Download University Physics AI No. 3 Newton`s Laws of Motion

University Physics AI
No. 3 Newton’s Laws of Motion
Class
Number
Name
I．Choose the Correct Answer
1. Which statement is most correct?
( C )
(A) Uniform circular motion causes a constant force toward the center.
(B) Uniform circular motion is caused by a constant force toward the center.
(C) Uniform circular motion is caused by a constant magnitude net force toward the center.
(D) Uniform circular motion is caused by a constant magnitude net force away the center.
Solution: For uniform circular motion, the speed or the angular speed of the particle is a constant.
Fc = mac = m
According to
v2
= mω 2 r ,
r
the magnitude of the force toward the center should be constant.
r
2. Suppose the net force F on an object is a nonzero constant. Which of the following could also
be constant?
(A) Position.
( D )
(B) Speed.
(C) Velocity.
(D) Acceleration.
r
r
Solution: According to Newton’s second law F = ma , the answer is (D).
3. Which of the graph in Figure 1 best shows the velocity-time graph for an object launched
vertically into the air when air resistance is given by D=bv? The dashed line shows the velocity
graph if there was no air resistance.
( B )
v
v
v
t
(A)
v
t
(B)
t
(C)
t
(D)
Solution: During the course of the object moves up, the acceleration of the object thinking about air
resistance is more than that no air resistance, that is that the tangent of the graph should be more
than that no air resistance. Thus we remove the answer (D).
During the course of the object moves down, when t → ∞ , a → 0 , v → constant . Thinking
about this, the right answer is (B)
II. Filling the Blanks
r
1. A 2.50 kg system has an acceleration a = ( 4.00 m/s ) iˆ . One of the forces acting on the system
2
is (3.00 N ) iˆ − (6.00 N ) ˆj . Is this the total force on the system?
No
. If not, the other force
acting on the system is 7.00 iˆ + 6.00 ˆj ( N) .
v
r
Solution: The total force is F = ma = 2.5 × 4.00 iˆ = 10.0 iˆ( N)
v
v
The other force is F = Ftotal − (3.00 iˆ − 6.00 ˆj ) = 7.00 iˆ + 6.00 ˆj ( N )
2. Two masses, m1 and m2, hang over an ideal pulley and the system is free to
move (see Figure 1). Such an arrangement is called an Atwood’s machine. The
m2 − m1
g.
m1 + m2
r
magnitude of the acceleration a of the system of two masses is
The magnitude of the tension in the cord is
2m1m2
g.
m1 + m2
m1
m2
Fig.1
Solution: The second law force diagrams of two mass are shown in figure.
Assume the mass m2 moves downward. Apply Newton’s second law of motion, we have
⎧m2 g − T = m2 a
⎨
⎩T − m1 g = m1a
v
T
Solving it, we can get
a=
m2 − m1
g
m1 + m2
T=
2m1m2
g
m1 + m2
3. Wahoo! You are swinging a mass m at speed v
around on a string in circle of radius r whose plane is
1.00 m above the ground (see Figure 2). The string
makes an angle θ with the vertical direction. (a) Make
a second law force diagram about the mass and
indicate the direction to the center of its circular path.
(b) The direction of the acceleration of the mass is
pointing to O . (c) Apply Newton’s second law to
the horizontal and vertical direction to calculate the
v
T
m1
m2
v
m1 g
v
m2 g
v
a
y
v
T r
m
v
mg
θ
O
1.00m
O′
Fig. 2
x
v2
angle θ is θ = arctan
.(d) If the angle θ = 47.4° and the radius of the circle is 1.50 m, the speed
gr
of the mass is
4.0m/s . (e) If the mass is 1.50 kg, the magnitude of the tension in the string is
21.7N . (f) The string breaks unexpectedly when the mass is moving exactly eastward. The
location the mass will hit the ground is 1.8m to the point O´ .
Solution: The second law force diagram is shown in figure.
(c) Apply Newton’s second law of motion, we have
⎧T cos θ − mg = 0
⎪
⎨
mv 2
⎪T sin θ =
r
⎩
Solving the equations, we have
(d)
θ = arctan
v2
gr
Then
v2
.
gr
⇒ v = gr tan θ = 9.8 × 1.5 × tan 47.4 o = 4.0 m/s
(e) T cos θ − mg = 0
(f) Using
θ = arctan
⇒ T=
1.5 × 9.8
mg
=
= 21.7 N
cos θ cos 47.4 o
⎧ x = vt
⎪
⎨
1 2 , in which y=1.0m.
y
gt
=
⎪⎩
2
x=v
2y
2 ×1
= 4×
= 1.8 m
g
9.8
III. Give the Solutions of the Following Problems
1. The static and kinetic coefficients of friction for a 50 kg
M=50kg
mass on a table surface are µs=0.20 and µk=0.15. See
Figure 3. The pulley is ideal. The system is at rest in
equilibrium. (a) For each mass, sketch a second law force
diagram indicating schematically all forces that are acting
on each mass. (b) How large can the mass m be and not
move the system? What the magnitude of the frictional
m
force on the 50 kg mass when this situation exists? (c) If m
Fig.3
is only half the maximum mass calculated in part (b), what
is the magnitude of the frictional force on the 50 kg block? Indicate whether this is a force of static
friction or a force kinetic friction.
