Transcript PPT from review

Aim: How can we apply Newton’s
2nd Law of Acceleration?
Do Now:
An object with mass m is
moving with an initial
velocity vo and speeds up
to a final velocity of v in
time t when an
unbalanced force F is
applied to it. From this
information, derive
Newton’s 2nd Law, F = ma
Newton’s 2nd Law
Law of Acceleration
Unbalanced forces produce acceleration
ΣF = ma
The key to solving these problems is in the
free-body diagram
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**1 minute**
1. Which of the following best indicates the direction of the net force, if any, on the
ball at point Q ?
Horizontal motion is constant
(A)
The only acceleration is vertical due to
gravity
(B)
Gravity only goes down
(C)
(D)
(E) There is no net force on the ball at point Q.
2. A ball falls straight down through the air under the
influence of gravity. There is a retarding force F on
the ball with magnitude given by F = bv, where v is
the speed of the ball and b is a positive constant. The
magnitude of the acceleration a of the ball at any time
is equal to which of the following?
(A) g - b
F = bv
(B) g - bv/m
(C) g + bv/m
(D) g/b
(E) bv/m
Fg = ma
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**1 min 15 sec**
A 2-kilogram block slides down a 30° incline
as shown at the right with an acceleration of
2 meters per second squared.
3. Which of the following diagrams best represents the
gravitational force W. the frictional force f, and the
normal force N that act on the block?
4. The magnitude of the frictional force along the plane is most nearly
(A)2.5 N
ΣF = ma
(2 kg)(10 m/s2)sin30 – FF = (2 kg)(2 m/s2)
(B)5 N
10 – FF = 4
F║ - FF = ma
(C)6 N
FF = 6 N
mgsinθ – FF = ma
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(D)10 N
(E)16 N
**2 min 30 sec**
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**2 min 30 sec**
5.
FNet = Fll
Sinθ = 10/20
FNet = mgsinθ
6.
FNet = (2 kg)(10 m/s2)(10/20)
v2 = v02 + 2ax
FNet = 10 N
v2 = 2ax
v2 = 2(FNet/m)x
v2 = 2(10 N/2 kg)20 m
v = 14 m/s
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**1 min 15 sec**
7. When the frictionless system shown above is
accelerated by an applied force of magnitude F, the
tension in the string between the blocks is
(A)2F
(B)F
Each block gets its own free-body
1 kg
T
T
F
2 kg
ΣF = ma
ΣF = ma
T = ma
F - T = ma
(D)1/2 F
T = 1a
F - T = 2a
(E)1/3 F
Add the two
equations together
to cancel out a
variable
(C)2/3 F
F – T = 2a
T = 1a
T = 1a
F = 3a
F = 3a
Therefore, T = 1/3 F
8. A helicopter holding a 70-kilogram package suspended
from a rope 5.0 meters long accelerates upward at a rate of
5.2 m/s2. Neglect air resistance on the package.
a. On the diagram below, draw and label all of the forces
acting on the package.
T
b. Determine the tension in the rope.
Fg
c. When the upward velocity of the
helicopter is 30 meters per second, the
rope is cut and the helicopter continues to
accelerate upward at 5.2 m/s2. Determine
the distance between the helicopter and the
package 2.0 seconds after the rope is cut.
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**7.5 min**
b.
ΣF = ma
T – mg = ma
T = ma + mg
T = m(a + g)
T = 70 kg(5.2 m/s2 + 9.8 m/s2)
T = 1050 N
c.
y = vot + ½ at2
yHelicopter = (30 m/s)(2 s) + ½ (5.2 m/s2)(2 s)2 = 70.4 m
yPackage = (30 m/s)(2 s) - ½ (9.8 m/s2)(2 s)2 = 40.4 m
yH – yP + 5 m = 30 m + 5 m = 35 m
9. Part of the track of an amusement park roller coaster is shaped as
shown above. A safety bar is oriented lengthwise along the top of each
car. In one roller coaster car, a small 0.10-kilogram ball is suspended
from this bar by a short length of light, inextensible string.
a. Initially, the car is at rest at point A.
i. On the diagram to the right, draw and
label all the forces acting on the ΣF = 0
0.10-kilogram ball.
T – mg = 0
T
mg
T = mg
ii. Calculate the tension in the
string.
T = (0.1 kg)(9.8 m/s2)
Calculator
T = 0.98 N
**12 min**
The car is then accelerated horizontally, goes up a 30°
incline, goes down a 30° incline, and then goes around
a vertical circular loop of radius 25 meters. For each of
the four situations described in parts (b) to (d), do all
three of the following. In each situation, assume that
the ball has stopped swinging back and forth.
1)Determine the horizontal component Th of the
tension in the string in newtons and record your
answer in the space provided. 2)Determine the
vertical component Tv of the tension in the string in
newtons and record your answer in the space
provided. 3)Show on the adjacent diagram the
approximate direction of the string with respect to the
vertical. The dashed line shows the vertical in each
situation.
b. The car is at point B moving horizontally 2 m/s to
the right with an acceleration of 5.0 m/s2
Th = 0.5 N
ΣFy = 0
Tv – mg = 0
Tv = 0.98 N
ΣFx = ma
Th = ma
Th = (0.1 kg)(5 m/s2)
Th = 0.5 N
Tv = mg
Tv = (0.1 kg)(9.8 m/s2)
Tv = 0.98 N
c. The car is at point C and is being pulled up the
30° incline with a constant speed of 30 m/s.
Th = 0 N
ΣF = 0
y
Tv = 0.98 N
Tv – mg = 0
Tv = mg
ΣFx = 0
Tv = (0.1 kg)(9.8 m/s2)
Th = 0 N
Tv = 0.98 N
Velocity is constant
Same as aii
d. The car is at point D moving down the incline with
an acceleration of 5.0 m/s2 .y-comp of tension counters only part of
the weight resulting in vertical
0.43 N
Th =
acceleration
Tv =
ΣFy = ma
-0.75 N
Tv = Fsinθ - mg
Th = masinθ - mg
Th = m(asinθ – g)
X-comp of tension is responsible
Th = 0.1 kg[(5 m/s2)(sin30) – 9.8 m/s2]
for the x-comp of the acceleration
ΣFx = ma
Th = -0.75 N
Fsinθ
Th = Fcosθ
Th = macosθ
Fcosθ
Th = (0.1 kg)(5 m/s2)cos30
Th = 0.43 N
mg