In Sample Prob. 5.01, for PART C, If another force, F4 = 3.0 N directed at angle 40 %3D degrees above the horizontal acts on the puck in addition to the F2 and F3, What is the acceleration of the puck?
In Sample Prob. 5.01, for PART C, If another force, F4 = 3.0 N directed at angle 40 %3D degrees above the horizontal acts on the puck in addition to the F2 and F3, What is the acceleration of the puck?
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Transcribed Image Text:In Sample Prob. 5.01, for PART C, If another
force, F4 = 3.0 N directed at angle 40
%3D
degrees above the horizontal acts on the
puck in addition to the F2 and F3, What is
the acceleration of the puck?
a. -5.7 m/s^2
b. 5.7 m/s^2
c. -5.8 m/s^2
d. 5.8 m/s^2
e. -17.2 m/s^2
f.
17.2 m/s^2
g. -6.0 m/s^2
h. 6.0 m/s^2
Transcribed Image Text:86
CHAPTER 5 FORCE AND MOTION-I
Sample Problem 5,01 One- and two-dimensional forces, puCk
The horizontal force
causes a horizontal
acceleration.
Here are examples of how to use Newton's second law for a
puck when one or two forces act on it. Parts A, B, and C of
Fig. 5,3 show three situations in which one or two forces act
on a puck that moves over frictionless ice along an x axis, in
one-dimensional motion. The puck's mass is m- 0.20 kg.
Forces F and F, are directed along the axis and have
magnitudes F=- 4.0 N and F, - 2,0 N. Force F is directed
at angle 0= 30) and has magnitude F
ation, what is the acceleration of the puck?
This is a free-body
diagram.
(a)
Puck
1.0 N. In each situ-
(6)
KEY IDEA
B
These forces compete.
Their net force causes
a horizontal acceleration.
In each situation we can relate the acceleration a to the net
force Fet acting on the puck with Newton's second law,
F= ma. However, because the motion is along only the x
axis, we can simplify each situation by writing the second
law for x components only:
(0)
This is a free-body
diagram.
Fet,= ma,.
(5-4)
The free-body diagrams for the three situations are also
given in Fig. 5-3, with the puck represented by a dot.
(d)
C
Situation A: For Fig. 5-3b, where only one horizontal force
acts, Eq. 5-4 gives us
Only the horizontal
component of Fa
competes with Fg.
F = ma,,
which, with given data, yields
(6)
F
a, =
m
4.0 N
This is a free-body
diagram.
- 20 m/s.
(Answer)
%3D
0.20 kg
The positive answer indicates that the acceleration is in the
positive direction of the x axis.
Figure 5-3 In three situations, forces act on a puck that moves
along an x axis. Free-body diagrams are also shown.
Situation B: In Fig. 5-3d, two horizontal forces act on the
puck, F, in the positive direction of x and F, in the negative
direction. Now Eq. 5-4 gives us
dimensional.) Thus, we write Eq. 5-4 as
F - F = ma,,
F-F2 ma,.
From the figure, we see that F = F, cos 0. Solving for the
which, with given data, yields
(5-5)
acceleration and substituting for F, yield
F- F;
a, =
4.0 N 2.0 N
= 10 m/s?.
%3D
F-F F cos 0-F,
a,
0.20 kg
(Answer)
Thus, the net force accelerates the puck in the positive direc-
m
(1.0 N)(cos 30) -2.0 N
tion of the x axis.
0.20 kg
-5.7 m/s.
Situation C: In Fig. 5-3f, force F, is not directed along the
direction of the puck's acceleration; only x component F
is. (Force F, is two-dimensional but the motion is only one-
(Answer)
Thus, the net force accelerates the puck in the negative di-
rection of thex axis.
PLUS Additional examples, video, and practice available at WileyPLUS
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