homework 05 – JACOBS, AMBER – Due: Feb 26 2008, 1:00 am
1
Question 1, chap 8, sect 6.
part 1 of 1
10 points
The following graph represents a hypothet
ical potential energy curve for a particle of
mass
m
.
r
U
(
r
)
O
3
U
0
2
U
0
U
0
r
0
2
r
0
If the particle is released from rest at posi
tion
r
0
, its speed
bardbl
vectorv
bardbl
at position 2
r
0
is most
nearly
1.
bardbl
vectorv
bardbl
=
radicalbigg
U
0
m
.
2.
bardbl
vectorv
bardbl
=
radicalbigg
U
0
6
m
.
3.
bardbl
vectorv
bardbl
=
radicalbigg
8
U
0
m
.
4.
bardbl
vectorv
bardbl
=
radicalbigg
2
U
0
m
.
5.
bardbl
vectorv
bardbl
=
radicalbigg
U
0
2
m
.
6.
bardbl
vectorv
bardbl
=
radicalbigg
U
0
8
m
.
7.
bardbl
vectorv
bardbl
=
radicalbigg
U
0
4
m
.
8.
bardbl
vectorv
bardbl
=
radicalbigg
4
U
0
m
.
correct
9.
bardbl
vectorv
bardbl
=
radicalbigg
6
U
0
m
.
Explanation:
The total energy of the particle is con
served.
So the change of the potential en
ergy is converted into the kinetic energy of
the particle, which gives
1
2
m v
2
= 3
U
0
−
U
0
Therefore
v
=
radicalbigg
4
U
0
m
.
Question 2, chap 8, sect 6.
part 1 of 2
10 points
Given:
The positive
x
direction is to the
right.
In part (a) of the figure, an air track cart
attached to a spring rests on the track at
the position
x
eq
and the spring is relaxed.
(a)
x
st
x
eq
(b)
In (b), the cart is pulled to the position
x
st
and released. It then oscillates about
x
eq
.
Which graph correctly represents the po
tential energy of the spring as a function of
the position of the cart?
1.
x
st
x
eq
2.
x
st
x
eq
3.
x
st
x
eq
4.
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homework 05 – JACOBS, AMBER – Due: Feb 26 2008, 1:00 am
2
x
st
x
eq
correct
5.
x
st
x
eq
6.
x
st
x
eq
Explanation:
The car starts at
x
st
with no kinetic energy,
and so the spring’s potential energy is a max
imum. Once released, the cart accelerates to
the right and its kinetic energy increases as
the potential energy of the spring is converted
into kinetic energy of the cart.
As the cart
passes the equilibrium position, its kinetic en
ergy is a maximum and so the spring’s poten
tial energy is a minimum. Once to the right of
x
equ
, the car starts to compress the spring and
it slows down as its kinetic energy is converted
back to potential energy of the recompressed
spring. At the rightmost point it reaches, the
cart reverses its direction of travel. At that in
stant, it has no kinetic energy and the spring
again has maximum potential energy.
Question 3, chap 8, sect 6.
part 2 of 2
10 points
The forces
vector
F
st
and
vector
F
eq
acting at
x
st
and
x
eq
are related as follows
1.
vector
F
st
>
0 ˆ
ı ,
vector
F
eq
<
0 ˆ
ı .
2.
vector
F
st
>
0 ˆ
ı ,
vector
F
eq
>
0 ˆ
ı .
3.
vector
F
st
<
0 ˆ
ı ,
vector
F
eq
>
0 ˆ
ı .
4.
vector
F
st
>
0 ˆ
ı ,
vector
F
eq
= 0 ˆ
ı .
correct
5.
vector
F
st
<
0 ˆ
ı ,
vector
F
eq
<
0 ˆ
ı .
6.
vector
F
st
= 0 ˆ
ı ,
vector
F
eq
>
0 ˆ
ı .
7.
vector
F
st
= 0 ˆ
ı ,
vector
F
eq
<
0 ˆ
ı .
8.
vector
F
st
<
0 ˆ
ı ,
vector
F
eq
= 0 ˆ
ı .
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 Spring '07
 KOPP
 Physics, Energy, Kinetic Energy, Mass, Potential Energy, Work, kg, xst, xst xeq

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