Solutions to Week 6 Assignment

1. W = F Drcosq = Tvtcosq  = (240N)(1.5m/s)(10.0s)cos30.0° = 3.1kJ.
2. W = DK  = 1/2m(v² - v₀²) = 1/2(5.00kg)[(2.50m/s)² - 0] = 15.6J.
3. W = 1/2(2.0N)(1.0m ) + 1/2(1.0N)(1.0m) + 1/2(-1.0N)(1.0m) = 0.5J.
4. (a) U = mgh = (100kg)(9.8m/s²)(12,000m) = 2.9J
   (b) U = mgDy = mg[(v² - v₀²)/2>g] =1/2(100kg)(250m/s)² = 3MJ.
5. Find k. Use Newton's second law.
   SF_y = kx₁ - mg = 0, so
     k = mg/x₁
   Find v_max
         K_i + K_f = U_i + U_f
   0 + 1/2mv_max² = 1/2 kx_max² + 0
       v_max = x_maxÖk/m
               = x_maxÖg/x₁
               =(0.100m - 0.050m)Ö[(9.8m/s²)/(0.060m - 0.050m)]
               = 1.6m/s
6.                 DK =  -DU
   1/2mv_f² - 1/2mv_i² == -mg Dy
         v_f² - v_i² = -2<mg Dy
                    v_f  = Ö(v_i² -2mg Dy)
                         = Ö(4.0m/s) - 2(9.8m/s²)[0 - (2.0m)sin30.0°]
                         = 6.0m/s
7. (a) P_av = DE/Dt = Dk/Dt =1/2m(v_f² - v_i² )/Dt = (1000.0kg)[(40.0m/s)² - 0]/2(10.0s) = 80.0kW.
   (b) mechanical energy required/chemical energy provided = k (efficiency)/46MJ/L = 1/2(1000.0kg)(40.0m/s)²/(46 x10J/L)(0.22) = 0.079L
                                      per liter
8. (a) Find the speed at the bottom of the incline.
               DK = W_C + W_NC
        1/2mv² - 0 = mgh - fd
              v = Ö2gh -2fd/m
        Use Newton's second law with +y perpendicular to the incline and +x down the incline.
        SF_y = N - mgcosq = 0, so N = mgcosq
        Now, d = 0.85m, h = dsinq,  f = mN = mmgcosq
        Substitute.
        v = Ö2gdsinq - 2 mmgdcosq = Ö2gd(sinq - mcosq)
        Find the maximum compression.
         K_i + U_i = K_f + U_f
        0 +1/2kx² = 1/2mv² + 0
                 x = Öm/k
                    = Ö2mgd/k(sinq - mcosq)
                    = Ö2[(0.50kg)(9.8m/s²)(0.85m)/35N/m](sin30.0° -0.25cos30.0°)
                    = 26cm.
   

   (b) When the block is accelerated by the spring, it attains its previous kinetic energy and speed. Find the distance along the incline, d'.
                  DK = W_C + W_NC
         0 - 1/2mv² = -mgh - fd'
                       1/2mv² = mgd'sinq  + mmgd'cosq
                                   = d'[mg(sinq  + mcosq)]
                                d' = v²/2g(sinq  + mcosq)
                                    =2gd(sinq - mcosq)/2g(sinq  + mcosq)
                                    = (85cm)[(sin30.0° - 0.25cos30.0°)/(sin30.0° + 0.25cos30.0°)
                                    = 34cm.