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Practice Problems, Chapter 4
Name: ___________________________


F
  ma
Weight  Fg  mg
Use separate sheets/ notebook to answer the questions. Show your work clearly
1. Why do you lunge forward when your car suddenly comes to a halt? Because your body wants to
continue to move in a straight line at constant speed. When the car comes to a sudden halt, the upper
part of the body continues forward (as predicted by Newton's first law) if the force exerted by the
lower back muscles is not great enough to give the upper body the same deceleration as the car. The
lower portion of the body is held in place by the force of friction exerted by the car seat and the floor
Why are you pressed backwards against a seat when your car rapidly accelerates? In your
explanations refer to Newton’s Laws. When the car rapidly accelerates, the upper part of the body
tries to remain at a constant velocity (again as predicted by Newton's first law). If the force provided
by the lower back muscles is not great enough to give the upper body the same acceleration as the car,
the upper body appears to be pressed backward against the seat as the car moves forward.
2. The net external force acting on an object is zero. Is it possible for the object to be traveling with a
velocity that is not zero? If your answer is yes, state what conditions must be placed on the magnitude
and direction of the velocity. If your answer is no, provide a reason for your answer.
YES. If the net external force acting on an object is zero, it is possible for the object to be traveling
with a nonzero velocity. According to Newton’s 1st and 2nd laws, if the net external force Fnet is zero,
the acceleration a is also zero. If the acceleration is zero, the velocity must be constant, both in
magnitude and in direction. Thus, an object can move with a constant nonzero velocity when the net
external force is zero.
3. Newton’s second Law indicates that when a net force acts on an object, it must accelerate. Does this
mean that when two or more forces are applied to an object simultaneously, it must accelerate?
Explain. No, if there are multiple forces acting on an object, they could all balance each other out or
all add (vectorally) to zero. If the NET force were zero, then the acceleration would be zero.

4. A 7.00 kg bowling ball experiences a net force of 5.00 N. What will be its
F  ma
acceleration? Acceleration would be in same direction as the force

a
 F  5  0.71m / s
m
7
5. An astronaut applies a force of 500 N to an asteroid and it accelerates at 7.00 m/s2. What is the

asteroid’s mass?
F  ma

m
 F  500  71kg
a
7
6. Acceleration due to gravity on the moon’s surface is 1/6th that on Earth. An astronaut’s life support
backpack weighs 300. lb on Earth. In pounds, what does it weigh on the moon? Because weight =
mg, it is proportional to the acceleration due to gravity. On the moon, where g is 1/6th that of Earth, a
person weighing 300lb on Earth would weight 1/6th of that on the moon – 50lb)
2
7. Determine the net force (magnitude and direction) acting on the objects below. Each object has a
mass of 5 kg. Determine the acceleration (magnitude and direction) if there is one.
10 N
10 N
Net Force: ______0_________
Acceleration: ____0_________
5N
10 N
Net Force: ____5N, left _______
Acceleration: __1 m/s2, left ______
7N
10 N
10 N
Acceleration: __4/5 m/s2 up ___
3N
60o
Net Force: ___4 N up ______
7.5 N
Net Force: ____15N right ______
Acceleration: ___3 m/s2 right _
13 N
8. Two ropes are attached to a 40.0 kg object. The first rope applies a force of 25.0 N and the second a
force of 40.0 N. If the two ropes are perpendicular to each other, what is the resultant acceleration of
the object? (Hint: Draw a force diagram showing the forces acting on the object. Find the net force
by adding the two forces by components. Use the net force to find the magnitude and direction of the
acceleration. Remember that the direction of the acceleration is in the same direction as the net force.)
Fnet
F2=40N
Direction

 
Fnet  F1  F2
tanq 
Fnet  F12  F22  47.2 N
q
40kg
Magnitude
F1=25N
q  tan 1

 F  ma
 F  47.2  1.18m / s
a
m
40
F2
F1
2
F2
 58 o
F1
9. An 8.00 kg rock is rolled in the sand. It starts at 5.00 m/s, moves in a straight line for a distance of
3.00 m, and then stops. What is its average acceleration? What was the average net force acting on the
rock as it slowed down? If we assume the acceleration of the rock is constant, we can use kinematics
to find a
v0  5.00m / s
v 2  v02  2ax
v0
x  3.00m
ax
 v02  25
a

