Physics Solution Manual for 1100 and 2101

8 for the kinetic frictional force to write the

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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 upward-acting 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 free-body 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 Free-body 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.

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