The Laws of Motion - Department of PhysicsThe Laws of Motion SOLUTIONS TO PROBLEMS Section 4.3 Mass...
Transcript of The Laws of Motion - Department of PhysicsThe Laws of Motion SOLUTIONS TO PROBLEMS Section 4.3 Mass...
107
The Laws of Motion
SOLUTIONS TO PROBLEMS Section 4.3 Mass P4.2 Since the car is moving with constant speed and in a straight line, the resultant force on it must be
zero regardless of whether it is moving
(a) toward the right or (b) the left.
Section 4.4 Newton’s Second Law⎯The Particle Under a Net Force P4.3
m = 3.00 kgra = 2.00i + 5.00 j( ) m s2
rF = mra = 6.00i + 15.0 j( ) N!rF! = 6.00( )2 + 15.0( )2 N = 16.2 N
108 The Laws of Motion
P4.4 (a) !
rF =
rF1 +
rF2 = "9.00i + 3.00 j( ) N
Acceleration
ra = ax i + ay j = !rF
m=
"9.00i + 3.00 j( ) N2.00 kg
= "4.50i + 1.50 j( ) m s2
Velocity
rv f = vx i + vy j = rv i +
rat =rat
rv f = !4.50i + 1.50 j( )m s2( ) 10 s( ) = !45.0i + 15.0 j( ) m s
(b) The direction of motion makes angle θ with the x-direction.
! = tan"1 vy
vx
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15.0 m s45.0 m s
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! = "18.4° + 180° = 162° from + x-axis
(c) Displacement: x-displacement = x f ! xi = vxit +
12
axt2 =12!4.50 m s2( ) 10.0 s( )2 = !225 m
y-displacement = y f ! yi = vyit +
12
ayt2 =12+1.50 m s2( ) 10.0 s( )2 = +75.0 m
!
rr = "225i + 75.0 j( ) m
(d) Position:
rrf =
rri + !rr
rrf = !2.00i + 4.00 j( ) + !225i + 75.0 j( ) = !227i + 79.0 j( ) m
P4.7 (a)
rF! =
rF1 +
rF2 = 20.0i + 15.0 j( ) N
rF! = mra : 20.0i + 15.0 j = 5.00ra
ra = 4.00i + 3.00 j( ) m s2
or
a = 5.00 m s2 at ! = 36.9° (b)
F2x = 15.0cos 60.0° = 7.50 NF2y = 15.0sin 60.0° = 13.0 NrF2 = 7.50i + 13.0 j( ) NrF! =
rF1 +
rF2 = 27.5i + 13.0 j( ) N = m
ra = 5.00ra
ra = 5.50i + 2.60 j( ) m s2 = 6.08 m s2 at 25.3°
FIG. P4.7
Chapter 4 109
P4.8
rF! = mra reads
!2.00i + 2.00 j + 5.00i ! 3.00 j ! 45.0i( ) N = m 3.75 m s2( ) a
where a represents the direction of
ra
!42.0i ! 1.00 j( ) N = m 3.75 m s2( ) a
rF! = 42.0( )2 + 1.00( )2 N at
tan!1 1.00
42.0"#$
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below the –x-axis
rF! = 42.0 N at 181° = m 3.75 m s2( ) a .
For the vectors to be equal, their magnitudes and their directions must be equal.
(a) ! a is at 181° counterclockwise from the x-axis
(b) m =
42.0 N3.75 m s2 = 11.2 kg
(d)
rv f =
rv i +rat = 0 + 3.75 m s2 at 181°( )10.0 s so
rv f = 37.5 m s at 181°
rv f = 37.5 m s cos181°i + 37.5 m s sin181° j so
rv f = !37.5i ! 0.893 j( ) m s
(c) rv f = 37.52 + 0.8932 m s = 37.5 m s
Section 4.5 The Gravitational Force and Weight P4.9 (a) Fg = mg = 120 lb = 4.448 N lb( ) 120 lb( ) = 534 N
(b) m =
Fg
g=
534 N9.80 m s2 = 54.5 kg
P4.11 Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just
canceled out by a glass of tomato juice. By subtraction,
Fg( )p = mgp and
Fg( )C = mgC give
!Fg = m gp " gC( ) .
For a person whose mass is 88.7 kg, the change in weight is
!Fg = 88.7 kg 9.809 5 " 9.780 8( ) = 2.55 N . A precise balance scale, as in a doctor’s office, reads the same in different locations because it
compares you with the standard masses on its beams. A typical bathroom scale is not precise enough to reveal this difference.
110 The Laws of Motion
P4.13 (a) F! = ma and v f2 = vi
2 + 2ax f or a =
v f2 ! vi
2
2x f.
Therefore,
F! = mv f
2 " vi2( )
2x f
F! = 9.11 # 10"31 kg7.00 # 105 m s2( )2 " 3.00 # 105 m s2( )2$
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2 0.050 0 m( )= 3.64 # 10"18 N .
(b) The weight of the electron is
Fg = mg = 9.11 ! 10"31 kg( ) 9.80 m s2( ) = 8.93 ! 10"30 N
The accelerating force is
4.08 ! 1011 times the weight of the electron. P4.14 We find acceleration:
rrf !rri =
rv it +12
rat2
4.20 mi ! 3.30 mj=0+ 12
ra 1.20 s( )2 = 0.720 s2rara = 5.83i ! 4.58 j( ) m s2 .
Now
rF! = mra becomes
rFg +
rF2 = mrarF2 = 2.80 kg 5.83i ! 4.58j( ) m s2 + 2.80 kg( ) 9.80 m s2( ) jrF2 = 16.3i + 14.6 j( ) N .
