Question

look at the attachement and asnwer the questions

Answer 1

8.
       Normal force on B by the floor = F
       Force exerted by block A on B:  F1
         F - m_B g -  F1 = - m_B * 2        =>  F = F1 + m_B g - 2 m_B
         F1 - m_A  g = - m_A * 2            =>  F1 = g/2 - 1  = 4  Newtons
            
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9.  component of weight/acceleration due to gravity along the inclined slope
             = g Sin 30 = g/2

     Component of weight/acceleration along the groove OA = g/2 Sin 30 = g/4

     s = u t + 1/2 a t²    => t² = 2 s / a
               t² = 2 * 5 /  (g/4) = 4              =>           t = 2 sec.
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10.      m1 = 100 kg          m2 = 60 kg

  a1 = acceleration of 100 kg upwards.  Rope is also moving with the same acceleration.
       T - 100 kg * g = 100 kg * a1        => T = 1000+ 100 a1          ---- (1)
       
  Acceleration of the person = a
     relative acceleration of the man wrt rope  = a + a1  = 5g/4
             => a = 5g/4 - a1  = 12.5 - a1      upwards
     T - 60 kg * g = 60 kg * a
       T = 600 + 60 a = 600 + 60 (12.5 - a1) = 1350 - 60 a1      ----- (2)

       Solving the two equations          a1 = 350/160  m/sec/sec
       T = 1000 + 100 * 35/16 = 1218  N
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11
         F = buoyancy force upwards  remains same
         F - M g  = M a        => F = M (g+a)
         F - (M-m) g = (M-m) a'          =>  a' = [ M g + Ma - Mg + mg ] / (M-m)
                 a' = (M a + m g)/(M-m)
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12.        T - 5 g = 5 a
             10 g - T = 10 a
           a = g/3
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13.      F - m g Sin 37  = m a          along the slope
           5.5 - 0.50 * g * 0.6  = 0.50 a      =>  a = 5 m/sec/sec
              
          v^2 = 2 a s = 2 * 5 * 10  = 100
             v = 10 m/s    at the top of the incline.
         angle of projection = 37 deg.
            y initial = 10 Sin 37 = 6 meters

           Time to reach the top point :    u Sin 37 / g = 10 * 0.6 / 10 = 0.60 sec
           height reached = s  =  u sin 37 * 0.60 - 1/2 * 10 * 0.60^2
                                         = 10  * 0.6 * 0.60 - 5 * 0.60^2  = 1.80 meters
           total height at the top of the flight = 6 + 1.80 = 7.80 m

           time to reach the ground = t  => t^2 = 2s / g =>        t = 2 * 7.80 / 10 = 1.56 sec

       total time of the flight = 2.16 sec.
          total  horizontal distance travelled after projection from the ramp
                   =  u cos 37 * 1.56 sec
                   = 12.48 meters