RECITATION
CHAPTER 4
4-5 Workmen are trying to free an SUV stuck in the mud. To extricate the vehicle, they
use three horizontal ropes, producing the force vectors shown in the following figure.
(a) Find the x and y components of each of the three pulls. (b) Use the components to
find the magnitude and direction of the resultant of the three pulls.
Let F1  985N , F2  788N , F3  411N . The angles θ that each force makes with
the +x axis are θ1 =31°, θ2 = 122°, and θ3 = 233°.
(a) The x and y components of these three forces are:
F1x  F1 cos 1  844 N ,
F1 y  F1 sin 1  507 N
F2 x  F2 cos  2  418 N ,
F2 y  F2 sin  2  668 N
F3 x  F3 cos  3  247 N
F3 y  F3 sin  3  328 N
(b) x and y components of the resultant pull are:
Fx  F1x  F2 x  F3 x  844 N  418 N  247 N  179 N
Fy  F1 y  F2 y  F3 y  507 N  668 N  328 N  847 N
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The magnitude and direction of the resultant pull are as follows:
F  Fx2  Fy2 
 Fy
 Fx
  tan 1 
179 N 2  847 N 2
 866 N

847 N 
  tan 1 
  78.1
 179 N 

4-11 A dock worker applies a constant horizontal force of 80.0 N to a block of ice on a
smooth horizontal surface. The frictional force is negligible. The block starts from rest
and moves 11.0 m in the first 5.00 s. What is the mass of the block of ice?
Acceleration can be calculated using constant acceleration equation:
1
axt 2
2
211.0m 
x  x0  v0 x t 
2x

 0.88m / s 2
2
2
t
5.00s 
Next, using Newton’s second law equation we can calculate mass.
ax 
F
x
m
 ma x
F
x
ax

80.0 N
 90.9kg
0.88m / s 2
4-20. At the surface of Jupiter’s moon Io, the acceleration due to gravity is 1.81 m/s2. If
a piece of ice weights 44.0 N at the surface of the earth, (a) What is its mass on the
earth’s surface? (b) What are its mass and weight on the surface of Io?


The weight w  mg of an object depends on its location. The mass of an object is an
intrinsic property of the object and doesn’t depend on the location.
(a) Since we are given the object’s weight on the surface of the earth, using g =
9.80 m/s2 we can calculate the mass of the object.
w  mg
m
w
44.0 N

 4.49kg
g 9.80m / s 2
(b) Mass is the same on the Jupiter’s moon Io, m  4.49kg .
Weight is wIO  mg IO  4.49kg 1.81m / s 2   8.13N
2
4-23. The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly
applies the brakes. Due to a slippery pavement, the friction of the road on the tires of
the car, which is what slows down the car, is 25% of the weight of the car. (a) What is
the acceleration of the car? (b) How many meters does it travel before stopping under
these conditions?
Let +x be the direction of motion of the car.
(a) The friction f of the road on the tires of the car is 25% of the weight of the car, i.e.,
f  0.25mg and is in the –x direction.
Fx  max gives  f  ma x
ax 
 f  0.25mg

 0.25 g  2.45m / s 2
m
m
The direction of this acceleration is opposite to the direction of motion.
(b) v0 x  110km / h  30.6m / s , v x  0 and a x  2.45m / s 2 .
Using constant acceleration equation, we can find how many meters the car
traveled before stopping.
v x2  v02x  2a x x OR
x 
v x2  v02x 0  30.6m / s 2

