homework 12 – ALIBHAI, ZAHID – Due: Apr 18 2007, 4:00 am
1
Question 1
part 1 of 1
10 points
A block of mass 4 kg is hung from a spring,
causing it to stretch 8 cm at equilibrium, as
shown below.
The 4 kg block is then replaced by a 6 kg
block, and the new block is released from
the position shown below, at which point the
spring is unstretched.
The acceleration of gravity is 9
.
8 m
/
s
2
.
8 cm
y
max
4 kg
equilbrium
6 kg
How far will the 6 kg block fall before its
direction is reversed?
Hint:
For the new
block case, the total distance of fall is twice
the amplitude of the oscillation.
Correct answer: 24 cm (tolerance
±
1 %).
Explanation:
Let :
m
4
= 4 kg
,
m
6
= 6 kg
,
y
4
= 8 cm
,
and
g
= 9
.
8 m
/
s
2
.
When the 4 kg block is replaced by 6 kg
block, the equilibrium point is
y
6
=
z
4
m
4
m
6
=
(8 cm) (4 kg)
(6 kg)
= 12 cm
.
below the unstretched point.
However, the point at which the block re
verses its direction is at its maximum am
plitude in the downward direction, which is
two times the amplitude from its equilibrium
point
y
amplitude
= 12 cm
.
Thus, the block will
go
y
max
= 2 (12 cm)
=
24 cm
,
before its direction is reversed.
Alternative solution:
The equilibrium
position for the 4 kg is 8 cm, as given.
The
maxium stretched position for 6 kg after re
leased from rest is
y
max
.
The spring constant is
k
=
m
4
g
x
4
.
Conservation of energy implies that
m
6
g y
max
=
1
2
k y
2
max
so
y
max
=
2
m
6
g
k
= 2
m
6
m
4
x
4
= 2
(6 kg)
(4 kg)
(8 cm)
=
24 cm
.
Question 2
part 1 of 1
10 points
Quantum concepts are critical in explaining
all of the following EXCEPT
1.
Compton scattering.
2.
Bohr’s theory of the hydrogen atom.
3.
Rutherfold’s
scattering
experiments.
correct
4.
the blackbody spectrum.
5.
the photoelectric effect.
Explanation:
Rutherfold’s
scattering
experiments
re
vealed that the atoms consist of a positively
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
homework 12 – ALIBHAI, ZAHID – Due: Apr 18 2007, 4:00 am
2
charged heavy nucleus and surrounding light
electrons.
Classical electromagnetic theory
can explain the results without referring to
quantum concepts.
In all of the other four choices, quantum
concepts are necessary to explain the phe
nomena.
Question 3
part 1 of 2
10 points
A 1
.
24 kg block at rest on a tabletop is at
tached to a horizontal spring having constant
16
.
1 N
/
m. The spring is initially unstretched.
A constant 27
.
7 N horizontal force is applied
to the object causing the spring to stretch.
The acceleration of gravity is 9
.
8 m
/
s
2
.
27
.
7 N
1
.
24 kg
16
.
1 N
/
m
Find the speed of the block after it has
moved 0
.
26 m from equilibrium if the surface
between block and tabletop is frictionless.
Correct answer: 3
.
27695 m
/
s (tolerance
±
1
%).
Explanation:
Given :
m
= 1
.
24 kg
,
x
f
= 0
.
26 m
,
k
= 16
.
1 N
/
m
,
and
F
= 27
.
7 N
.
Applying the work kinetic energy theorem,
F x
f
=
1
2
m v
2
f
+
1
2
k x
2
f
m v
2
f
= 2
F x
f
−
k x
2
f
v
2
f
=
2
F x
f
−
k x
2
f
m
=
2 (27
.
7 N) (0
.
26 m)
1
.
24 kg
−
(16
.
1 N
/
m) (0
.
26 m)
2
1
.
24 kg
= 10
.
7384 m
2
/
s
2
so that
v
f
=
radicalBig
10
.
7384 m
2
/
s
2
=
3
.
27695 m
/
s
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Energy, Mass, Work, Correct Answer, Rutherfold

Click to edit the document details