Chapter 6
CHAPTER 6 - Gravitation and Newton's Synthesis 1. Because the spacecraft is 2 Earth radii above the surface, it is 3 Earth radii from the center. The gravitational force on the spacecraft is F = GMEM/r2 = (6.67 1011 N m2 /kg2 )(5.98 102
Chapter 4
1
CHAPTER 4 - Dynamics: Newton's Laws of Motion 1. We convert the units: #( lb = (0.25 lb)(4.45 N/lb) 1 N. If we select the bike and rider as the object, we apply Newton's second law to find the mass: F = ma; 255 N = m(2.20 m/s2 ), which
Chapter 3
CHAPTER 3 - Kinematics in Two Dimensions; Vectors 1.
D We choose the west and south coordinate system shown. 2x For the components of the resultant we have W D D 2 RW = D1 + D2 cos 45 2y = (200 km) + (80 km) cos 45 = 257 km; RS = 0 + D2 =
Chapter 2
CHAPTER 2 - Describing Motion: Kinematics in One Dimension 1. We find the time from average speed = d/t; 15 km/h = (75 km)/t , which gives
t = 5.0 h. 88 km/h.
2. 3.
We find the average speed from average speed = d/t = (280 km)/(3.2 h) =
Chapter 11
CHAPTER 11 General Rotation 1. (a) For the magnitudes of the vector products we have i i = i i sin 0 = 0; j j = j j sin 0 = 0; k k k = k k sin 0 = 0. (b) For the magnitudes of the vector products we have i j = i j sin 90 = (1)(1)(1) = 1;
Chapter 9
CHAPTER 9 - Linear Momentum and Collisions 1. We find the force on the expelled gases from F = p/t = (m/t)v = (1200 kg/s)(50,000 m/s) = 6.0 107 N. An equal, but opposite, force will be exerted on the rocket: 6.0 107 N, up. For the moment
Chapter 8
CHAPTER 8 - Conservation of Energy 1. The potential energy of the spring is zero when the spring is not compressed (x= 0). For the stored potential energy, we have U = !kxf2 0; 35.0 J = !(82.0 N/m)xf2 0, which gives xf = 0.924 m. For the
Chapter 7
CHAPTER 7 - Work and Energy 1. The displacement is in the direction of the gravitational force, thus W = Fh cos 0 = mgh = (250 kg)(9.80 m/s2 )(2.80 m) = 6.86 103 J. The displacement is opposite to the direction of the retarding force, thu
Chapter 5
CHAPTER 5 - Further Applications of Newton's Laws 1. The friction is kinetic, so Ffr = k FN. With constant velocity, the acceleration is zero. Using the force diagram for the crate, we can write F = ma: x-component: F k FN = 0; y-componen