Solutions to Week 10 Assignment

 

  1. The stress is proportional to the strain.
      Y = FL/ADL = [(1.00 x10N)(5.00m)]/[(0.100cm2)(1m/100cm)2(6.50 x10-3m) = 7.69 x 1010Pa.
  2. The stress on the copper wire must be less than its tensile strength.
       F/A < tensile strength
       F < pr2(tensile strength)
       F < p(0.0010m)2(2.0 x108Pa)
       F < 1300N.
  3. Use Hooke's law for volume deformations.
       DV/V = -DP/B
           DV = -VDP/B = -[(0.230m3)(1.75 x106Pa)]/(60.0 x109Pa) = -6.7cm3.
  4. (a) am = w2A
           A = am/w2 = (98m/s2)/4p2(120Hz)2 = 1.7 x 10-4m.
    (b) vm = wA = w(am/w2) = am/w = (98m/s2)/2p(120Hz) = 0.13m/s.
    (c) According to Newton's second law, F = mam = (5.24kg)(98m/s2) = 510N.
  5. (a) The average speed is the total distance traveled divided by the time of travel.
         vav = Dx/Dt = 4A/T = 4A/(2p/w) = (2/p)wA
    (b) The maximum speed for SHM is vm = wA
    (c)  vav/vm = (2/p)wA/wA = 2/p
    (d)
      
  6. Find the length using equation for the period of a pendulum.
                T = 2pÖL/g
       T2/4p2 = L/g
               L = gT2/4p2
                  = (9.8m/s2)(1.0s)2/4p2
                  = 0.25m.