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Physics Solution Manual for 1100 and 2101

# This gravitational force is equal in magnitude to the

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Unformatted text preview: 5 kg) (9.80 m/s 2 ) = 1.13 × 103 N b. In interplanetary space where there are no nearby planetary objects, the gravitational force exerted on the space traveler is zero and g = 0 m/s2. Therefore, the weight is W = 0 N . Since the mass of an object is an intrinsic property of the object and is independent of its location in the universe, the mass of the space traveler is still m = 115 kg . ____________________________________________________________________________________________ Chapter 4 Problems 22. REASONING With air resistance neglected, only two forces act on the bungee jumper at this instant (see the free-body diagram): the bungee cord pulls up on her with a force B, and the earth pulls down on her with a gravitational force mg. Because we know the jumper’s mass and acceleration, we can apply Newton’s second law to this free-body diagram and solve for B. 175 B + − mg SOLUTION We will take the direction of the jumper’s acceleration (downward) as negative. Then, the net force acting on the bungee jumper is ΣF = B − mg. With Newton’s second law (ΣF = ma), this becomes ma = B − mg. We now solve for B: Free-body diagram of jumper ( ) B = ma + mg = m ( a + g ) = ( 55 kg ) −7.6 m/s2 + 9.80 m/s2 = 120 N As indicated in the free-body diagram, the direction of the force applied by the bungee cord is upward . 23. REASONING The earth exerts a gravitational force on the raindrop, and simultaneously the raindrop exerts a gravitational force on the earth. This gravitational force is equal in magnitude to the gravitational force that the earth exerts on the raindrop. The forces that the raindrop and the earth exert on each other are Newton’s third law (action–reaction) forces. Newton’s law of universal gravitation specifies the magnitude of both forces. SOLUTION a. The magnitude of the gravitational force exerted on the raindrop by the earth is given by Equation 4.3: Fraindrop = = Gmearth mraindrop ( 2 rearth 6.67 × 10−11 N ⋅ m 2 / kg 2 5.98 × 1024 kg 5.2 × 10−7 kg )( ( 6.38 × 10 m ) 6 )( 2 )= 5.1 × 10−6 N b. The magnitude of the gravitational force exerted on the earth by the raindrop is Fearth = = Gmearth mraindrop ( 2 rearth 6.67 × 10−11 N ⋅ m 2 / kg 2 5.98 × 1024 kg 5.2 × 10−7 kg )( ( 6.38 × 10 m ) 6 )( 2 )= 5.1 × 10−6 N ______________________________________________________________________________ 176 FORCES AND NEWTON'S LAWS OF MOTION 24. REASONING Newton’s law of universal gravitation indicates that the gravitational force that each uniform sphere exerts on the other has a magnitude that is inversely proportional to the square of the distance between the centers of the spheres. Therefore, the maximum gravitational force between two uniform spheres occurs when the centers of the spheres are as close together as possible, and this occurs when the surfaces of the spheres are touching. Then, the distance between the centers of the spheres is the sum of the two radii. SOLUTION When the bowling ball and the billiard ball are touching, the distance between their centers is r = rBowling + rBilliard. Using this expression in Newton’s law of universal gravitation gives F= GmBowling mBilliard GmBowling mBilliard = ( rBowling + rBilliard ) ( 6.67 ×10−11 N ⋅ m2 / kg2 ) ( 7.2 kg )( 0.38 kg ) = 9.6 ×10−9 N = r2 2 ( 0.11 m + 0.028 m )2 25. SSM REASONING AND SOLUTION a. Combining Equations 4.4 and 4.5, we see that the acceleration due to gravity on the surface of Saturn can be calculated as follows: gSaturn = G M Saturn 2 rSaturn ( = 6.67 ×10 –11 N ⋅ m /kg 2 2 (5.67 ×1026 kg ) = ) (6.00 ×107 m)2 10.5 m/s 2 b. The ratio of the person’s weight on Saturn to that on earth is WSaturn Wearth = mgSaturn mg earth = gSaturn gearth = 10.5 m/s 2 = 1.07 9.80 m/s 2 ____________________________________________________________________________________________ 26. REASONING As discussed in Conceptual Example 7, the same net force is required on the moon as on the earth. This net force is given by Newton’s second law as ΣF = ma, where the mass m is the same in both places. Thus, from the given mass and acceleration, we can calculate the net force. On the moon, the net force comes about due to the drive force and the opposing frictional force. Since the drive force is given, we can find the frictional force. SOLUTION Newton’s second law, with the direction of motion taken as positive, gives ΣF = ma or (1430 N ) – f ( )( = 5.90 × 103 kg 0.220 m/s 2 ) Chapter 4 Problems 177 Solving for the frictional force f , we find ( )( ) f = (1430 N ) – 5.90 × 103 kg 0.220 m/s 2 = 130N ____________________________________________________________________________________________ 27. SSM REASONING AND SOLUTION According to Equations 4.4 and 4.5, the weight of an object of mass m at a distance r from the center of the earth is mg = GM E m r2 In a circular orbit that is 3.59 ×107 m above the surface of the earth ( radius = 6.38 ×106 m , mass = 5.98 ×1024 kg ), the total distance from the center of the earth is r = 3.59 ×107 m + 6.38 ×106 m . Thus the acceleration g due to gravity is g= GM E r 2 = (6.67 × 10...
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