Solutions to Week 2 Assignment

1. Using Hooke's law, the following proportion is obtained
   k = F₁/x₁ = F₂/x₂
   Solve for x₂
   x₂ = (F₂/F₁)x₁ = (7.0N/5.0N)(3.5cm) = 4.9cm
   Substitute known values for F₁ and x₁ to find k
   k = F₁/x₁= 5.0N/3.5cm = 1.4N/cm
2. Let the subscripts be the following:
   c = car     e = Earth      w = wall
   
3. The vertically directed forces balance, so the net force is due to the difference in the east-west forces.
   7kN east + 5kN west = 7kN east - 5kN east = 2kN east.
4. (a) Multiply the extension per mass by the mass to find the maximum extension required.
     [1.0mm/25g](5.0kg)[1000g/1kg][1m/1000mm] = 0.20m
   (b) Set the weight of the mass equal to the force in Hooke's law:
        W = F
      mg = kx
        k = mg/x
           = [(5.0kg)(9.8N/kg)]/(0.20m)
           = 250N/m.
5. This is the same as asking, "At what altitude is the gravitational field strength half of its value at the surface of the Earth?
      g =GM/R², so let the new field strength be g' = ng = GM/r² where n =1/2. Determine r in term of R.
      g'/g = ng/g = n = (GM/r²)/(GM/R²) = r²/R², so r = R/Ön.
   Find and expression for the altitude, h.
   h = r - R = R/Ön - R = R(1/Ön - 1). so h = (6.371 x 10^{3km)[1/Ö2 - 1] = 2639km.}
6. (a) Since the sleigh is moving with constant speed, the net force acting on the sleigh is zero.
   (b) The force of magnitude T must be equal the force of kinetic friction, since F_net = 0 (constant speed).
      T = f
         = m_kmg
      m_k = T/mg.
7. (a) All forces a collinear. The magnitude of the force of static friction on block A due to the floor must be equal to the magnitude of the tension in the cord.
          T = f_sA = m_AN = m_Amg
        The magnitude of the applied force must be equal to the magnitude of the tension in the cord plus the magnitude of the force of static friction on block B due to the floor.
          F = T + f_sB = m_Amg + m_Bmg = mg(m_{A +} m_B) = (2.0kg)[9.8N/kg](0.45 + 0.30) = 15N.
   (b) T = m_Amg (0.45(2.0kg)[9.8N/kg] = 8.8N.
8. (a) Let the subscripts be the following:
        c = computer     d = desk     e = Earth
   
   
   (b) Since the only forces acting on the computer are i9n the vertical direction, the frictional force is zero.
   (c) Find the maximum force of staic friction on the computer due to the desk: this is the horizontal force necessary to make it begin to slide.
           F = f_s = m_sN = m_sW = 0.60(87N) = 52N.