Section 4.6 Newton’s Third Law P4.17 (a) 15.0 lb up
(b) 5.00 lb up
(c) 0
Chapter 4 111
Section 4.7 Applications of Newton’s Laws P4.22 (a) Isolate either mass
T ! mg = ma = 0T = mg .
The scale reads the tension T, so
T = mg = 5.00 kg 9.80 m s2( ) = 49.0 N . (b) Isolate the pulley
rT2 + 2
rT1 = 0T2 = 2 T1 = 2mg = 98.0 N .
(c)
rF! =
rn +rT + mrg = 0
Take the component along the incline
rnx +
rTx + m
rgx = 0 or
0 +T ! mg sin 30.0° = 0
T = mg sin 30.0° = mg2
=5.00 9.80( )
2= 24.5 N .
FIG. P4.22(a)
FIG. P4.22(b)
FIG. P4.22(c)
112 The Laws of Motion
P4.24 (a) First construct a free body diagram for the
5 kg mass as shown in the Figure 4.24a. Since the mass is in equilibrium, we can require T3 ! 49 N = 0 or T3 = 49 N . Next, construct a free body diagram for the knot as shown in Figure 4.24a. Again, since the system is moving at constant velocity, a = 0 and applying Newton’s second law in component form gives
Fx! = T2 cos 50° "T1 cos 40° = 0
Fy! = T2 sin 50° +T1 sin 40° " 49 N = 0
Solving the above equations
simultaneously for T1 and T2 gives
T1 = 31.5 N and T2 = 37.5 N and above
we found T3 = 49.0 N . (b) Proceed as in part (a) and construct a free
body diagram for the mass and for the knot as shown in Figure 4.24b. Applying Newton’s second law in each case (for a constant-velocity system) we find:
T3 ! 98 N = 0T2 !T1 cos 60° = 0T1 sin 60° !T3 = 0
Solving this set of equations we find:
T1 = 113 N T2 = 56.6 N and T3 = 98.0 N
FIG. P4.24(a)
FIG. P4.24(b)
Chapter 4 113 P4.26 The two forces acting on the block are the normal force, n, and the
weight, mg. If the block is considered to be a point mass and the x-axis is chosen to be parallel to the plane, then the free body diagram will be as shown in the figure to the right. The angle θ is the angle of inclination of the plane. Applying Newton’s second law for the accelerating system (and taking the direction up the plane as the positive x direction) we have
Fy! = n " mg cos# = 0 : n = mg cos!
Fx! = "mg sin# = ma : a = !g sin"
FIG. P4.26 (a) When ! = 15.0°
a = !2.54 m s2 (b) Starting from rest
v f2 = vi
2 + 2a x f ! xi( ) = 2ax f
v f = 2ax f = 2 !2.54 m s2( ) !2.00 m( ) = 3.18 m s
P4.27 Choose a coordinate system with i East and j North.
rF! = mra = 1.00 kg 10.0 m s2( ) at 30.0°
5.00 N( ) j +rF1 = 10.0 N( )!30.0° = 5.00 N( ) j + 8.66 N( ) i
!F1 = 8.66 N East( )
FIG. P4.27 P4.28 First, consider the block moving along the horizontal. The only
force in the direction of movement is T. Thus, Fx! = ma
T = 5 kg( ) a (1) Next consider the block that moves vertically. The forces on it
are the tension T and its weight, 88.2 N. We have Fy! = ma
88.2 N !T = 9 kg( ) a (2)
FIG. P4.28
Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be
added to give 88.2 N = 14 kg( ) a . Then
a = 6.30 m s2 and T = 31.5 N .
114 The Laws of Motion
P4.30 m1 = 2.00 kg , m2 = 6.00 kg , ! = 55.0° (a) Fx! = m2 g sin" #T = m2a and
T ! m1g = m1a
a = m2 g sin" ! m1gm1 + m2
= 3.57 m s2
(b) T = m1 a + g( ) = 26.7 N
FIG. P4.30
(c) Since vi = 0 , v f = at = 3.57 m s2( ) 2.00 s( ) = 7.14 m s .
Additional Problems P4.44 (a) Following the in-chapter Example about a block on a frictionless incline, we have
a = g sin! = 9.80 m s2( )sin 30.0°
a = 4.90 m s2
(b) The block slides distance x on the incline, with sin 30.0° = 0.500 m
x
x = 1.00 m : v f2 = vi
2 + 2a x f ! xi( ) = 0 + 2 4.90 m s2( ) 1.00 m( )
v f = 3.13 m s after time ts =
2x f
v f=
2 1.00 m( )3.13 m s
= 0.639 s .
(c) Now in free fall y f ! yi = vyit +
12
ayt2 :
!2.00 = !3.13 m s( )sin 30.0°t ! 12
9.80 m s2( )t2
4.90 m s2( )t2 + 1.56 m s( )t ! 2.00 m = 0
t =!1.56 m s ± 1.56 m s( )2 ! 4 4.90 m s2( ) !2.00 m( )
9.80 m s2
Only one root is physical
t = 0.499 s
x f = vxt = 3.13 m s( )cos 30.0°[ ] 0.499 s( ) = 1.35 m
(d) total time = ts + t = 0.639 s + 0.499 s = 1.14 s (e) The mass of the block makes no difference.
Chapter 4 115 P4.47 F! = ma For m1 : T = m1a For m2 : T ! m2 g = 0 Eliminating T,
a = m2 g
m1
For all 3 blocks:
FIG. P4.47
F = M + m1 + m2( ) a = M + m1 + m2( ) m2 g
m1
!
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