Download PH201 Recitation Problem Set 8

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PH201 Recitation Problem Set 8
PH201R-H5 & PH201-N5 (2009 Spring)
A spring lies on a horizontal table, and the left end is attached to a wall. The other end is connected to a
box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the
table. When the box is released, it does not move.
(a) Is the direction of the restoring force of the spring the same as or opposite to the direction of the
static frictional force? Why?
(b) How is the magnitude of the restoring force related to the magnitude of the static frictional force?
(c) What factors determine the maximum static frictional force that can be applied to the box just
before it begins to move?
(d) A 0.80-kg box is attached to a horizontal spring whose spring constant is 59 N/m. The coefficient of
static friction between the box and the table on which it rests is µs = 0.74. How far can the spring be
stretched from its unstrained position without the box moving?
(a) The restoring force of the spring and the static frictional force
point in opposite directions. Since the block is in equilibrium, the
net force in the horizontal direction is zero. Thus, these two forces
must point in opposite directions.
(b) The two forces are equal in magnitude. Since the block is in equilibrium, the net force in the horizontal direction is zero. Thus,
the two forces must point in opposite directions and have equal
magnitudes.
(c) The maximum static frictional force is determined by the magnitude of the normal force exerted on the box and the coefficient of
static friction (see Equation 4.7).
(d) The drawing at the right shows the four forces that act on the block: its weight mg, the normal force
FN , the restoring force Fx exerted by the spring, and the maximum static frictional force fsMAX .
Since the box is not moving, it is in equilibrium. Let the x axis be parallel to the table top.
According to Equation 4.9a, the net force ΣFx in the x direction must be zero, ΣFx = 0.
The restoring force is given by Equation 10.2 as Fx = kx, where k is the spring constant and x is the
displacement of the spring (assumed to be in the +x direction). The magnitude of the maximum
static frictional force is given by Equation 4.7 as fsMAX = µs FN , where µs is the coefficient of static
friction and FN is the magnitude of the normal force. Thus, the condition for equilibrium can be
written as
µs Fn
−kx + µs FN = 0 → x =
k
We can determine FN by noting that the box is also in vertical equilibrium, so that
−mg + FN = 0
→
FN = mg
The distance that the spring is stretched from its unstrained position is
x=
µs FN
µs mg
(0.74)(0.80 kg)(9.80 m/s2)
=
=
= 9.8 × 10−2 m
k
k
59 N/m
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A vertical ideal spring is mounted on the floor. A block is placed on the unstrained spring in two different
ways.
(a) The block is placed on the spring and not released until it rests stationary on the spring in its
equilibrium position. As the block rests on the spring in equilibrium, what determines how much the
spring is compressed? Explain.
(b) Immediately after being placed on the unstrained spring, the block is released from rest and falls
downward until it comes to a momentary halt. What determines how much the spring is compressed
in this case, which is not an example of equilibrium? Account for your answer.
(c) In which of the situations described in (a) and (b) is the compression of the spring greater? Give a
reason for your answer.
(d) The spring constant of the spring is 170 N/m, and the mass of the block is 0.64kg. Determine the
amount by which the spring is compressed in case (a) and in case (b). Verify that your answers are
consistent with your answer to (c).
(a) As the block rests stationary in its equilibrium position, it has no acceleration. According to Newtons
second law, the net force acting on the block is, therefore, zero. This means that the
downward-directed weight of the block must be balanced by an upward-directed force. This upward
force is the restoring force of the spring and is produced because the spring is compressed. The
amount of compression must be enough for the spring to exert on the block a restoring force that has
a magnitude equal to the blocks weight.
(b) As the block falls downward after being released, its speed is changing in the manner characteristic of
simple harmonic motion. The block is not in equilibrium, and the forces acting on it do not balance
to zero. Instead of thinking about forces, we may think about mechanical energy and its
conservation. When the block is released from rest, the energy of the spring/block system is all in the
form of gravitational potential energy. Being at rest, the block has no initial kinetic energy. It also
has no initial elastic potential energy, since the spring is unstrained initially. When the block comes
to a momentary halt at the lowest point in its fall, the energy is all in the form of elastic potential
energy. Since the block is again at rest, it again has no kinetic energy. The spring has been
compressed, and gravitational potential energy has been converted entirely into elastic potential
energy. The amount by which the spring is compressed is determined by the amount of gravitational
potential energy that must be converted into elastic potential energy. The amount must be enough
that the elastic potential energy equals the gravitational potential energy.
(c) The compression of the spring is greater in the non-equilibrium case than in the equilibrium case.
The reason is that in the non-equilibrium case, the block has been allowed to move, and its inertia
carries it beyond its stationary equilibrium position on the spring. The compression of the spring
must increase beyond that corresponding to the stationary equilibrium position in order to produce
the force that is needed to decelerate the block to a momentary halt.
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(d)
a. As the block rests stationary on the spring, the downward-directed weight balances the
upward-directed restoring force from the spring. The magnitude of the weight is mg, and the
magnitude of the restoring force is given by Equation 10.2 without the minus sign as kx. Thus,
we have
(0.64 kg)(9.80 m/s2
mg
=
= 0.037 m
mg = kx → x =
k
170 N/m
b. The conservation of mechanical energy states that the final total mechanical energy Ef is equal
to the initial total mechanical energy E0 . The expression for the total mechanical energy for an
object on a spring is given by Equation 10.14, so that we have
1
1
1
1
mv 2 + mghf + kx2f = mv02 + mgh0 + kx20
2 f
2
2
2
Since the block comes to a momentary halt on the spring and is released from rest, the
translational speeds vf and v0 are both zero. Because the spring is initially unstrained, the
initial displacement x0 of the spring is likewise zero. Thus, the above expression can be
simplified as follows:
1
mghf + kx2f = mgh0
2
→
1 2
kx = mg(h0 − hf )
2 f
The term h0 − hf is the amount by which the spring has compressed, or h0 − hf = xf . Making
this substitution into the simplified energy-conservation equation gives
1 2
kx = mgxf
2 f
→
1
xf ( kxf − mg) = 0
2
Since xf 6= 0,
1
kxf − mg = 0
2
→
xf =
2mg
2(0.64 kg)(9.80 m/s2)
=
= 0.074 m
k
170 N/m
As expected, the spring compresses more in the non-equilibrium situation.
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