This preview shows pages 1–4. Sign up to view the full content.
P207  Fall 2007
Solutions Assignment 6
1. Chapter 7, Problem 13
(a) The work done by the forces is Δ
x
∑
F
x
+ Δ
y
∑
F
y
and we know that
Δ
y
= 0 so the forces in the ydirection make no contribution. In the
xdirection
s
F
x
=

v
F
2

cos
θ
− 
v
F
1

= (9 N) cos 60
◦
−
5 N =
−
0
.
5 N
Then
W
= Δ
x
s
F
x
= (
−
3 m)(
−
0
.
5 N) = 1
.
5 J
(b) The work on the trunk is translated into kinetic energy. Δ
KE
=
W
=
1
.
5J
2. (a)
0
5
10
15
20
25
30
velocity [m/s]
0
1
2
3
4
5
6
force [kN]
0
50
100
150
200
8
7
6
5
4
3
2
1
0
power [kW]
time [seconds]
(b) The car is accelerating due to the force of static friction that the road is
exerting on the tires.
F
=
ma
= (1500 kg)(30 m
/
s)
/
(8 s) = 5625 N.
(c) The acceleration
a
=
V/t
= (30 m
/
s)
/
(8 s) = 3
.
75 m
/
s
2
. Since the
acceleration is constant, the average velocity over an interval Δ
t
is given
by the average of initial and ±nal velocities for the interval. Initial and
±nal velocities are
V
i
=
at
i
and
V
f
=
at
f
. Then
a
V
A
=
a
t
f
+
t
i
2
. During the
period
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document (i) 02 s,
a
v
A
=
a
1
2
(
t
f
+
t
i
) = (3
.
75 m
/
s
2
)(1 s) = 3
.
75 m
/
s
a
P
A
=
F
a
v
A
= (5625 N)(3
.
75 m
/
s) = 21kW
(ii) 24 s,
a
v
A
=
a
1
2
(
t
f
+
t
i
) = (3
.
75 m
/
s
2
)(3 s) = 11
.
25 m
/
s
a
P
A
=
F
a
v
A
= (5625 N)(11
.
25 m
/
s) = 63kW
(iii) 46 s,
a
v
A
=
a
1
2
(
t
f
+
t
i
) = (3
.
75 m
/
s
2
)(5 s) = 18
.
75 m
/
s
a
P
A
=
F
a
v
A
= (5625 N)(3
.
75 m
/
s) = 105kW
(iv) 68 s,
a
v
A
=
a
1
2
(
t
f
+
t
i
) = (3
.
75 m
/
s
2
)(7 s) = 26
.
25 m
/
s
a
P
A
=
F
a
v
A
= (5625 N)(3
.
75 m
/
s) = 147kW
(d) The work done by the engine is
W
=
Pt
= 240kJ. If all of that work
appears as the kinetic energy of the car, then
W
=
1
2
mv
2
→
v
=
r
2
W
m
=
p
2(240
,
000 J)
1500 kg
P
1
2
= 18 m
/
s
3. (a) The Fea stores enough energy to do the work necessary to achieve a height
of 25 cm.
W
=
Fd
=
mgd
= (10
−
6
kg)(10 m
/
s
2
)(0
.
25 m) = 2
.
5
×
10
−
6
J
(b) The energy stored in the spring is
U
=
1
2
kx
2
. Then
k
=
2
U
x
2
=
2(2
.
5
×
10
−
6
J)
(10
−
4
m)
2
= 500 N
/
m
4. Chapter 7, Problem 28
Let the applied force
F
= 80 N, that yields the initial displacement
x
i
=
−
2cm.
The spring constant
k
=

F/x
i

= (80 N)
/
(0
.
02 m) = 4000N
/
m. The work
done on a spring when changing its displacement from equilibrium from
x
i
to
x
f
=
x
i
+ Δ
x
is
W
=
1
2
k
(
x
2
f
−
x
2
i
) =
1
2
k
((
x
i
+ Δ
x
)
2
−
x
2
i
) =
1
2
k
(Δ
x
2
+ 2
x
i
Δ
x
)
(1)
There are two solutions for Δ
x
. We can do the work on the spring by continuing
to compress it beyond the original
x
i
of 2cm. Or we can stretch it in the other
2
direction, allowing it to relax to zero and then compress it in the positive
direction. The solutions to Equation 1 are
Δ
x
=
1
2
p
−
2
x
i
±
r
4
x
2
i
+ 8
W/k
P
=
1
2
±
−
2(
−
0
.
02 m)
±
R
4(
−
0
.
02 m)
2
+ 8(4 J)
/
(4000N
/
m)
²
= (0
.
02
±
0
.
05) m
The fnal position is
x
f
=
x
i
+ Δ
x
=
−
0
.
02 m + (0
.
02 m
±
0
.
05 m) =
±
0
.
05 m
5. Chapter 7, Problem 54
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.
This note was uploaded on 01/22/2009 for the course PHYS 207 taught by Professor Liepe during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 LIEPE
 Force, Work

Click to edit the document details