15.(a) For the motion from the bridge to the lowest point, we use energy conservation: K_{i}+ U_{grav i}+ U_{cord i}= K_{f}+ U_{grav f}+ U_{cord f}; 0 + 0 + 0 = 0 + mg(- h) + 1/2k(h - L_{0})^{2}; 0 = - (60 kg)(9.80 m/s^{2})(31 m) + 1/2k(31 m - 12 m)^{2}, which gives k = 1.0x10^{2}N/m. (b)The maximum acceleration will occur at the lowest point, where the upward restoring force in the cord is maximum: kx_{max}- mg = ma_{max}; (1.0x10^{2}N/m)(31 m - 12 m) - (60 kg)(9.80 m/s^{2}) = (60 kg)a_{max}, which gives a_{max}= 22 m/s^{2}.

27.(a)We find the normal force from the force diagram for the ski: y-component: F_{N1}= mg cos q; which gives the friction force: F_{fr1}= m_{k}mg cos q. For the work-energy principle, we have W_{NC}= DK + DU = (1/2mv_{f}^{2}- 1/2mv_{i}^{2}) + mg(h_{f}- h_{i}); - m_{k}mg cos q L = (1/2mv_{f}^{2}- 0) + mg(0 - L sin q); - (0.090)(9.80 m/s^{2}) cos 20^{o}(100 m) = 1/2v_{f}^{2}- (9.80 m/s^{2})(100 m) sin 20D, which gives v_{f}= 22 m/s. (b) On the level the normal force is F_{N2}= mg, so the friction force is F_{fr2}= m_{k}mg. For the work-energy principle, we have W_{NC}= DK + DU = (1/2mv_{f}^{2}- 1/2mv_{i}^{2}) + mg(h_{f}- h_{i}); - m_{k}mg D = (0 - 1/2mv_{i}^{2}) + mg(0 - 0); - (0.090)(9.80 m/s^{2})D = - 1/2(22.5 m/s)^{2}, which gives D = 2.9x10^{2}m.

32.(a)For the motion from A to B, there is no friction and the normal force does no work. For the work-energy principle, we have W_{NC}= (1/2mv_{B}^{2}- 1/2mv_{A}^{2}) + mg(h_{B}- h_{A}); 0 = 1/2mv_{B}^{2}- 0 + 0 - mgr; 0 = 1/2v_{B}^{2}- (9.80 m/s^{2})(2.0 m), which gives v_{B}= 6.3 m/s. (b)On the level the normal force is F_{N}= mg, so the friction force is F_{fr}= m_{k}mg. The work done by this force from B to C is W_{NC}= - m_{k}mg_{L}= - (0.25)(1.0 kg)(9.80 m/s^{2})(3.0 m) = - 7.4 J. (c)For the motion from B to C, the normal force does no work. For the work-energy principle, we have W_{NC}= (1/2mv_{C}^{2}- 1/2mv_{B}^{2}) + mg(h_{C}- h_{B}); - 7.4 J = 1/2(1.0 kg)v_{C}^{2}- 1/2(1.0 kg)(6.3 m/s)^{2}+ 0 - 0, which gives v_{C}= 4.9 m/s. (d)For the motion from C to the point where the block momentarily comes to rest, there is no friction. For the work-energy principle, we have W_{NC}= (1/2mv^{2}- 1/2mv_{C}^{2}) + mg(h - h_{C}) + (1/2kx^{2}- 1/2kx_{C}^{2}); 0 = 0 - 1/2(1.0 kg)(4.9 m/s)^{2}+ 0 - 0 + 1/2k(0.20 m)^{2}- 0, which gives k = 6.1x10^{2}N/m.

43. The escape velocity for a mass m is the speed required so the mass can get infinitely far away with essentially zero velocity. From energy conservation we have K_{1}+ U_{1}= K_{2}+ U_{2}; 1/2mv_{esc}^{2}- GM_{E}m/r = 0 - GM_{E}m/D, or v_{esc}^{2}= 2GM_{E}/r. Because the gravitational attraction provides the radial acceleration of the satellite, we have GmM_{E}/r^{2}= mv_{orbit}^{2}/r, or v_{orbit}^{2}= GM_{E}/r. For the ratio we get v_{esc}^{2}/v_{orbit}^{2}= (2GM_{E}/r )/(GM_{E}/r), or v_{esc}/v_{orbit}= (2)^{1/2}.

