Potential Energy and Energy Conservation
75
7.14.
I
DENTIFY
:
Only gravity does work, so apply Eq.(7.4). Use
m
=
∑
F
a
!
!
to calculate the tension.
S
ET UP
:
Let
0
y
=
at the bottom of the arc. Let point 1 be when the string makes a 45
°
angle with the vertical and
point 2 be where the string is vertical. The rock moves in an arc of a circle, so it has radial acceleration
2
rad
/
a
v
r
=
E
XECUTE
:
(a)
At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the
bottom of the circular arc) is
(1
cos ),
mgl
θ
−
where
l
is the length of the string and
θ
is the angle the string makes
with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so
2
1
2
(1
cos )
,
mgl
θ
mv
−
=
or
2
2
(1
cos
)
2(9 80 m/s
) (0 80 m) (1
cos45
)
2.1 m/s
v
gl
θ
.
.
=
−
=
−
° =
.
(b)
At 45
°
from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the radial
component of the weight, or
2
cos
(0.12 kg) (9.80 m/s
) cos 45
0.83 N.
mg
θ
=
° =
(c)
At the bottom of the circle, the tension is the sum of the weight and the mass times the radial acceleration,
2
2
(1
2(1
cos45
))
1.9 N
mg
mv
l
mg
+
=
+
−
°
=
E
VALUATE
:
When the string passes through the vertical, the tension is greater than the weight because the
acceleration is upward.
7.15.
I
DENTIFY
:
Apply
2
1
el
2
U
kx
=
.
S
ET UP
:
kx
F
=
, so
1
2
U
Fx
=
,where
F
is the magnitude of force required to stretch or compress the spring a
distance
x
.
E
XECUTE
:
(a)
(1 2)(800 N)(0.200 m)
80.0
J.
=
(b)
The potential energy is proportional to the square of the compression or extension;
2
(80.0 J) (0.050 m
0.200 m)
5.0 J.
=
E
VALUATE
:
We could have calculated
800 N
4000 N/m
0.200 m
F
k
x
=
=
=
and then used
2
1
el
2
U
kx
=
directly.
7.16.
I
DENTIFY
:
Use the information given in the problem with
F
kx
=
to find
k
. Then
2
1
el
2
U
kx
=
.
S
ET UP
:
x
is the amount the spring is stretched. When the weight is hung from the spring,
F
mg
=
.
E
XECUTE
:
2
(3.15 kg)(9.80 m/s
)
2205 N/m
0.1340 m
0.1200 m
F
mg
k
x
x
=
=
=
=
−
.
el
2
2(10.0 J)
0.0952 m
9.52 cm
2205 N/m
U
x
k
= ±
= ±
= ±
= ±
. The spring could be either stretched 9.52 cm or
compressed 9.52 cm. If it were stretched, the total length of the spring would be 12.00 cm
9.52 cm
21.52 cm
+
=
.
If it were compressed, the total length of the spring would be 12.00 cm
9.52 cm
2.48 cm
−
=
.
E
VALUATE
:
To stretch or compress the spring 9.52 cm requires a force
210 N
F
kx
=
=
.
7.17.
I
DENTIFY
:
Apply
2
1
el
2
U
kx
=
.
S
ET UP
:
2
1
0
0
2
U
kx
=
.
x
is the distance the spring is stretched or compressed.
E
XECUTE
:
(a)
(i)
0
2
x
x
=
gives
2
2
1
1
el
0
0
0
2
2
(2
)
4(
)
4
U
k
x
kx
U
=
=
=
. (ii)
0
/ 2
x
x
=
gives
2
2
1
1
1
el
0
0
0
2
4
2
(
/ 2)
(
)
/4
U
k x
kx
U
=
=
=
.
(b)
(i)
0
2
U
U
=
gives
2
2
1
1
0
2
2
2(
)
kx
kx
=
and
0
2
x
x
=
. (ii)
0
/2
U
U
=
gives
2
2
1
1
1
0
2
2
2
(
)
kx
kx
=
and
0
/
2
x
x
=
.
E
VALUATE
:
U
is proportional to
2
x
and
x
is proportional to
U
.
7.18.
I
DENTIFY
:
Apply Eq.(7.13).
S
ET UP
:
Initially and at the highest point,
0
v
=
, so
1
2
0
K
K
=
=
.
other
0
W
=
.
E
XECUTE
:
(a)
In going from rest in the slingshot°s pocket to rest at the maximum height, the potential energy
stored in the rubber band is converted to gravitational potential energy;
3
2
(10
10
kg)(9.80 m/s
) (22.0 m)
2.16 J.
U
mgy
−
=
=
×
=
(b)
Because gravitational potential energy is proportional to mass, the larger pebble rises only 8.8 m.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics, Energy, Gravity, Potential Energy, Work, Wother

Click to edit the document details