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Unformatted text preview: find the magnitude and direction of the net
force that produces the deceleration of the car.
SOLUTION The average acceleration of the car is, according to Equation 2.4, a= v − v0
t = 17.0 m/s − 27.0 m/s
= − 1.25 m/s 2
8.00 s 210 FORCES AND NEWTON'S LAWS OF MOTION where the minus sign indicates that the direction of the acceleration is opposite to the
direction of motion; therefore, the acceleration points due west.
According to Newton's Second law, the net force on the car is
∑ F = ma = (1380 kg)(–1.25 m/s 2 ) = –1730 N
The magnitude of the net force is 1730 N . From Newton's second law, we know that the
direction of the force is the same as the direction of the acceleration, so the force also points due west .
____________________________________________________________________________________________ 74. REASONING In the absence of air resistance, the two forces
acting on the sensor are its weight W and the tension T in the
towing cable (see the freebody diagram). We see that Tx is the
only horizontal force acting on the sensor, and therefore
Newton’s second law ΣFx = max (Equation 4.2a) gives Tx = max.
Because the vertical component of the sensor’s acceleration is
zero, the vertical component of the cable’s tension T must
balance the sensor’s weight: Ty = W = mg. We thus have
sufficient information to calculate the horizontal and vertical
components of the tension force T, and therefore to calculate its
magnitude T from the Pythagorean theorem: T 2 = Tx2 + Ty2 . +y
T
Ty
+x
Tx = max
W
Freebody diagram
of the sensor SOLUTION Given that Tx = max and that Ty = mg, the Pythagorean theorem yields the
magnitude T of the tension in the cable: T = Tx2 + Ty2 =
= (129 kg) 75. ( ma )2 + ( mg )2 = m2a 2 + m2 g 2 = m a2 + g 2 ( 2.84 m/s2 ) + (9.80 m/s2 )
2 2 = 1320 N SSM REASONING AND SOLUTION
a. Each cart has the same mass and acceleration; therefore, the net force acting on any one
of the carts is, according to Newton's second law ∑ F = ma = (26 kg)(0.050 m/s 2 ) = 1.3 N Chapter 4 Problems 211 b. The fifth cart must essentially push the sixth, seventh, eight, ninth and tenth cart. In
other words, it must exert on the sixth cart a total force of ∑ F = ma = 5(26 kg)(0.050 m/s 2 )= 6.5 N
____________________________________________________________________________________________ 76. REASONING AND SOLUTION
Newton's second law applied to
object 1 (422 N) gives Object 1 Object 2
N1 T = m1a1 T
T Similarly, for object 2 (185 N)
W1 T – m2g = m2a2 W 2 If the string is not to break or go
slack, both objects must have
accelerations of the same magnitude.
Then a1 = a and a2 = –a. The above equations become
T = m1a
T – m2g = – m2a (1)
(2) a. Substituting Equation (1) into Equation (2) and solving for a yields a= m2 g
m1 + m2 The masses of objects 1 and 2 are m1 = W1 / g = ( 422 N ) / ( 9.80 m/s 2 ) = 43.1 kg
m2 = W2 / g = (185 N ) / ( 9.80 m/s 2 ) = 18.9 kg
The acceleration is
a= m2 g
(18.9 kg ) ( 9.80 m/s 2 )
=
= 2.99 m/s 2
m1 + m2
43.1 kg + 18.9 kg b. Using this value in Equation (1) gives
T = m1a = ( 43.1 kg ) ( 2.99 m/s 2 ) = 129 N
____________________________________________________________________________________________ 212 FORCES AND NEWTON'S LAWS OF MOTION 77. F N
SSM WWW REASONING The speed of the skateboarder
at the bottom of the ramp can be found by solving Equation 2.9
2
( v 2 = v0 + 2ax , where x is the distance that the skater moves
mgsin θ
mgcos θ
down the ramp) for v. The figure at the right shows the freebody diagram for the skateboarder. The net force ΣF, which
θ
accelerates the skateboarder down the ramp, is the component of
the weight that is parallel to the incline: ∑ F = mg sinθ . Therefore, we know from
Newton's second law that the acceleration of the skateboarder down the ramp is a= ΣF mg sin θ
=
= g sin θ
m
m SOLUTION Thus, the speed of the skateboarder at the bottom of the ramp is
2
2
v = v0 + 2ax = v0 + 2 gx sin θ = (2.6 m/s) 2 + 2(9.80 m/s 2 )(6.0 m) sin18° = 6.6 m/s ______________________________________________________________________________
78. REASONING AND SOLUTION From Newton's second law and the equation: v = v0 + at,
we have
v − v0
F = ma = m
t
a. When the skier accelerates from rest (v0 = 0 m/s) to a speed of 11 m/s in 8.0 s, the
required net force is
F =m v − v0
t = (73 kg) (11 m/s) − 0 m/s
= 1.0 × 102 N
8.0 s b. When the skier lets go of the tow rope and glides to a halt (v = 0 m/s) in 21 s, the net
force acting on the skier is F =m v − v0
t = (73 kg) 0 m/s − (11 m/s)
= −38 N
21 s The magnitude of the net force is 38 N .
____________________________________________________________________________________________ Chapter 4 Problems 213 79. REASONING The only horizontal force acting on the boat and trailer is the tension in the
hitch; therefore, it is the net force. According to Newton’s second law, the tension (or the net
force) equals the mass times the acceleration. The mass is known, and the acceleration can
be found by applyin...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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