Download Grade Level Physics Dynamics Review Quiz

KEY for Forces Review
1. A(n) __NET___ force causes an object to start moving, stop moving, or change directions, which means
it causes the object to __ACCELERATE_______.
2. An unbalanced force is also known as a(n) _NET__ force.
3. Label the variable used for each force.
a. Frictional Force - Ff
d. Applied Force- Fa
b. Gravitational Force - Fg
e. Weight- Fw
c. Force of Tension - FT
f. Normal Force - Fn
4. The gravitational force always points (up / down / left / right).
5. Objects that are at rest or moving with constant velocity have acceleration equal to _ZERO_ m/s2 and
therefore the objects are in __EQUILIBRIUM__.
6. _INERTIA__ is the tendency of an object to resist changes in motion or the tendency to maintain its
current motion.
7. If an object is moving at constant velocity, an additional force of _ZERO_ N is needed to bring it into
equilibrium because an object in equilibrium must be (accelerating / moving at constant velocity) and
for an object to be in equilibrium the net force acting on it must be equal to _ZERO_ N.
8. Newton’s (1st / 2nd / 3rd) law is the relationship between force, mass, and acceleration.
9. Newton’s (1st / 2nd / 3rd) law says that for every action there is an equal and opposite reaction.
10. Newton’s (1st / 2nd / 3rd) law says that an object at rest will stay at rest until an unbalanced force acts on
it.
11. A rolling soccer ball will continue to roll in a straight line until an unbalanced force acts on it is an
example of Newton’s (1st / 2nd / 3rd) Law.
12. By pushing on it, you are able to accelerate a Volkswagen Rabbit more than you can a Cadillac. This is
an example of Newton’s (1st / 2nd / 3rd) Law.
13. When you step off of a boat and onto the pier, the boat moves away from the pier. This is an example of
Newton’s (1st / 2nd / 3rd) Law.
14. If an object is moving to the left, then the frictional force will be to the ___RIGHT___.
15. A bug hits the windshield of the car. The magnitude of the force the bug experiences is ( < / > / = ) to
the magnitude of the force the windshield experiences.
16. As an apple falls from a tree, the earth exerts a force on the apple and the _APPLE _ exerts an equal but
opposite force on the _EARTH _ according to Newton’s (1st / 2nd / 3rd) Law.
17. An object with (more / less) mass has less inertia.
18. Which has more inertia? (loaded dump truck going 30 mph / Volkswagen rabbit going 80 mph)
19. The weight of a 2 kg object is __19.6___ Newtons.
20. Newton’s (1st / 2nd / 3rd) law explains motion when only balanced forces are acting.
21. Newton’s (1st / 2nd / 3rd) law explains motion when unbalanced forces are acting.
Use the following velocity time graph to answer questions 22 and 25.
Velocity
(m/s)
D
B
C
A
Time (sec)
22. In which section(s) of the velocity time graph is the object moving at a constant velocity? _C__
23. In which section(s) of the velocity time graph are balanced forces acting? _C___
24. In which section(s) of the velocity time graph does the object have a nonzero acceleration? _A, B, D_
25. In which section(s) of the velocity time graph is Newton’s 2nd Law being illustrated? __ A, B, D__
Use the following diagram to answer question 26 - 27.
30 N
100 N
2 kg
60 N
26. The net force acting on the 2 kg object shown above is _10_ N to the (left / right).
27. The acceleration of the 2 kg object shown above is _5__ m/s2 to the (left / right).
Use the following diagram to answer question 28.
FN
FW
28. For the object shown in the free body diagram above to accelerate upward, the normal force (FN) must
be (greater than / less than / equal to) the weight (FW) of the object.
29. According to Newton’s 2nd Law, as the amount of net force acting on an object increases the
acceleration (increases / decreases / remains the same).
30. The sliding friction is always (greater than / less than / equal to) the static friction.
Match the following free body diagrams with the situations that match them and tell if they are in
equilibrium (yes or no).
