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Unformatted text preview: TON'S LAWS OF MOTION c. When the elevator accelerates downward, a = –1.20 m/s2. Then, f k = µ k FN = µ k ( mg + ma ) = µ k m ( g + a ) ( )( ) = ( 0.360 )( 6.00 kg ) 9.80 m/s 2 + –1.20 m/s 2 = 18.6 N ____________________________________________________________________________________________ 47. REASONING The magnitude of the kinetic frictional force is given by Equation 4.8 as the
coefficient of kinetic friction times the magnitude of the normal force. Since the slide into
second base is horizontal, the normal force is vertical. It can be evaluated by noting that
there is no acceleration in the vertical direction and, therefore, the normal force must
balance the weight.
To find the player’s initial velocity v0, we will use kinematics. The time interval for the
slide into second base is given as t = 1.6 s. Since the player comes to rest at the end of the
slide, his final velocity is v = 0 m/s. The player’s acceleration a can be obtained from
Newton’s second law, since the net force is the kinetic frictional force, which is known from
part (a), and the mass is given. Since t, v, and a are known and we seek v0, the appropriate
kinematics equation is Equation 2.4 (v = v0 + at).
SOLUTION
a. Since the normal force FN balances the weight mg, we know that FN = mg. Using this
fact and Equation 4.8, we find that the magnitude of the kinetic frictional force is ( ) f k = µk FN = µk mg = ( 0.49 ) ( 81 kg ) 9.8 m/s 2 = 390 N
b. Solving Equation 2.4 (v = v0 + at) for v0 gives v0 = v − at. Taking the direction of the
player’s slide to be the positive direction, we use Newton’s second law and Equation 4.8 for
the kinetic frictional force to write the acceleration a as follows:
a= ΣF − µk mg
=
= −µk g
m
m The acceleration is negative, because it points opposite to the player’s velocity, since the
player slows down during the slide. Thus, we find for the initial velocity that ( ) v0 = v − ( − µk g ) t = 0 m/s − − ( 0.49 ) 9.8 m/s 2 (1.6 s ) = +7.7 m/s Chapter 4 Problems 191 48. REASONING AND SOLUTION The deceleration produced by the frictional force is
a=– fk
m = – µ k mg
m = – µk g The speed of the automobile after 1.30 s have elapsed is given by Equation 2.4 as v = v0 + at = v0 + ( − µk g ) t = 16.1 m/s − ( 0.720 ) ( 9.80 m/s 2 ) (1.30 s ) = 6.9 m/s
____________________________________________________________________________________________ 49. SSM REASONING Let us assume that the skater is moving horizontally along the +x
axis. The time t it takes for the skater to reduce her velocity to vx = +2.8 m/s from
v0x = +6.3 m/s can be obtained from one of the equations of kinematics:
v x = v0 x + a x t (3.3a) The initial and final velocities are known, but the acceleration is not. We can obtain the
acceleration from Newton’s second law ( ΣFx = ma x , Equation 4.2a ) in the following
manner. The kinetic frictional force is the only horizontal force that acts on the skater, and,
since it is a resistive force, it acts opposite to the direction of the motion. Thus, the net force
in the x direction is ΣFx = − f k , where fk is the magnitude of the kinetic frictional force.
Therefore, the acceleration of the skater is ax = ΣFx /m = − f k / m .
The magnitude of the frictional force is f k = µk FN (Equation 4.8), where µ k is the
coefficient of kinetic friction between the ice and the skate blades and FN is the magnitude
of the normal force. There are two vertical forces acting on the skater: the upwardacting
normal force FN and the downward pull of gravity (her weight) mg. Since the skater has no
vertical acceleration, Newton's second law in the vertical direction gives (taking upward as
the positive direction) ΣFy = FN − mg = 0 . Therefore, the magnitude of the normal force is
FN = mg and the magnitude of the acceleration is ax = − f k − µk FN − µk m g
=
=
= − µk g
m
m
m SOLUTION
Solving the equation vx = v0 x + a x t for the time and substituting the expression above for
the acceleration yields t= vx − v0 x
ax = vx − v0 x
− µk g = 2.8 m/s − 6.3 m/s
= 4.4 s
− ( 0.081) ( 9.80 m/s 2 ) 192 FORCES AND NEWTON'S LAWS OF MOTION ____________________________________________________________________________________________ 50. REASONING We assume the car accelerates in the
+x direction. The air resistance force fA opposes the
car’s motion (see the freebody diagram). The
frictional force is static, because the tires do not slip,
and points in the direction of the car’s acceleration.
The reason for this is that without friction the car’s
wheels would simply spin in place, and the car’s
acceleration would be severely limited. The frictional
force has its maximum value fsMAX because we seek
the maximum acceleration before slipping occurs.
Applying Newton’s second law (ΣFx = max, Equation
4.2a) to the horizontal motion gives +y FN
fsMAX fA +x D W Freebody diagram of the car fsMAX − f A = ma x (1) where ax is the maximum acceleration we seek. The air resistance force fA is given, and we ( ) will find the maximum static frictional force...
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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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