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Week 9 homework
IMPORTANT NOTE ABOUT WEBASSIGN:
In the WebAssign versions of these problems, various details have been changed, so that
the answers will come out differently.
The method to find the solution is the same, but
you will need to repeat part of the calculation to find out what your answer should have
been.
WebAssign Problem 1:
A 10.1kg uniform board is wedged into a corner and held by a
spring at a 50.0° angle, as the drawing shows. The spring has a spring constant of 176
N/m and is parallel to the floor. Find the amount by which the spring is stretched from its
unstrained length.
REASONING AND SOLUTION
In order to find the amount the spring stretches we
need to calculate the force that acts on the spring. The magnitude of this force is
F
. Since
the board is in equilibrium, the net torque acting on it is zero. Taking the axis of rotation
to be at the corner and assuming the board has a length
L
, the net torque is
Σ
τ
=
FL
sin 50.0°

mg
(
L
/2) cos 50.0° = 0
Solving for
F
gives
cos 50.0
2 sin 50.0
2 tan 50.0
mg
mg
F
°
=
=
°
°
The amount
x
by which the spring stretches is equal to the magnitude
F
of the
force applied to it divided by the spring constant
k
(see Equation 10.1). Thus,
2
(10.1 kg)(9.80 m/s )
0.236 m
2 tan 50.0
2(176 N/m) tan 50.0
F
mg
x
k
k
=
=
=
=
°
°
WebAssign Problem 2: Concept Simulation 10.3
at
www.wiley.com/college/cutnell
illustrates the concepts pertinent to this problem. An 0.80kg object is attached to one end
of a spring, as in Figure
10.6
, and the system is set into simple harmonic motion. The
displacement
x
of the object as a function of time is shown in the drawing. With the aid
of these data, determine (a) the amplitude
A
of the motion, (b) the angular frequency
ω
(c) the spring constant
k
, (d) the speed of the object at
t
= 1.0 s, and (e) the magnitude of
the object’s acceleration at
t
= 1.0 s.
REASONING AND SOLUTION
a.
Since the object oscillates between
±
0.080 m, the amplitude of the
motion is
0.080 m
.
b.
From the graph, the period is
T
=
4.0 s.
Therefore, according to Equation
10.4,
ϖ
=
2
π
T
=
2
4.0 s
=
1.6 rad/s
c.
Equation 10.11 relates the angular frequency to the spring constant:
=
k
/
m
.
Solving for
k
we find
k
=
2
m
=
(1.6 rad/s)
2
(0.80 kg)
=
2.0 N/m
d.
At
t
=
1.0 s, the graph shows that the spring has its maximum displacement. At
this location, the object is momentarily at rest, so that its speed is
0 m/s
v
=
.
e.
The acceleration of the object at
t
=
1.0 s is a maximum, and its magnitude is
a
max
=
A
2
=
(0.080 m)(1.6 rad/s)
2
= 0.20 m/s
2
WebAssign Problem 3:
A 3.0kg block is between two horizontal springs. Neither
spring is strained when the block is at the position labeled
x
= 0 m in the drawing.
The block is then displaced a distance of 0.070 m from the position where
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