1. Aim: How can we apply Newton’s 2nd Law of Acceleration?
HW #5
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

2. 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

3. No Calculator **1 minute**
1. Which of the following best indicates the direction of the net force, if any, on the ball at point Q ?
(A)
(B)
(C)
(D)
(E) There is no net force on the ball at point Q.
Horizontal motion is constant
The only acceleration is vertical due to gravity
Gravity only goes down

4. 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
(B) g - bv/m
(C) g + bv/m
(D) g/b
(E) bv/m
F = bv
Fg = ma
No Calculator
**1 min 15 sec**

5. 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
2.5 N
5 N
6 N
10 N
16 N
ΣF = ma
F║ - FF = ma
mgsinθ – FF = ma
(2 kg)(10 m/s2)sin30 – FF = (2 kg)(2 m/s2)
10 – FF = 4
FF = 6 N
No Calculator
**2 min 30 sec**

6. No Calculator **2 min 30 sec** FNet = Fll FNet = mgsinθ5.
FNet = Fll
FNet = mgsinθ
FNet = (2 kg)(10 m/s2)(10/20)
FNet = 10 N
Sinθ = 10/20
v2 = v02 + 2ax
v2 = 2ax
v2 = 2(FNet/m)x
v2 = 2(10 N/2 kg)20 m
v = 14 m/s 6.

7. No Calculator
**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 2F F
2/3 F
1/2 F
1/3 F
Each block gets its own free-body
1 kg T T
2 kg F
ΣF = ma
T = ma
T = 1a
ΣF = ma
F - T = ma
F - T = 2a
T = 1a
F = 3a
Therefore, T = 1/3 F
Add the two equations together to cancel out a variable
F – T = 2a
T = 1a
F = 3a

8. b. Determine the tension in the rope.
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.
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. Fg
Calculator
**7.5 min**

9. b. ΣF = ma
T – mg = ma
T = ma + mg
T = m(a + g)
T = 70 kg(5.2 m/s 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

10. a. Initially, the car is at rest at point A.
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 0.10‑kilogram ball. T
ΣF = 0
T – mg = 0
T = mg
T = (0.1 kg)(9.8 m/s2)
T = 0.98 N mg
Calculator
**12 min**
ii. Calculate the tension in the string.

11. 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.

12. b. The car is at point B moving horizontally 2 m/s to the right with an acceleration of 5.0 m/s2
ΣFy = 0
Tv – mg = 0
Tv = mg
Tv = (0.1 kg)(9.8 m/s2)
Tv = 0.98 N
Th =
Tv =
0.5 N
0.98 N
ΣFx = ma
Th = ma
Th = (0.1 kg)(5 m/s2)
Th = 0.5 N

13. c. The car is at point C and is being pulled up the 30° incline with a constant speed of 30 m/s.
Tv =
0 N
ΣFy = 0
Tv – mg = 0
Tv = mg
Tv = (0.1 kg)(9.8 m/s2)
Tv = 0.98 N
Same as aii
0.98 N
ΣFx = 0
Th = 0 N
Velocity is constant

14. d. The car is at point D moving down the incline with an acceleration of 5.0 m/s2 .
Tv =
y-comp of tension counters only part of the weight resulting in vertical acceleration
ΣFy = ma
Tv = Fsinθ - mg
Th = masinθ - mg
Th = m(asinθ – g)
Th = 0.1 kg[(5 m/s2)(sin30) – 9.8 m/s2]
Th = N
0.43 N
-0.75 N
X-comp of tension is responsible for the x-comp of the acceleration
ΣFx = ma
Th = Fcosθ
Th = macosθ
Th = (0.1 kg)(5 m/s2)cos30
Th = 0.43 N
Fsinθ
Fcosθ mg