49. The total energy of the satellite in an orbit of radius r is E = K + U = 1/2mv^{2}- GM_{E}m/r = 1/2mGM_{E}/r - GM_{E}m/r = - 1/2mGM_{E}/r. Thus the work required is W = DE = - 1/2mGM_{E}[(1/r_{2}) - (1/r_{1})] = - 1/2mGM_{E}[(1/3r_{E}) - (1/2r_{E})] = GmM_{E}/12r_{E}.

79. We choose the reference level for the gravitational potential energy at the lowest point. The tension in the cord is always perpendicular to the displacement and thus does no work. (a)With no air resistance during the fall, we have 0 = DK + DU = (1/2mv_{1}^{2}- 1/2mv_{0}^{2}) + mg(h_{1}- h_{0}), or 1/2(v_{1}^{2}- 0) = - g(0 - L), which gives v_{1}= (2gL)^{1/2}. (b)For the motion from release to the rise around the peg, we have 0 = DK + DU = (1/2mv_{2}^{2}- 1/2mv_{0}^{2}) + mg(h_{2}- h_{0}), or 1/2(v_{2}^{2}- 0) = - g[2(L - h) - L] = g(2h - L) = 0.60gL, which gives v_{2}= (1.2g_{L})^{1/2}.

20.(a)The work done against gravity is the increase in the potential energy: W = mgh = (75.0 kg)(9.80 m/s^{2})(92.0 m) = 6.76x10^{4}J. (b)If this work is done by the force on the pedals, we need to find the distance that the force acts over one revolution of the pedals and the number of revolutions to climb the hill. We find the number of revolutions from the distance along the incline: N = (h/ sin q)/(5.10 m/revolution) = [(92.0 m)/ sin 9.50^{o}]/(5.10 m/revolution) = 109 revolutions. Because the force is always tangent to the circular path, in each revolution the force acts over a distance equal to the circumference of the path: pD. Thus we have W = NFpD; 6.76x10^{4}J = (109 revolutions)Fp(0.360 m), which gives F = 547 N.

29. On the level the normal force is F_{N}= mg, so the friction force is F_{fr}= m_{k}mg. For the work-energy principle, we have W_{NC}= DK + DU = (1/2mv_{f}^{2}- 1/2mv_{i}^{2}) + mg(h_{f}- h_{i}); F(L_{1}+ L_{2}) - m_{k}mg L_{2}= (1/2mv_{f}^{2}- 0) + mg(0 - 0); (350 N)(15 m + 15 m) - (0.25)(90 kg)(9.80 m/s^{2})(15 m) = 1/2(90 kg)v_{f}^{2}, which gives v_{f}= 13 m/s.

42.(a)Because the gravitational attraction provides the radial acceleration of the satellite, we have GmM_{E}/r^{2}= mv^{2}/r, or v^{2}= GM_{E}/r. The total energy of the satellite is E = K + U = 1/2mv^{2}- GM_{E}m/r = 1/2mGM_{E}/r - GM_{E}m/r = - 1/2GmM_{E}/r. (b)From the expression for the total energy, we see that a decrease in E (E becomes more negative) means a decrease in r. Because the kinetic energy is positive, a decrease in r means an increase in kinetic energy.

73. We choose the potential energy to be zero at the initial level of the center of mass (y = 0). We find the minimum speed by ignoring any frictional forces. Energy is conserved, so we have E = K_{i}+ U_{i}= K_{f}+ U_{f}; 1/2mv_{i}^{2}+ mgy_{i}= 1/2mv_{f}^{2}+ mgy_{f}; 1/2mv_{i}^{2}+ m(9.80 m/s^{2})(0) = 1/2m(6.5 m/s)^{2}+ m(9.80 m/s^{2})(1.1 m), which gives v_{i}= 8.0 m/s. Note that the initial velocity will not be horizontal, but will have a horizontal component of 6.5 m/s.

