rathbun (rar2954) – HW #6 – Antoniewicz – (57380)
1
This
printout
should
have
21
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 3) 4.0 points
Starting from rest, a(n)13 kg block slides 9 m
down a frictionless ramp (inclined at 30
◦
from
the floor) to the bottom.
The block then
slides an additional 23 m along the floor before
coming to stop.
The acceleration of gravity is 9
.
8 m
/
s
2
.
9 m
13 kg
13 kg
30
◦
23 m
Find the speed of the block at the bottom
of the ramp.
Correct answer: 9
.
39149 m
/
s.
Explanation:
L
m
m
θ
d
Given :
m
= 13 kg
,
θ
= 30
◦
,
L
= 9 m
,
and
d
= 23 m
.
Let
P
= 0 at the bottom of the ramp.
Using conservation of energy along the ramp,
we have
U
i
+
K
i
=
U
f
+
K
f
mgL
sin
θ
+ 0 = 0 +
1
2
mv
2
bottom
where L is the length moved along the ramp.
v
bottom
=
2
gL
sin
θ
=
2(9
.
8 m
/
s
2
)(9 m) sin 30
◦
=
9
.
39149 m
/
s
.
002 (part 2 of 3) 3.0 points
Find the coe
ffi
cient of kinetic friction between
block and floor.
Correct answer: 0
.
195652.
Explanation:
Let
d
be the distance moved along the floor.
The frictional force is given by
f
=
μ
k
N
=
μ
k
mg .
Applying the workkinetic energy theorem,
we have
W
f
= (
K

U
g
)
f

(
K
+
U
g
)
i
,
where
K
i
is the kinetic energy at the bottom
of the ramp and
K
f
= 0.
μ
k
m g d
= (0 + 0)

0 +
1
2
m v
2
i
μ
k
=
v
2
i
2
g d
=
(9
.
39149 m
/
s)
2
2(9
.
8 m
/
s
2
)(23 m)
=
0
.
195652
.
003 (part 3 of 3) 3.0 points
Find the magnitude of the mechanical energy
lost due to friction.
Correct answer: 573
.
3 J.
Explanation:
All of the initial potential energy is lost due
to friction along the floor, so is
m g y
i
=
m g L
sin
θ
= (13 kg) (9
.
8 m
/
s
2
) (9 m) sin 30
◦
=
573
.
3 J
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
rathbun (rar2954) – HW #6 – Antoniewicz – (57380)
2
004
10.0 points
The spring has a constant of 29 N
/
m and
the frictional surface is 0
.
4 m long with a
coe
ffi
cient of friction
μ
= 1
.
44
.
The 4 kg
block depresses the spring by 24 cm, then is
released.
The first drop is 2
.
3 m and the
second is 1 m.
The acceleration of gravity is 9
.
8 m
/
s
2
.
μ
h
h
L
m
k
x
1
2
s
How far from the bottom of the cli
ff
does it
land?
Correct answer: 2
.
6422 m.
Explanation:
Basic concepts
The potential energy of a
spring is
U
s
=
1
2
k x
2
=
1
2
(29 N
/
m) (24 cm)
2
= 0
.
8352 J
.
Gravitational potential energy is
U
g
=
m g h
= (4 kg) (9
.
8 m
/
s
2
) (2
.
3 m)
= 90
.
16 J
.
Work done against friction on a flat surface is
W
fr
=
μ m g L
= (1
.
44) (4 kg) (9
.
8 m
/
s
2
) (0
.
4 m)
= 22
.
5792 J
.
Kinetic energy is
K
=
1
2
m v
2
.
The height of the cli
ff
determines how long it
takes for the mass to reach the bottom.
Energy considerations give us the horizon
tal velocity as the mass leaves the cli
ff
top.
The mass gets its initial kinetic energy from
the spring, then it gains kinetic energy by
dropping a distance
h
1
, then it loses kinetic
energy by doing work against friction.
U
s
+
U
g

W
fr
=
K
f
=
1
2
m v
2
Since
v
2
=
2 (
U
s
+
U
g

W
fr
)
m
=
2 (0
.
8352 J + 90
.
16 J

22
.
5792 J)
4 kg
= 34
.
208 m
2
/
s
2
then
v
=
34
.
208 m
2
/
s
2
= 5
.
84876 m
/
s
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Physics, Force, Potential Energy, Correct Answer, rathbun

Click to edit the document details