Answers to homework problems:

Chapter 5.
 7) W = U = mgh = 2000 kg * 9.8 m/s^2 * 15 m = 294000 J
 9) Force component responsible for motion is parallel to the ground.
    F(parallel) = F * cos(30) 
    Work: W = F * d * cos (30) 
           --> d = W/(F * cos (30)) = 1440 J/(250 N * cos(30)) = 6.65 m
23) Total work is given by W = 1/2 * k * x^2
    The total displacement is due to streaching x = 17cm - 4cm = 0.13 m
    In order to find the spring constant k we use the force equation
    F = -k * (x-x_0)
    The value of the force is F = m*g = 0.075 kg * 9.8 m/s^2, x = 7 cm,
        x_0 = 4 cm
    ----> k = F/(x-x_0) = 0.075 kg * 9.8 m/s^2 /(0.07 m - 0.04 m) =
            24.5 kg/s^2
    W = 1/2 * 24.5 kg/s^2 * (0.17 m - 0.04 m)^2 = 0.21 J
34) Several ways of answering this problem.
                                 small car    large car            
     mass, m                         m            2m
     velocity                       2v             v
     kinetic energy,K             4mv^2/2       2mv^2/2
     friction,F                    mu*mg         mu*2mg
     deceleration, a=F/m            mu*g          mu*g
     Energy conservation K=F*x
           ---> x = K/F      4mv^2/(2*mu*mg)   2mv^2/(2mu*2mg)
          ----->small car breaking distance 4 times as long as large car
     Deceleration consideration: v^2=v_0^2+2ax since v=0 --> x=v_0^2/(2a)
         --> breaking distance      2v/(mu*g)     v/(2*mu*g)
          ----->small car breaking distance 4 times as long as large car
48)  U = m * g * h
       = 1000kg * 9.8 m/s^2 * 1400ft/(3.28 ft/m) =4182927 J  
56) Speed of a pendulum is greatest when ALL of its potential energy is
    transformed into kinetic energy i.e., at the bottom of its swing. 
62) Conservation of energy tells us that the velocity is greatest at the bottom
    of the swing when all potential energy is transformed into kinetic energy.
    The potential energy relative to the bottom of the swing is U = m * g * h
    The hight of the swing h = maximum - minimum = 2m - 0.5m = 1.5 m
     U = K  --> m* g * h = m * v^2/2 --> v = sqrt(2*g*h)
     v= sqrt( 2 * 9.8 m/s^2 * 1.5 m) = 5.43 m/s
63)  Potential energy U = m * g * h. Since mass and gravitatinal acceleration
     is not affected by the efficiency, the hight after each bounce is
     proportional to the efficiency.
   a) 1.25 m * (1.00 - 0.18) = 1.03 m
   b) 1.25 m * (1.00 - 0.18) * (1.00 - 0.18) = 0.84 m
   c) The inital kinetic energy K=mv^2/2 has to be added to the potential
      energy before the bounce efficiency is applied. After the bounce the
      total remaining energy is transformed into potential energy when reaching
      the initial height.
      Total initial energy  E= m*g*h + m*v^2/2
      Total energy after the bounce E*(1.00-0.18) = m*g*h
            (m*g*h + m*v^2/2) * 0.82 = m*g*h
        ---->  0.82 * m*v^2/2 = m*g*h*(1.00-0.82) = 0.18*m*g*h
        ---->   v = sqrt(0.18*m*g*h*2/0.82) = 2.32 m/s