Physics Solution Manual for 1100 and 2101

60 ms 2 0 ms 2 v 2 v02 f ma m 73 kg

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Unformatted text preview: d law of motion gives the relationship between the net force ΣF and the acceleration a that it causes for an object of mass m. The net force is the vector sum of all the external forces that act on the object. Here the external forces are the drive force, the force due to the wind, and the resistive force of the water. SOLUTION We choose the direction of the drive force (due west) as the positive direction. Solving Newton’s second law ( ΣF = ma ) for the acceleration gives a= ΣF +4100 N − 800 N − 1200 N = = +0.31 m/s 2 m 6800 kg The positive sign for the acceleration indicates that its direction is due west . 101. REASONING AND SOLUTION The acceleration needed so that the craft touches down with zero velocity is a= 2 v 2 − v0 2s 2 − (18.0 m/s ) = = 0.982 m/s 2 ( −165 m ) 2 Newton's second law applied in the vertical direction gives F – mg = ma Then F = m(a + g) = (1.14 × 104 kg)(0.982 m/s2 + 1.60 m/s2) = 29 400 N ____________________________________________________________________________________________ 102. REASONING AND SOLUTION The apparent weight is FN = mw(g + a) Chapter 4 Problems 229 We need to find the acceleration a. Let T represent the force applied by the hoisting cable. Newton's second law applied to the elevator gives T – (mw + me)g = (mw + me)a Solving for a gives a= 9410 N T −g = − 9.80 m/s2 = 0.954 m/s2 mw + me 60.0 kg + 815 kg Now the apparent weight is FN = (60.0 kg)(9.80 m/s2 + 0.954 m/s2) = 645 N ____________________________________________________________________________________________ 103. SSM REASONING We can use the appropriate equation of kinematics to find the acceleration of the bullet. Then Newton's second law can be used to find the average net force on the bullet. SOLUTION According to Equation 2.4, the acceleration of the bullet is a= v − v0 t = 715 m/s − 0 m/s = 2.86 ×105 m/s 2 2.50 × 10 –3 s Therefore, the net average force on the bullet is ∑ F = ma = (15 × 10−3 kg)(2.86 ×105 m/s 2 ) = 4290 N ____________________________________________________________________________________________ 104. REASONING The magnitude ΣF of the net force acting on the kayak is given by Newton’s second law as ΣF = ma (Equation 4.1), where m is the combined mass of the person and kayak, and a is their acceleration. Since the initial and final velocities, v0 and v, and the displacement x are known, we can employ one of the equations of kinematics from Chapter 2 to find the acceleration. ( ) SOLUTION Solving Equation 2.9 v 2 = v0 2 + 2ax from the equations of kinematics for the acceleration, we have a= v 2 − v0 2 2x Substituting this result into Newton’s second law gives ( 0.60 m/s )2 − ( 0 m/s )2 v 2 − v02 ΣF = ma = m = ( 73 kg ) = 32 N 2 ( 0.41 m ) 2x 230 FORCES AND NEWTON'S LAWS OF MOTION ______________________________________________________________________________ 105. REASONING AND SOLUTION a. According to Equation 4.4, the weight of an object of mass m on the surface of Mars would be given by GM M m W= 2 RM where MM is the mass of Mars and RM is the radius of Mars. On the surface of Mars, the weight of the object can be given as W = mg (see Equation 4.5), so mg = GM M m 2 RM or g= GM M 2 RM Substituting values, we have g= (6.67 ×10−11N ⋅ m 2 /kg 2 )(6.46 ×1023 kg) = 3.75 m/s 2 6 2 (3.39 ×10 m) b. According to Equation 4.5, W = mg = (65 kg)(3.75 m/s2) = 2.4 × 102 N ____________________________________________________________________________________________ 106. REASONING Each particle experiences two gravitational forces, one due to each of the remaining particles. To get the net gravitational force, we must add the two contributions, taking into account the directions. The magnitude of the gravitational force that any one particle exerts on another is given by Newton’s law of gravitation as F = Gm1m2 / r 2 . Thus, for particle A, we need to apply this law to its interaction with particle B and with particle C. For particle B, we need to apply the law to its interaction with particle A and with particle C. Lastly, for particle C, we must apply the law to its interaction with particle A and with particle B. In considering the directions, we remember that the gravitational force between two particles is always a force of attraction. SOLUTION We begin by calculating the magnitude of the gravitational force for each pair of particles: Chapter 4 Problems FAB = GmAmB FBC = GmB mC FAC = Gm AmC r2 r2 r2 231 ( 6.67 × 10–11 N ⋅ m2 / kg 2 ) (363 kg )(517 kg ) = 5.007 × 10–5 N = ( 0.500 m )2 ( 6.67 × 10–11 N ⋅ m2 / kg 2 ) (517 kg )(154 kg ) = 8.497 × 10–5 N = ( 0.500 m )2 ( 6.67 × 10–11 N ⋅ m2 / kg 2 ) (363 kg )(154 kg ) = 6.629 × 10–6 N = ( 0.500 m )2 In using these magnitudes we take the direction to the right as positive. a. Both particles B and C attract particle A to the right, the net force being FA = FAB + FAC = 5.007 × 10 –5 N + 6.629 × 10 –6 N = 5.67 × 10 –5 N, right b. Particle C attracts particle B to the right, while particle A attracts particle B to the left, the net force bei...
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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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