Download QUICK QUIZZES 1. Newton`s second law says that the acceleration

QUICK QUIZZES
1.
Newton’s second law says that the acceleration of an object is directly proportional to the resultant (or net)
force acting on. Recognizing this, consider the given statements one at a time: (a) True— If the resultant
force on an object is zero (either because no forces are present or the vector sum of the forces present is zero),
the object can still move with constant velocity. (b) False — An object that remains at rest has zero
acceleration. However, any number of external forces could be acting on it, provided that the vector sum of
these forces is zero. (c) True — When a single force acts, the resultant force cannot be zero and the object
must accelerate. (d) True — When an object accelerates, a set containing one or more forces with a nonzero
resultant must be acting on it. (e) False — Many external forces could be acting on an object with zero
acceleration, provided that the vector sum of these forces is zero. (f) False — If the net force is in the positive
x-direction, the acceleration will be in the positive x-direction. However, the velocity of an object does not
have to be in the same direction as its acceleration (consider the motion of a projectile).
2.
(b). The newton is a unit of weight, and the quantity (or mass) of gold that weighs 1 newton is m   1 N  g .
Because the value of g is smaller on the Moon than on the Earth, someone possessing 1 newton of gold on the
Moon has more gold than does a person having 1 newton of gold on Earth.
3.
(c) and (d). Newton’s third law states that the car and truck will experience equal magnitude (but oppositely
directed) forces. Newton’s second law states that acceleration is inversely proportional to mass when the
force is constant. Thus, the lower mass vehicle (the car) will experience the greater acceleration.
4.
(c). In case (i), the scale records the tension in the rope attached to its right end. The section of rope in the
man’s hands has zero acceleration, and hence, zero net force acting on it. This means that the tension in the
rope pulling to the left on this section must equal the force F the man exerts toward the right on it. The scale
reading in this case will then be F. In case (ii), the person on the left can be modeled as simply holding the
rope tightly while the person on the right pulls. Thus, the person on the left is doing the same thing that the
wall does in case (i). The resulting scale reading is the same whether there is a wall or a person holding the
left side of the scale.
5.
(c). The tension in the rope has a vertical component that supports part of the total weight of the woman and
sled. Thus, the upward normal force exerted by the ground is less than the total weight.
6.
(b). Friction forces are always parallel to the surfaces in contact, which, in this case, are the wall and the
cover of the book. This tells us that the friction force is either upward or downward. Because the tendency of
the book is to fall downward due to gravity, the friction force must be in the upward direction.
7.
(b). The static friction force between the bottom surface of the crate and the surface of the truck bed is the net
horizontal force on the crate that causes it to accelerate. It is in the same direction as the acceleration, toward
the east.
8.
(b). It is easier to attach the rope and pull. In this case, there is a component of your applied force that is
upward. This reduces the normal force between the sled and the snow. In turn, this reduces the friction force
between the sled and the snow, making it easier to move. If you push from behind, with a force having a
downward component, the normal force is larger, the friction force is larger, and the sled is harder to move.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1.
Newton’s second law gives the net force acting on the crate as


Fnet  95.0 N  fk   60.0 kg  1.20 m s2  72.0 N
This gives the kinetic friction force as fk  23.0 N , so choice (a) is correct.
2.
Since the pedestal is in equilibrium, the normal force pressing upward on its base supports the total weight of
the man and pedestal. Therefore,
or
 
n  Fg
man
 
 Fg
pedestal




 mman  m pedestal g   97.0 kg  9.80 m s2  951 N
showing that (e) is the correct choice.
3.
The tension, Fupper , in the vine at the point above the upper monkey must support the weight of both

monkeys  i.e., Fupper  2 Fg

 

single 
monkey 
. However, the tension in the vine between the two monkeys supports

only the weight of the lower monkey  Flower  Fg

 

single 
monkey 
, meaning that Fupper Flower  2 and choice (d)
is correct.
4.
Because the block has zero vertical acceleration, Newton’s second law says that
Fy  F sin  30.0   mg  n  0
or


n    70.0 N  sin   30.0     8.00 kg  9.80 m s2   35.0 N  78.4 N  113 N
and we see that (c) is correct.
5.
From Newton’s law of universal gravitation, the force Earth exerts on an object on its surface is
Fg  GM E m RE2  mg , or the acceleration of gravity at Earth’s surface is g  GM E RE2 . If both the mass
and radius of the Earth should double, so M E  2 M E and RE  2 RE , the acceleration of gravity at the
surface would then be
g  G
M E
 RE 
2
 2 ME 
1 M 
g
9.80 m s2
 G
  G 2E  

 4.90 m s2

2
2  RE 
2
2
 4 RE 
meaning that (b) is the correct answer.
6.
When the crate is held in equilibrium on the incline as shown in the sketch, Newton’s second law requires


that Fx  Fy  0 . From  Fx   F g  f s  0 , the magnitude of the friction force equals the
x
component of gravitational force acting down the incline, or choice (e) is correct. Note that


fs  fs
  s n only when the crate is on the verge of starting to slide.
max
7.
According to Newton’s third law, the force of reaction (in this case, the force exerted on the rock by the
glass) is always equal in magnitude to the force of action (in this case, the force the rock exerts on the glass).
Thus, (b) is the correct choice.
8.
The box will accelerate in the direction of the resultant force acting on it. The only horizontal forces present
are the force exerted by the manager and the friction force. Since the acceleration is given to be in the
direction of the force applied by the manager, the magnitude of this force must exceed that of the opposing
friction force, however the friction force is not necessarily zero. Thus, choice (b) is correct.
9.
Choose a coordinate system with the positive x-direction being east and the positive y-direction being north.
Then the components of the four given forces are:
Ax   40 N, Ay  0
C x  70 N, C y  0
Bx  0, By  50 N
Dx  0, Dy   90 N
The components of the resultant (or net) force are Rx  Ax  Bx  C x  Dx  30 N and
Ry  Ay  By  C y  Dy  40 N . Therefore, the magnitude of the net force acting on the object is
R 
Rx2  Ry2 
 30 N 2   40 N 2
 50 N ,
or choice (a) is correct.
10.


