dang (ttd73) – HW05 – Tsoi – (58020)
1
This
printout
should
have
23
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
(part 1 of 3) 10.0 points
A block of mass
m
is pushed a distance
D
up
an inclined plane by a horizontal force
F
. The
plane is inclined at an angle
θ
with respect to
the horizontal. The block starts from rest and
the coefficient of kinetic friction is
μ
k
.
m
D
μ
k
F
θ
If
N
is the normal force, what is the work
done by friction?
1.
W
= +
μ
k
N
D
2.
W
=

μ
k
(
N 
m g
cos
θ
)
D
3.
W
= +
μ
k
(
N 
m g
cos
θ
)
D
4.
W
= +
μ
k
(
N
+
m g
cos
θ
)
D
5.
W
=

μ
k
(
N
+
m g
cos
θ
)
D
6.
W
=

μ
k
N
D
correct
7.
W
= 0
Explanation:
The
force
of
friction
has
a
magnitude
F
friction
=
μ
k
N
.
Since it is in the direc
tion opposite to the motion, we get
W
friction
=

F
friction
D
=

μ
k
N
D.
002
(part 2 of 3) 10.0 points
What is the work done by the normal force
N
?
1.
W
= (
N 
m g
cos
θ

F
sin
θ
)
D
2.
W
= (
m g
cos
θ
+
F
sin
θ
 N
)
D
3.
W
=
N
D
cos
θ
4.
W
=
N
D
sin
θ
5.
W
=
N
D
6.
W
= 0
correct
7.
W
=
N
D
8.
W
= (
N
+
m g
cos
θ
+
F
sin
θ
)
D
Explanation:
The normal force makes an angle of 90
◦
with the displacement, so the work done by it
is zero.
003
(part 3 of 3) 10.0 points
What is the final speed of the block?
1.
v
=
radicalbigg
2
m
(
F
cos
θ
+
m g
sin
θ
)
D
2.
v
=
radicalbigg
2
m
(
F
sin
θ
+
μ
k
N
)
D
3.
v
=
radicalbigg
2
m
(
F
sin
θ

μ
k
N
)
D
4.
v
=
radicalbigg
2
m
(
F
cos
θ

m g
sin
θ
+
μ
k
N
)
D
5.
v
=
radicalbigg
2
m
(
F
cos
θ

m g
sin
θ

μ
k
N
)
D
correct
6.
v
=
radicalbigg
2
m
(
F
cos
θ

μ
k
N
)
D
7.
v
=
radicalbigg
2
m
(
F
cos
θ

m g
sin
θ
)
D
8.
v
=
radicalbigg
2
m
(
F
cos
θ
+
m g
sin
θ

μ
k
N
)
D
Explanation:
The work done by gravity is
W
grav
=
m g D
cos(90
◦
+
θ
)
=

m g D
sin
θ .
The work done by the force
F
is
W
F
=
F D
cos
θ .
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
dang (ttd73) – HW05 – Tsoi – (58020)
2
From the workenergy theorem we know that
W
net
= Δ
K ,
W
F
+
W
grav
+
W
friction
=
1
2
m v
2
f
.
Thus
v
f
=
radicalbigg
2
m
(
F
cos
θ

m g
sin
θ

μ
k
N
)
D .
004
(part 1 of 2) 10.0 points
A 500N crate needs to be lifted 1 meter ver
tically in order to get it into the back of a
pickup truck.
What gives the crate a greater potential
energy?
1.
Either
correct
2.
Unable to determine
3.
lift it straight up into the truck
4.
slide it up a frictionless inclined plane
Explanation:
The change in height is the same for either
method, so the new potential energies are the
same.
005
(part 2 of 2) 10.0 points
What is the advantage of using the inclined
plane?
1.
less distance
2.
less force
correct
3.
less total energy
4.
more power
Explanation:
An
inclined
plane
can
exchange
an in
creased distance for less force.
006
(part 1 of 2) 10.0 points
Consider a bungee cord of unstretched length
L
0
= 35 m.
When the cord is stretched to
L > L
0
it behaves like a spring and obeys
Hooke’s law with the spring constant
k
=
22 N
/
m.
However, unlike a spring,
the cord
folds instead of becoming compressed
when
the distance between its ends is less than the
unstretched length: For
L < L
0
the cord has
zero tension and zero elastic energy.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Kaplunovsky
 Energy, Force, Friction, Mass, Correct Answer, chanical Energy, Ucord

Click to edit the document details