Two blocks are sliding to the right across a horizontal surface, as the drawing

Question

Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In case A the mass of each block is 4.9kg. In case the mass of block 1(the block behind) is 9.8 kg, and the mass of block 2 is 4.9 kg. No frictional force acts on block 1 in either case A or case B. .however, a kinetic frictional force 9.6 N does act on block 2 in both cases and opposes the motion. For both case A and case B determine ( a ) the magnitude of the forces with which the blocks push against each other and ( b ) the magnitude of the acceleration of the blocks. Case A P = case B P = case A a= case B a=

Explanation / Answer

            Let the blocks will accelerate towards the right with acceleration a.
      Let F1 be the push force acting on the block 2 by block 1,       then according to Newton's third law of motion, the same force will       be acted on the block 1 by the block 2.
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      The frictional force acting on block 2, f = 9.6 N
Case 1:
      The mass of the blok's, m1 = m2 = 4.9 kg
      The net forces acting on block 1 is                  -F1 = m1a
      The net forces acting on block 2 is                 F1-f = m2a
      Adding these two equations, we get                    -f = (m1+m2)a       The acceleration of the blocks is             a = -f/(m1+m2)                = -(9.6 N)/(4.9 kg+4.9 kg)                = -0.979 m/s2               = -0.98 m/s2   (blocks are getting slower)       The push force, F1= -(4.9 kg)(-0.98 m/s2)                                      = 4.802 N _________________________________________________________________               = -0.98 m/s2   (blocks are getting slower)       The push force, F1= -(4.9 kg)(-0.98 m/s2)                                      = 4.802 N _________________________________________________________________
Case 2:
            The mass of the block 1, m1 = 9.8 kg                The mass of the block 2, m2 = 4.9 kg
      The net forces acting on block 1 is                      -F1 = m1a
      The net forces acting on block 2 is                     F1-f = m2a
      Adding these two equations, we get                  -f = (m1+m2)a       The acceleration of the blocks is             a = -f/(m1+m2)                = -(9.6 N)/(9.8 kg+4.9 kg)                = -0.653 m/s2   (blocks are getting slower)       The push force is               F1 = -(9.8 kg)(-0.653 m/s2)                    = 6.399 N                    = 6.4 N ____________________________________________________________ a)       The magnitude of the forces,       Case AP = 4.802 N
      Case BP = 6.4 N _____________________________________________________________ _____________________________________________________________
b)       Case Aa = -0.98 m/s2
      Case Ba = -0.653 m/s2       The magnitude of the accelerations,         Case Aa = 0.98 m/s2
        Case Ba = 0.653 m/s2         The magnitude of the accelerations,         Case Aa = 0.98 m/s2
        Case Ba = 0.653 m/s2  

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