B
A
C
D
31. A box being pulled by a horizontal force across level ground at a constant velocity (assume friction). A
yes
32. A bowling ball rolling down the frictionless lane at constant velocity. C yes
33. A ping pong ball flying through the air toward the target (with out friction). D no
34. A tire hanging from a rope on a tree. C yes
35. A parachutist falling toward the earth at constant velocity. C yes
36. A crate is being accelerated across frictionless ice by a horizontal force. B no
37. The gravitation force of weight always points down (direction).
38. The tension in a rope or chain points (towards / away from) the object.
Newton’s Law Problems KEY
39. When a karate expert strikes a wooden block, the hand undergoes an acceleration of -6500 m/s2.
Medical data indicates the mass of the forearm and hand to be about 0.7 kg. What is the force exerted
on the hand by the blocks? What is its direction?
Fnet  ma
Fnet  (0.7)( 6500)
4550 N, upward
Fnet  4550 N
40. A 60 kg boy and a 40 kg girl use an elastic rope while engaged in a tug of war on an icy frictionless
surface. If the acceleration of the girl toward the boy is 3 m/s2, determine the magnitude of the
acceleration of the boy toward the girl.
Fnet (boy)   Fnet ( girl)
m boy a boy  m girl a girl
(60)a boy  (40)(3)
The magnitude of the boy’s acceleration is 2 m/s2.
60a boy  120
a boy  2m / s 2
41. A 95 kg boxer has his first match in the Canal Zone (g = 9.782 m/s2) and his second match at the North
Pole (g = 9.832 m/s2). What is his mass in the Canal Zone? What is his weight in the Canal zone?
What is the mass at the North Pole? What is his weight at the North Pole?
Mass is independent of location so is mass is 95 kg in the Canal Zone and in at the North Pole.
Fnet  ma
Fnet  ma
Fnet  (95)(9.782)
Fnet  (95)(9.832)
Fnet  929 N
Fnet  934 N
42. The acceleration of a 925 kg car while braking is -12.2 m/s2. What is the frictional force that brings the
car to a stop?
(925)(9.8)
Fnet  ma
11285 N
Right  Left  ma
0  FF  (925)( 12.2)
(925)(9.8)
 FF  11285
FF  11285 N
43. After a day of testing race cars, you decide to take your own 1550 kg car onto the test track. While
moving down the track at 10 m/s, you suddenly accelerate to 30 m/s in 10 sec. What is the average net
force that you have applied to the car in the 10 second interval?
FN = (1550)(9.8)
a
3100 N
v f  vi
t
30  10
a
10
a  2m / s 2
Fnet  ma
Right  Left  ma
Fapplied  0  (1550)( 2)
Fapplied  3100 N
FW = (1550)(9.8)
44. A 4500 kg helicopter accelerates upward at 2 m/s2. What lift force is exerted by the air on the
propellers?
Fapplied = 53100 N
Fnet  ma
Up  Down  ma
Fapplied  (4500)(9.8)  (4500)( 2)
Fapplied  (4500)( 2)  (4500)(9.8)
FW = (4500)(9.8)
Fapplied  53100 N
45. If you use a horizontal force of 30 N to slide a 12 kg wooden crate across a floor at a constant velocity,
what is the frictional force between the crate and the floor?
Fnet  ma
(12)(9.8)
Right  Left  ma
FF = 30 N
30 N
30  FF  12(0)
30  FF  0
 FF  30
(12)(9.8)
FF  30 N
46. You are driving a 2500 kg car at a constant speed of 14 m/s along an icy, but straight and level road.
You slam on the brakes and the wheels lock. Sending you sliding and skidding to a halt in a distance of
25 m. What is the magnitude of the frictional force between your tires and the icy road?
v 2f  vi2
a
(2500)(9.8)
Fnet  ma
2 d
0 2  14 2
2(25)
 196
a
50
a  3.92m / s 2
a
FF = 9800 N
(2500)(9.8)
Right  Left  ma
0  FF  (2500)( 3.92)
 FF  9800
FF  9800 N
47. A 200 kg crate is pushed horizontally with a force of 700 N. If the frictional force is 392 N, calculate
the acceleration of the crate.
Fnet  ma
(200)(9.8)
700  392  200a
308  200a
FF = 392 N
Fapplied = 700 N
(200)(9.8)
308
a
200
1.54m / s 2  a