Solution:
(a) The second law force diagrams of two mass are shown in figure.
v
f
v
N
v
T
v
T
x
v
Mg
v
mg
y
(b) Assume that the system does not move, so we have
⎧mg − T = 0
⎪T − f = 0
⎪
s
⎨
⎪ f s = Nµ s
⎪⎩Mg − N = 0
(c) If m ′ =
⎧m = µ s M = 0.2 × 50 = 10(kg )
⎪
⇒ ⎨ f s = Mgµ s = 50 × 9.8 × 0.2 = 98.0( N)
⎪or f = T = mg = 10 × 9.8 = 98.0( N)
s
⎩
1
m = 5kg , assume the system does not move, then
2
⎧m′g − T = 0
⎨
⎩T − f s′ = 0
⇒ T = f s′ =
1
1
mg = × 10 × 9.8 = 49.0 ( N)
2
2
As the maximum static friction force is
f s max = Mgµ s = 50 × 9.8 × 0.2 = 98.0( N)
From the above, we know
T < f s max
Thus the magnitude of the frictional force on the 50 kg block is 49N. The system does not move, so
the friction force is a static friction force.
2. In the system shown in Fig.4, a block (of mass
m1=9.5kg) slides on a frictionless plane inclined
at an angle
θ = 34 o . The block is attached by a
m1
string to a second block (of mass m2=2.6kg). The
m2
system is released from rest. (a) Find the
acceleration and the tension in the string. (b) If
taking into account a frictional force between
block 1 and the plane, what is the acceleration of
the tension in the string? Use the values
µs=0.24 and µk=0.15
34o
Fig.4
Solution:
(a) The second law force diagrams of two mass are
shown in figure. Set the coordinate system as in
figure.
v
Assuming the acceleration is a shown in figure.
Apply Newton’s second law of motion, we have
y
v
N
⎧T − m1 g sin θ = m1 a
⎪
⎨ N − m1 g cos θ = 0
⎪m g − T = m a
2
⎩ 2
v
T
v
T
x
m1
m2
v
m1 g
v
m2 g
v
a
Solving it, we get
a=
m2 − m1 sin θ
2.6 − 9.5 × sin 34 o
g=
= −2.2 m/s 2
m1 + m2
9.5 + 2.6
T=
m1 m2 g
9.5 × 2.6 × sin 34 o
(1 + sin θ ) =
× (1 + sin 34 o ) = 31 N
9.5 + 2.6
m1 + m2
(b) The second law force diagrams of two mass are
shown in figure. Set the coordinate system as in
figure.
v
Assuming the acceleration is a shown in figure.
Apply Newton’s second law of motion, we have
⎧m1 g sin θ − T − f = m1 a
⎪ N − m g cos θ = 0
⎪
1
⎨
⎪T − m2 g = m2 a
⎪⎩ f = µ k N
y
v
N
m1
v x
T
v
f av
v
m1 g
v
T
v
a
m2
v
m2 g
Solving the equations, we get
a=
m1 (sin θ − µ k cos θ ) − m2
9.5 × (sin 34 o + 0.15 cos 34 o ) − 2.6
g=
= 1.2 m/s 2
m1 + m2
9.5 + 2.6
m1m2 g
9.5 × 2.6 × sin34o
(1 + sinθ − µk cosθ ) =
× (1 + sin34o − 0.15× cos34o ) = 29 N
T=
9.5 + 2.6
m1 + m2
3. An object is drop from rest. Find the terminal speed assuming that the drag
force is given by D = bv .
Solution:
The forces acting on the object and the coordinate system are shown in figure.
Apply Newton’s second law of motion, we have
2
v
D
ĵ
v
mg
D − mg = − ma
⇒ bv 2 − mg = − m
1
⇒
v2 −
⇒
∫
mg
b
dv = −
1
v
dv
dt
b
dt
m
dv = −
b t
dt
m ∫0
mg
b
vy
1
b t
⇒ ∫
dv = − ∫ dt
0
mg
m 0
v2 −
b
0
v2 −
⇒
mg
1
b
b
ln(
)=− t
m
mg
mg
2
v+
b
b
⇒
mg
−2
b
=e
mg
v+
b
v−
v−
bg
t
m
mg
b
When t → ∞ , the terminal speed of the object is v =
4. Assume the kinetic frictional force on a falling mass m is proportional to its speed v, with a
proportionality constant β. Choose ĵ to be vertically downward.
(a) Show that Newton’s second law of motion yields m
(b)When the mass reaches its terminal sped, what is
(c) Show that the terminal speed is vterm =
mg
β
dv y
dv y
dt
dt
= mg − βv y .
?
.
(d) If the mass is dropped from rest, show that v y (t ) = vterm (1 − e
−β t / m
)
Solution:
The forces acting on the mass are shown in figure. Apply Newton’s second
law of motion, we have
v
f
(a) Apply Newton’s second law of motion mg − f = ma , we have
mg − βv y = ma y = m
m
That is
dv y
ĵ
dv y
v
mg
dt
= mg − β v y
dt
(b) When the mass reaches its terminal speed, the acceleration does not change, so
mg − f = ma = 0 , thus mg − β v y = m
dv y
(c) according to the answer in (b), we know
vterm =
The terminal speed is
mg
β
dt
dv y
dt
= 0 , that is
= 0 . So mg − βv y = 0 .
.
(d) If the mass is dropped from rest, then
m
dv y
dt
⇒∫
vy
0
= mg − β v y
d(mg - βv y )
mg - βv y
mdv y
⇒
=−
mg − βv y
β
t
dt
m∫
0
⇒ ln(mg - β v y ) − ln(mg ) = − βt
⇒ ln(1 −
⇒ 1−
β
β
mg
vy ) = − β t
vy = e
−β t
= dt
m
m
m
mg
mg
−β t
−β t
⇒ vy =
(1 − e m ) = vterm (1 − e m )
β
dv y
dt
=0.