Answers to homework problems:

Chapter 10 

5) Relation between the force and the displacement due to a sheer is
    F = S * Delta(X)/L * A
    ===> Delta(X)=F*L/(S*A)
   Here,
    Force = 1/4 of the total applied force =1.4 * 32 N = 8 N
    S = 2.6 10^6 N/m^2
    L = 0.03 m
    A(surface of the block) = 1.2 10^-3 m^2
        Delta(X) = 8 N * 0.03 m / (2.6 10^6 N/m^2 * 1.2 10^-3 m^2) = 7.7 10^-5 m

11) Tension will be equal to the applied force needed to change the length of the wire
    by two circumferences of the peg.
    Delta(L)=2*2*pi*r = 2*2*3.14*0.0018 m = 0.0226 m
    tension = F = Y * Deltat(L)/L * A =
                =2.0 10^10 * 0.0226/0.76 * (pi*0.0008^2) = 1197 N

27) The length of the unstrained spring = 0.30 m
    The length of the streatched spring = hight of the room - length of the board =
                                        = 2.44 m - 1.98 m = 0.46 m
    The Hooke's law F=k*x
          ===> k = F/x = 102 N / (0.46 m - 0.30 m) = 637.5 N

36) v = A*omega*sin(omega*t)  ===> maximum is for sin(omega*t)=1
    The amplitude, A, is equal to the radius of the record
    v(max) = A * omega = 0.152 m * 3.49 rad/s = 0.53 m/s
    a(max) = A * omega^2 = 0.152 m * (3.49 rad/s)^2 = 1.85 m/s^2

47) Elastic potential energy = 1/2 * k * x^2 =
                             = 0.5 * 425 N/m * (0.47 m)^2 = 46.9 N*m
    This energy is transformed into kinetic energy of the arrow E=mv^2/2
              46.9 = m*v^2/2  ==> v = sqrt(2*46.9/0.03) m/s = 55.9 m/s

63) From Example 13
     L = T^2 * g /(4*pi^2) = 2^2 * 9.8 /(4*3.14^2) = 0.993 m

64) From Example 13
     L = T^2 * g /(4*pi^2) = 9.2^2 * 9.8 /(4*3.14^2) = 21 m (about 7 storeys)