Constant velocity means zero acceleration. From Newton’s second law,  F  ma , so the total (or resultant)
force acting on the object must be zero if it moves at constant velocity. This means that (d) is the correct
choice.
11.



An object in equilibrium has zero acceleration a  0 , so both the magnitude and direction of the object’s
velocity must be constant. Also, Newton’s second law states that the net force acting on an object in
equilibrium is zero. The only untrue statement among the given choices is (d), untrue because the value of the
velocity’s constant magnitude need not be zero.
12.
As the trailer leaks sand at a constant rate, the total mass of the vehicle (truck, trailer and remaining sand)
decreases at a steady rate. Then, with a constant net force present, Newton’s second law states that the
magnitude of the vehicle’s acceleration  a  Fnet m  will steadily increase. Choice (b) is the correct answer.
13.
When the truck accelerates forward, its natural tendency is to slip from beneath the crate, leaving the crate
behind. However, friction between the crate and the bed of the truck acts in such a manner as to oppose this
relative motion between truck and crate. Thus, the friction force acting on the crate will be in the forward
horizontal direction and tend to accelerate the crate forward. The crate will slide only when the coefficient of
static friction is inadequate to prevent slipping. The correct response to this question is (c).
14.
The mass of an object is the same at all locations in space (e.g., on Earth, the Moon, or space station).
However, the gravitational force the object experiences  weight, w  mg  does vary, depending on the
acceleration of gravity g at the object’s current location in space. It is the gravitation attraction of the Earth
that holds the space station (and all its contents, including astronauts) in orbit around Earth. The only correct
choice listed is (b).
15.
Assuming that the cord connecting m1 and m2 has constant length, the two masses are a fixed distance
(measured along the cord) apart. Thus, their speeds must always be the same, which means that their
accelerations must have equal magnitudes. The magnitude of the downward acceleration of m2 is given by
Newton’s second law as
a2 
 Fy
m2

 T 
m2 g  T
 g
 g
m2
 m2 
where T is the tension in the cord, and downward has been chosen as the positive direction. Therefore, the
only correct statement among the listed choices is (a).
16.
Only forces which act on the object should be included in the free-body diagram of the object. In this case,
these forces are (1) the gravitational force (acting vertically downward), (2) the normal force (perpendicular
to the incline) exerted on the object by the incline, and (3) the friction force exerted on the object by the
incline, and acting up the incline to oppose the motion of the object down the incline). Choices (d) and (f) are
forces exerted on the incline by the object. Choice (b) is the resultant of forces (1), (2), and (3) listed above,
and its inclusion in the free-body diagram would duplicate information already present. Thus, correct answers
to this question are (b), (d), and (f).
ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUSTIONS
2.
w = mg and g decreases with altitude. Thus, to get a good buy, purchase it in Denver. If gold were sold by
mass, it would not matter where you bought it.
4.
If it has a large mass, it will take a large force to alter its motion even when floating in space. Thus, to avoid
injuring himself, he should push it gently toward the storage compartment.
6.
The barbell always exerts a downward force on the lifter equal in magnitude to the upward force that she
exerts on the barbell. Since the lifter is in equilibrium, the magnitude of the upward force exerted on her by
the scale (that is, the scale reading) equals the sum of her weight and the downward force exerted by the
barbell. As the barbell goes through the bottom of the cycle and is being lifted upward, the scale reading
exceeds the combined weights of the lifter and the barbell. At the top of the motion and as the barbell is
allowed to move back downward, the scale reading is less than the combined weights. If the barbell is moving
upward, the lifter can declare she has thrown it just by letting go of it for a moment. Thus, the case is
included in the previous answer.
8.
The net force acting on the object decreases as the resistive force increases. Eventually, the resistive force
becomes equal to the weight of the object, and the net force goes to zero. In this condition, the object stops
accelerating, and the velocity stays constant. The rock has reached its terminal velocity.
10.
While the engines operate, their total upward thrust exceeds the weight of the rocket, and the rocket
experiences a net upward force. This net force causes the upward velocity of the rocket to increase in
magnitude (speed). The upward thrust of the engines is constant, but the remaining mass of the rocket (and
hence, the downward gravitational force or weight) decreases as the rocket consumes its fuel. Thus, there is
an increasing net upward force acting on a diminishing mass. This yields an acceleration that increases in
time.
12.
→
→
T
R
→
R
→
R
→
f
→
mg
→
→
mg
(a)
(b)
n
→
mg
(c)


In the free-body diagrams give above, R represents a force due to air resistance, T is a force due to the



thrust of the rocket engine, n is a normal force, f is a friction force, and the forces labeled m g are
gravitational forces.
14.
If the brakes lock, the car will travel farther than it would travel if the wheels continued to roll, because the
coefficient of kinetic friction is less than that of static friction. Hence, the force of kinetic friction is less than
the maximum force of static friction.