 191m
2a x
2  2.45m / s 2


The stopping distance is proportional to the square of the initial speed.
4-34 A factory worker pushes horizontally on a 250 N crate with a force of 75 N on a
horizontal rough floor. A 135 N crate rests on top of the one being pushed and moves
along with it. Make a free-body diagram of each crate if the friction force is less than
the worker’s push.
There is a friction force between the floor and the bottom crate and a different friction
force between the two crates. Call the bottom crate A and the top crate B. Weight of
crate A is wA = 250 N and weight of crate B is wB = 135 N. Assume that the crates
accelerate to the right. The crates will accelerate since the total friction force from the
floor is less than the worker’s push.
The free-body diagram for the top crate is given below in figure (a), and for the bottom
crate is given in figure (b).
fA and f A' are the friction forces between the crates and nA and n A' are the normal
forces the crates exert on each other.
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fA and f A' are a Newton’s 3rd law pair and are equal in magnitude and opposite in
direction. The same is true for nA and n A' . ff and nf are the friction and normal forces
exerted on block A by the floor.
4-41. A 22-caliber rifle bullet of mass 1.80 g, traveling at 350 m/s, strikes a block of soft
wood, which it penetrates to a depth of 0.130 m. The block of wood is clamped in place and
doesn’t move. Assume a constant retarding force. (a) Make a free-body diagram of the bullet
while it is slowing down in the wood. (b) How much time is required for the bullet to stop?
(c) What force, in newtons, does the wood exert on the bullet?
Use a constant acceleration equation to find the stopping time and acceleration. Let
in the direction the bullet is traveling. F is the force the wood exerts on the bullet.
(a) The free-body diagram is given in Figure below.
(b) Initial and final velocities are given: v0 x  350m / s, v x  0 . This change in the
velocity takes place in a distance of 0.130 m.
Since the acceleration is constant, average velocity is vav 
be written as
x  x0 v0 x  v x 
.

t  t0
2
4
v0 x  v x  .
2
This can
x
be
Letting t0= 0 and solving for time gives,
t
2x  x0  20.130m

 7.43  10 4 s
v0 x  v x  350m / s
(c) To calculate force using Newton’s second law, we must first find acceleration.
v x2  v02x  2a x x OR
v x2  v02x 0  350m / s 2

 4.71  10 5 m .
2x
20.130m 
ax 
Since the bullet has decelerated, the acceleration direction is opposite to the
direction of motion of the bullet.
Using Newton’s second law, we find the force exerted by the wood causing this
deceleration as follows:
F
x
 max
F  max  (1.80  10 3 kg)( 4.71  10 5 m / s 2 )  848 N
4-52. You leave the doctor’s office after your annual checkup and recall that you
weighted 683 N in her office. You then get into an elevator that, conveniently, has a
scale. Find the magnitude and direction of the elevator’s acceleration if the scale reads
(a) 725 N, (b) 595 N.
w  683 N is the force of gravity exerted on you, independent of your motion. Your
mass is m  w / g  69.7 kg. Use coordinates with  y upward. n is the scale
reading, the force the scale exerts on you. You and the elevator have the same
acceleration.
nw
.
m
725 N  683 N
 0.603 m/s 2 .
(a) n  725 N, so a y 
69.7 kg
Fy  ma y gives n  w  ma y so a y 
a y is positive so the acceleration is upward.
(b) n  595 N, so a y 
595 N  683 N
  1.26 m/s 2 .
69.7 kg
a y is negative so the acceleration is downward.
5
4-58 An electron (mass = 9.11 x 10-31 kg) leaves one end of a TV picture tube with
zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 cm
away. It reaches the grid with a speed of 3.00 x 106m/s. If the accelerating force is
constant, compute (a) the acceleration of the electron, (b) the time it takes the electron
to reach the grid, and (c) the net force that is accelerating the electron, in newtons.
(You can ignore the gravitational force on the electron.)
Let +x be in the direction of motion of the electron.
(a) v0 x  0, x  x0   1.80  10 2 m, v x  3.00  10 6 m / s
The constant acceleration equation v x2  v02x  2a x  x  x0  gives


2
v x  v02x
3.00  10 6 m / s  0
ax 

 2.50  1014 m / s 2 .
2
2x  x0 
2 1.80  10 m


Since there is an increase in the speed of the electron, direction of
acceleration is the same as the direction of motion of the electron.
(b) v x  v0 x  a x t gives t 
v x  v0 x 3.00  10 6 m / s  0

 1.2  10 8 s.
14
2
ax
2.50  10 m / s
(c) Using Newton’s second law, the net force that accelerates the electron is
F
x



 max  9.1110 31 kg 2.50 1014 m / s 2  2.28 10 16 N.
6