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```aguilar (fa6754) – hk7 – opyrchal – (11106)
This print-out should have 10 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
001 10.0 points
A(n) 561 kg elevator starts from rest. It moves
upward for 4.73 s with a constant acceleration
until it reaches its cruising speed of 1.97 m/s.
The acceleration of gravity is 9.8 m/s2 .
Find the average power delivered by the
elevator motor during the period of this acceleration.
1
002 10.0 points
A block of mass 0.21 kg is placed on a vertical spring of constant 3227 N/m and pushed
downward, compressing the spring 0.15 m.
After the block is released it leaves the spring
and continues to travel upward.
The acceleration of gravity is 9.8 m/s2 .
What height above the point of release will
the block reach if air resistance is negligible?
Explanation:
Explanation:
Given :
Let : m = 561 kg ,
t = 4.73 s , and
v = 1.97 m/s ,
The height y is
y = vave t
1
= vt
2
1
= (1.97 m/s) (4.73 s)
2
= 4.65905 m .
Since the elevator starts from rest, the power
P supplied by the motor, totally transferred
into Kinetic Energy K and Potential Energy
U , where E = K + U , is
U +K
E
=
t
t
1
m g y + 2 m v2
=
t
1
m g 2 v t + 12 m v 2
=
t
mv
=
(g t + v)
2t
(561 kg) (1.97 m/s)
=
2 (4.73 s)
× [(9.8 m/s2 ) (4.73 s) + 1.97 m/s]
= 5645.48 W
= 5.64548 kW .
P =
m = 0.21 kg ,
x = 0.15 m ,
k = 3227 N/m ,
g = 9.8 m/s2 .
and
Choose Ug = 0 at the level of the release
point.
Ki = Kf = 0, so from conservation of energy,
(Ug + Us )i = (Ug + Us )f
0 + Us,i = Ug,f + 0
1
k x2 = m g h
2
k x2
h=
2mg
(3227 N/m)(0.15 m)2
=
2(0.21 kg)(9.8 m/s2 )
= 17.6403 m .
003 (part 1 of 2) 10.0 points
A pendulum consists of a sphere of mass 1.7 kg
attached to a light cord of length 11.1 m as in
the figure below. The sphere is released from
rest when the cord makes a 41.9◦ angle with
the vertical, and the pivot at P is frictionless.
The acceleration of gravity is 9.8 m/s2 .
aguilar (fa6754) – hk7 – opyrchal – (11106)
2
For the velocity at the lowest point, we obtain
11
.1
m
p
vb = 2 g L (1 − cos θi )
h
= 2 (9.8 m/s2 ) (11.1 m)
i1/2
× (1 − cos 41.9◦ )
1.7 kg
41.9◦
= 7.45839 m/s ,
2
9.8 m/s
Find the speed of the sphere when it is at
the lowest point.
Explanation:
Let : m = 1.7 kg ,
ℓ = 11.1 m ,
θi = 41.9◦ , and
g = 9.8 m/s2 .
The only force that does work on the sphere
is the force of gravity, since the force of tension is always perpendicular to each element
of the displacement and hence does no work.
Since the force of gravity is a conservative
force, the total mechanical energy is constant.
Therefore, as the pendulum swings, there is a
continuous transfer between potential and kinetic energy. At the instant the pendulum is
released, the energy is entirely potential energy. At the lowest point, the pendulum has
kinetic energy but has lost some potential energy. The pendulum swings to the opposite
side and stops at the same height as it started.
The pendulum has regained its potential energy and its kinetic energy is again zero. If we
measure the y coordinates from the center of
rotation, then
ya = −L cos θ
004 (part 2 of 2) 10.0 points
What is the tension of the cord at the lowest
point?
and
yb = −L .
and
Ub = −m g L .
Explanation:
Since the force of tension does no work,
it cannot be determined using the energy
method. To find Tb , we can apply Newton’s
second law to the radial direction. First, recall that the centripetal acceleration of a particle moving in a circle of radius L is equal
v2
directed toward the center of rotation.
to
L
Therefore,
X
m vb2
Fr = Tb − m g = m ar =
.
L
Substituting the expression for vb obtained in
the first part of the problem, into the Newton’s second law, we find for the tension at
point the lowest point
Tb = m g + 2 m g (1 − cos θi )
= m g (3 − 2 cos θi )
= (1.7 kg) (9.8 m/s2 ) (3 − 2 cos 41.9◦ )
= 25.1795 N .
Therefore, i
Ua = −m g L cos θi
Applying the principle of constancy of mechanical energy gives
K a + Ua = K b + Ub
1
0 − m g L cos θi = m vb2 − m g L .
2
005 10.0 points
A spring with a spring-constant 2.6 N/cm
is compressed 24 cm and released. The 9 kg
mass skids down the frictional incline of height
43 cm and inclined at a 24◦ angle.
The acceleration of gravity is 9.8 m/s2 .
aguilar (fa6754) – hk7 – opyrchal – (11106)
The path is frictionless except for a distance of 0.5 m along the incline which has a
coefficient of friction of 0.4 .
k = 2.6 N/cm
9 kg
43 cm
24 cm
0.
24◦ µ
=
0.
vf
Figure: Not drawn to scale.
What is the final velocity vf of the mass?
Kf = Us + Ul − Wf r
= (7.488 J) + (37.926 J)
− (16.1149 J)
= 29.2991 J .
Kf =
2
Let : g = 9.8 m/s = ,
k = 2.6 N/cm = 260 N/m ,
x = 24 cm = 0.24 m ,
µ = 0.4 ,
ℓ = 0.5 m ,
h = 0.43 m ,
m = 9 kg , and
θ = 24◦ ,
1
m v2 .