87. We choose the potential energy to be zero at the lowest point (y = 0). (a)Because the tension in the rope does no work, energy is conserved, so we have K_{i}+ U_{i}= K_{f}+ U_{f}; 1/2mv_{0}^{2}+ 0 = 0 + mgh = mg(L - L cos q) = mgL(1 - cos q); 1/2m(5.0 m/s)^{2}= m(9.80 m/s^{2})(10.0 m)(1 - cos q) which gives cos q = 0.872, or q = 29^{o}. (b)The velocity is zero just before he releases, so there is no centripetal acceleration. There is a tangential acceleration which has been decreasing his tangential velocity. For the radial direction we have F_{T}- mg cos q = 0; or F_{T}= mg cos q = (75 kg)(9.80 m/s^{2})(0.872) = 6.4x10^{2}N. (c)The velocity and thus the centripetal acceleration is maximum at the bottom, so the tension will be maximum there. For the radial direction we have F_{T}- mg = mv_{0}^{2}/L, or F_{T}= mg + mv_{0}^{2}/L = (75 kg)[(9.80 m/s^{2}) + (5.0 m/s)^{2}/(10.0 m)] = 9.2x10^{2}N.

97.(a)At point B the distance of the satellite from the center of the Earth is r_{B}= [(1/2a)^{2}+ b^{2}]^{1/2}= [(8,000 km)^{2}+ (13,900 km)^{2}]^{1/2}= 16,000 km. For energy conservation for the motion from A to B, we have K_{A}+ U_{A}= K_{B}+ U_{B}; 1/2mv_{A}^{2}- GM_{E}m/r_{A}= 1/2mv_{B}^{2}- GM_{E}m/r_{B}, or v_{B}^{2}= v_{A}^{2}+ 2GM_{E}(1/r_{B}- 1/r_{A}); v_{B}^{2}= (8650 m/s)^{2}+ 2(6.67x10^{-11}N * m^{2}/kg^{2}) x (5.98x10^{24}kg)[(1/16.0x10^{6}m) - (1/8.0x10^{6}m)] which gives v_{B}= 5.00x10^{3}m/s. (b)For energy conservation for the motion from A to C, we have K_{A}+ U_{A}= K_{C}+ U_{C}; 1/2mv_{A}^{2}- GM_{E}m/r_{A}= 1/2mv_{C}^{2}- GM_{E}m/r_{C}, or v_{C}^{2}= v_{A}^{2}+ 2GM_{E}[1/(*a) - 1/(1/2a)]; v_{C}^{2}= (8650 m/s)^{2}+ 2(6.67x10^{-11}N * m^{2}/kg^{2})(5.98x10^{24}kg)[(1/24.0x10^{6}m) - (1/8.0x10^{6}m)] which gives v_{C}= 2.89x10^{3}m/s.

53. The escape speed from the surface of the Earth, ignoring the Sun, is found from v_{E}^{2}= 2GM_{E}/r_{E}= 2(6.67x10^{-11}N * m^{2}/kg^{2})(5.98x10^{24}kg)/(6.38x10^{6}m), which gives v_{E}= 1.12x10^{4}m/s = 11.2 km/s. The escape speed from the Sun when at the Earth's orbit, ignoring the Earth, is found from v_{S}^{2}= 2GM_{S}/r_{SE}= 2(6.67x10^{-11}N * m^{2}/kg^{2})(2.0x10^{30}kg)/(1.50x10^{11}m), which gives v_{S}= 4.22x10^{4}m/s = 42.2 km/s. The orbital speed of the Earth is v_{O}= r_{SE}w = (1.50x10^{11}m)(2p rad/yr)/(3.17x10^{7}s/yr) = 2.98x10^{4}m/s = 29.8 km/s. (a)In the reference frame of the Earth, if the spacecraft leaves the surface of the Earth with speed v, we find the speed v' at a distance where the gravitational attraction is negligible from energy conservation: K_{1}+ U_{1}= K_{2}+ U_{2}; 1/2mv^{2}- GM_{E}m/r_{E}= 1/2mv'^{2}- 0, or v^{2}= v'^{2}+ 2GM_{E}/r_{E}= v'^{2}+ v_{E}^{2}. The reference frame of the Earth is orbiting the Sun with speed v_{O}. In the reference frame of the Sun, the speed far from the Earth is v_{S}= v' + v_{O}. When we use this in the previous result, we get v^{2}= (v_{S}- v_{O})^{2}+ v_{E}^{2}, or v = [v_{E}^{2}+ (v_{S}- v_{O})^{2}]^{1/2}= [(11.2 km/s)^{2}+ (42.2 km/s - 29.8 km/s)^{2}]^{1/2}= 16.7 km/s. (b)The required kinetic energy is K = 1/2mv^{2}, so we have K/m = 1/2v^{2}= 1/2(1.67x10^{4}m/s)^{2}= 1.40x10^{8}J/kg.