 4.17m / s 2
2x 2(3)

F
  ma  8(4.17)  33.4 N
t
10. Rita accelerates a 0.400 kg ball from rest to 9.00 m/s during the 0.150 s in which her foot is in contact
with the ball. What average force does she apply to the ball during the kick? To find Fnet, we need
acceleration. If we assume the acceleration of the rock is constant, we can use kinematics to find a
v0  0
v  v0  at
v  9.00m / s
x
ax
t  0.15s
v  v0
9

 60m / s 2
t
0.15
ax 
F
x
 ma x  (0.4)(60)  24 N
11. An elevator weighing 20,000. N (weight = Fg) is supported by a steel cable (see the free body
diagram at right). What is the tension in the cable, FT, when the elevator is being accelerated
upward at 3.00 m/s2? (Hint: Find Fnet by adding up the forces and use Newton’s 2nd Law)
Mass of elevator:
F
y
 ma y
FT
Fg  mg
FT  Fg  ma y
m  Fg / g  20000 / 9.8  2041kg
FT  Fg  ma y  mg  ma y
a
FT  20,000  2041(3)  26,123N
12. An automobile of mass 2000. kg moving at 30.0 m/s, is braked suddenly, with a constant
braking force of 10,000. N. How far does the car travel before stopping?

F
  ma
F
 Fbrake  10000
a


 5m / s 2
m
m
2000
v0  30m / s
v0
x
a x  5m / s2
t
Fbrake
v 2  v02  2ax
 v02  900
x 

 90m
2a
2(5)
Fg
13. When you jump up, does the world recoil downward? Explain. When you jump up, you push on the
world causing it to recoil in response to the push. However the mass/inertia of the Earth is so large,
that the acceleration of the Earth in response to the push is tiny (a=Fnet/m). The tiny acceleration
causes the Earth to recoil by an amount that is much
14. a) When a rifle is fired, how does the size of the force of the rifle on the bullet
compare to the force of the bullet on the rifle? The forces are equal in
magnitude and opposite in direction. The gun pushes the bullet and the bullet
pushes back on the gun with an equal and opposite force.
b) How do the accelerations of the rifle and bullet compare? Why? The
forces are equal but the mass of the rifle and bullet are NOT equal so the
accelerations will not be equal. The bullet has more acceleration because it has smaller mass
Frifle = Fbullet
Mrar=mbAb
15. a) If a bicycle and a massive truck have a head-on collision, upon which vehicle is the impact force
greater? The impact forces are equal in magnitude and opposite in
direction (Newton’s 3rd Law). The truck hits the bike and the
bike hits back with an equal and opposite force.
b) Which vehicle undergoes the greater acceleration or change
in velocity? Why? The bike undergoes the greater acceleration because it has smaller mass. The
impact forces are equal but the mass of the bike and truck are NOT equal so the accelerations will
not be equal.
16. Suppose two carts, one twice as massive as the other, fly apart when
the compressed spring that joins them is released. How fast does the
heavier cart roll compared to the lighter cart? Explain.
When they fly apart, they push on each other with equal and
opposite forces. Therefore the car with twice the mass will have
half the acceleration of the other car. Because they start from rest, the 2m car with half the
acceleration will have half the speed of the m car.
17. A semi-truck collides with a car with a force of 5000N. The semi-truck has a mass of 2000kg and the
car has a mass of 500kg.
a) What force did the car exert on the semi-truck at impact? Since the
truck exerted a force of 5000N on the car, the car
exerted an equal and opposite force on the truck.
FCT
b) What was the acceleration of the truck when it hit the
car? On impact, the only horizontal force acting on the truck
is the impact force on the truck by the car (FTC). See the
figure at right. That is the net force

 F  ma
FTC  mT aT
aT 
FTC 5000

 2.5m / s 2
mT 2000
FTC
c) How much did the car accelerate when it was hit? On impact, the only horizontal force acting on
the car is the impact force
on the car by the truck (FCT). See the figure at right. That is the net

force
F  ma

FCT  mC aC
aC 
FCT 5000

 10m / s 2
mC
500
d) If the action statement is ‘The semi-truck exerts a force on the car,’ what is the reaction statement?
The car exerts an equal and opposite for on the truck.