
mg
sin
θ
k
P
1
m
+
g
sin
θ
=
−
g
sin
θ
= 9
.
8 m
/
s
2
sin(40
◦
) =
−
6
.
3 m
/
s
2
Note that at
x
= 0, we see by inspection that
a
(0) =
g
sin
θ
. Then by
symmetry the acceleration at the other point of zero velocity,
x
=
x
0
,
that the acceleration
a
(
x
0
) =
−
a
(0) =
−
g
sin
θ
, which is exactly what we
found above.
(d) The acceleration is in the direction of the spring restoring force, up the
ramp.
2. Chapter 8, Problem 40
(a) The total mechanical energy
E
mech
=
U
+
1
2
mv
2
= 12 J+4 J = 16 J. The
speed of the particle at
x
= 3
.
5 m is
v
=
p
2(
E
mech
−
U
A
)
m
P
1
2
=
p
2(16 J
−
9 J)
0
.
2 kg
P
1
2
= 8
.
4 m
/
s
(b) The speed of the particle at
x
= 6
.
5 m is
v
=
p
2(
E
mech
−
0)
m
P
1
2
=
p
2(16 J)
0
.
2 kg
P
1
2
= 12
.
6 m
/
s
(c) The turning point on the right side is where the potential energy is equal to
the total mechanical energy. That would be at position
x
= 7 m+Δ
x
when
E
mech
= (
U
D
/
(1 m))Δ
x
or
x
= 7 m + (
E
mech
/U
D
)(1 m) = (16 J
/
24 J) =
7
.
67 m
(d) The turning point on the left side is where the potential energy is equal
to the total mechanical energy. Since the total energy is 16 J, the turning
point will be between
x
1
= 1 m and
x
3
= 3 m at some distance Δ
x
from
x
1
.
U

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