2
(3)
Multiplying by 2 and dividing by m gives us
2 Kf
= v 2 , so
m
r
2 Kf
v=
s m
=
Consider the kinetic energy of the mass.
The mass receives its initial kinetic energy
from the potential energy of the spring
(1)
It gains kinetic energy because of the potential
energy lost in moving down the incline
Kgained = Ulost
= mgh
(2)
2
= (9 kg) (9.8 m/s ) (0.43 m)
= 37.926 J .
and loses kinetic energy by doing work on the
frictional surface
Klost = Wf r
(3)
However, the final kinetic energy is
Explanation:
Ki = Uspring
1
= k x2
2
1
= (260 N/m) (0.24 m)2
2
= 7.488 J .
= µ m g ℓ cos θ
= (0.4) (9 kg) (9.8 m/s2 )
× (0.5 m) cos(24◦ )
= 16.1149 J .
Since energy is concerved, the final kinetic
energy is
5m
4
3
2 (29.2991 J)
(9 kg)
= 2.55165 m/s .
Alternate Explanation: The potential
energy at the top of the hill will be converted
into kinetic energy at the bottom of the hill
minus energy lost due to the nonconservative
friction force.
The potential energy at the top of the hill
consists of the gravitational potential energy
plus the potential energy contained in the
compressed spring.
Combining Eqs. 1, 2, 3, and 4, we have
s 2
1
2
v=
m g h + k x − µ m g ℓ cos θ
m
2
006 (part 1 of 2) 10.0 points
The spring constant of a toy dart gun is
2934 N/m. To cock the gun the spring is compressed 1.1 cm. The 10 g dart, fired straight
upward, reaches a maximum height of 23 m.
aguilar (fa6754) – hk7 – opyrchal – (11106)
4
For the return trip, k0 = 0 , vf = 0 , so
Wext = ∆K + ∆U + Wa = 0
23 m
1.1 cm
Determine the magnitude of the energy dissipated by air friction during the dart’s ascent.
The acceleration of gravity is 9.81 m/s2 .
1
m vf 2 − m g h + Wa = 0 .
2
Since
2 (m g h − Wa )
m
2 (0.01 kg) (9.81 m/s2 ) (23 m)
=
0.01 kg
2 (2.07879 J)
−
0.01 kg
= 35.5014 m2 /s2 ,
vf2 =
Explanation:
Let :
k = 2934 N/m ,
x = 1.1 cm = 0.011 m ,
m = 10 g = 0.01 kg ,
h = 23 m , and
g = 9.81 m/s2 .
Applying conservation of energy during the
dart’s ascent (k0 = kf = 0 , v0 = 0) ,
Wext = ∆K + ∆U + Wa = 0
1
− k x2 + m g h − Wa = 0
2
1
k x2
2
= (0.01 kg) (9.81 m/s2 ) (23 m)
1
− (2934 N/m) (0.011 m)2
2
= 2.07879 J .
vf =
q
35.5014 m2 /s2 = 5.95831 m/s .
008 (part 1 of 3) 10.0 points
A 18 kg block is dragged over a rough, horizontal surface by a constant force of 131 N
acting at an angle of 32 ◦ above the horizontal. The block is displaced 15.7 m, and the
coefficient of kinetic friction is 0.125.
The acceleration of gravity is 9.8 m/s2 .
13
Wa = m g h −
18 kg
µ = 0.125
Find the work done by the 131 N force.
and the energy dissipated is 2.07879 J .
007 (part 2 of 2) 10.0 points
What is the maximum speed the projectile
could have when it returns to its starting
point? Assume the difference in the frictional
energy loss due to the diminished velocity on
the way down compared to its velocity on the
way up is negligible.
Explanation:
Explanation:
1N
◦
32
Let : m = 18 kg ,
F = 131 N ,
sx = 15.7 m ,
µ = 0.125 , and
g = 9.8 m/s2 .
Consider the force diagram
aguilar (fa6754) – hk7 – opyrchal – (11106)
5
If the block was originally at rest, determine
its final speed.
n
F
θ
fk
Explanation:
The net work done on the block is equal to
the change in kinetic energy, so
mg
~ · ~s, where ~s is the distance
Work is W = F
traveled. In this problem ~s = 5ı̂ is only in the
x direction.
1
m v 2 − 0 = WF + Wµ
2
2 [WF + Wµ ]
v h m
i
u
u 2 (1744.18 J) + (−209.949 J)
t
=
(18 kg)
v=
⇒ WF = Fx sx
= F sx cos θ
= (131 N) (15.7 m) cos 32◦
= 1744.18 J .
009 (part 2 of 3) 10.0 points
Find the work done by the force of friction.
Explanation:
To find the frictional force, Ff riction = µ N ,
we need to find N from vertical force balance.
Note that N is in the same direction as the
y component of F and opposite the force of
gravity. Thus
F sin θ + N = m g
so that
N = m g − F sin θ .
Thus the friction force is
~ f riction = −µ N ı̂ = −µ (m g − F sin θ)ı̂ .
F
The work done by friction is then
~ f riction · ~s = −|fµ | |s|
Wµ = F
= −µ (m g − fµ sin θ) sx
= −(0.125) [(18 kg) (9.8 m/s2 )
−(131 N) sin 32◦ ] (15.7 m)
= −209.949 J .
010 (part 3 of 3) 10.0 points
r
= 13.0564 m/s .
```